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Russian
00:00:03
the more likes we get, the more00:00:06
people will see this video parameters that is00:00:08
18 problem is solved by only a few00:00:10
percent of people and the more difficult the parameter00:00:14
than for example problem 1300:00:15
why 13 will solve everything and the parameter00:00:18
units why I think because 1300:00:22
the task is quite typical00:00:27
there is an algorithm that can be solved00:00:29
most likely this is an electrical equation00:00:31
you decide there, well there are only a few00:00:33
types of these elementary equations in this00:00:36
solve plus or minus in similar ways00:00:38
always use limited00:00:40
there are several formulas there00:00:43
all the time in function b you can select by00:00:45
circles for example or double00:00:46
inequality as you are used to, that is00:00:48
is there a clear solution method yes you have00:00:51
a clear solution plan on the thirteenth00:00:53
template, roughly speaking, you decide which one00:00:55
you have it in your head that if I tell you00:00:58
I will say that I found a template solution00:01:02
18 tasks, well, not all 18 tasks yet, but00:01:07
huge type of tasks00:01:09
a huge popular type of problem where00:01:11
we need to find for what a the equations has00:01:14
exactly one root on the segment if y00:01:17
you have a big equation that has00:01:19
there are a lot of roots, maybe 23 roots00:01:21
so you need to find and at which00:01:24
out of these two, three will remain to feed00:01:25
only one if the condition is when00:01:29
equation with some muda00:01:30
has exactly one root on the segment because00:01:32
again the equation has the only one00:01:34
the root of the interval is00:01:35
the only root on the segment00:01:36
has exactly 1 course if the condition00:01:39
such problems are solved by00:01:40
a certain pattern and today I00:01:44
I’ll tell you about the template and we’ll sort it all out00:01:46
these tasks, that is, today you can00:01:48
solve all these problems and if you are at the game00:01:50
there will be a task with similar conditions and this00:01:52
super popular type of task then you are more likely to00:01:55
in total you will score 4 4 primary points00:01:57
that is, this task is the most significant in the game00:01:59
and you will score maximum points00:02:02
I'm recording a video course right now00:02:05
according to the parameters which will be released on March 31 and00:02:08
before recording the video solutions I made00:02:10
list of all tasks00:02:12
I'm writing it down and suddenly I realize that00:02:14
that's when I write down task number 20 of00:02:17
list I have already decided this many times00:02:21
in the same way as the twentieth is solved00:02:23
task and look and understand what00:02:26
the previous 20 tasks using this method you can00:02:29
I had to solve 8 problems 820 and I understand that00:02:32
this is a super popular type of parameters and00:02:36
everything is solved one way00:02:38
I need to record a video that's why I00:02:40
Now this is a video and I’m recording what it’s all about00:02:43
method means using a specific example here00:02:47
you have a court equation and you need00:02:48
find for which a there will be exactly one00:02:50
root on segment where to start right00:02:53
throw part to the left, well, great00:02:55
obviously started to bring out the general00:02:57
right side time to take out the general00:03:00
the product is equal to zero when at least00:03:02
one of the factors is zero + 1 for00:03:04
hang it and plus x should be in00:03:06
segment from zero to one all in one00:03:07
the system can be pushed in, so you solve these00:03:10
elementary equations you have x 1 x00:03:11
toroid00:03:12
question at what a x 1 is a root00:03:16
equations on the segment, that is, x 1 should00:03:19
satisfy now and00:03:21
lie on the segment from 0 to 1 by 1 2 and so on00:03:24
lies on the segment for this reason I00:03:25
I'm not including this here, I'm just being rude00:03:27
speaking of yes for for ics for x 1 I find not00:03:30
I understand that with these a x 1 is00:03:33
root of the equation on the interval a x 2 at00:03:36
what kind of00:03:37
is a root on the segment x 2 must00:03:39
lie on the segment and remove stabbed even00:03:41
I find for what a x 2 is a root00:03:44
on the segment then taking into account that in and to00:03:48
sides are a x 1 of can coincide with00:03:51
old at some point and I find at00:03:53
which a x 1 coincides with k old then00:03:56
there is when we have, like, two roots, but they00:03:57
they coincide they are the same here and then in00:04:00
at the end I make a number line and where00:04:03
I show at what and how many roots here00:04:05
with such and there are no roots for such there are no roots for00:04:08
there is only one root00:04:09
with such a there are two different roots here00:04:11
there are two different roots, but not00:04:13
coincide, you know there is one root00:04:15
and I write down at what time and it turned out only00:04:18
I circle them with one root and write them in00:04:20
answer end here is the solution algorithm00:04:24
it’s clear there the equation can be a little00:04:26
be different today I picked up 1400:04:30
tasks we will analyze 14 tasks video00:04:33
very long but super useful because00:04:35
if you get a task in the game where00:04:37
there will be an equation with a j and we need to find00:04:39
one root on the segment most likely she00:04:42
solved this way and you can dial 4 400:04:44
primary points even if you are earlier00:04:47
parameters of communication were decided by I think you00:04:48
you can understand this video and00:04:50
We’re not interested at all, so it turned out00:04:53
can you figure it out?00:04:54
then write in the comments you are with this00:04:57
as easily as I do now you decide00:04:59
such tasks or something didn’t work out00:05:01
write in the comments if possible00:05:04
someone managed to solve it for the first time00:05:05
parameter then write about it too00:05:07
interesting, did you like this?00:05:09
did you understand the explanation, the solution algorithm00:05:12
or not, find all the values for each00:05:16
of which the equation has exactly one00:05:17
root on the interval from 0 to 100:05:19
the obvious start is the right side to the left00:05:21
reroll remove general so root from00:05:25
2 x minus 1 times ln 4x minus a00:05:30
minus00:05:35
rewrite00:05:40
equals zero, then take the root beyond00:05:43
buying up00:05:45
brackets remains the difference of logarithms00:05:55
when the product is zero when although00:05:57
if one of the factors is equal to 0 200:05:58
each of the factors exists in general00:06:01
equate to zero plus add water00:06:03
for plus we say that x is from zero to one00:06:05
that is, 2x + 1 can be taken as the root equal to00:06:09
zero00:06:11
I will write in detail although it was already possible00:06:14
Of course some transformations need to be done00:06:17
well, you still need a few parameters00:06:19
more details equal to zero00:06:21
set + a to z we have what 2x00:06:24
minus 1 must be greater than or equal to 0 400:06:29
x minus a must be large or00:06:31
islands 5x plus a must be greater than 000:06:36
strict and x must be between 000:06:41
to one00:06:43
such a system we get the root00:06:47
is equal to zero when when the radical is equal to00:06:49
zero, that is, x is equal to 1 2 there and here00:06:54
it is also easy to solve the equation l n right00:06:56
throw Elena, remove sex00:06:59
leipzig with the right subject is shorter00:07:02
it turns out for us00:07:05
system can well the first equation is the root00:07:10
remove the root 2 x minus 1 equals zero 2x00:07:12
is equal to 1 x is equal to 1 2 is x 1 x 1 is 1 200:07:18
x 2 what is equal to l n to the right we remove Elena00:07:24
4x minus a equals 5x plus a00:07:26
and jambs to the left errors to the right x equals00:07:30
minus 2 and we are these x 2 this is -2 and so 2 x00:07:37
minus 1 large equals 0 x greater or00:07:40
equals than 1 to the right and divide by 2 1 2 well00:07:46
let's rewrite this 4x minus a should be00:07:48
greater than 0 5 x plus a must be greater than 000:07:53
x must be between zero and one00:07:57
so we have two roots, well, more and00:08:02
equals 1 2 and from zero to one find00:08:04
let's find the suppression now, that is00:08:06
it turns out we already have five lines00:08:09
busy00:08:11
x 1 equals 1 2 x 200:08:15
is minus 2 x must be greater or00:08:22
is equal to 1 2 and lies in the segment00:08:24
from zero to one if we find00:08:25
intersection then it turns out that x should00:08:27
be from 1 2 to one so well, plus y00:08:32
yes for 4 minus more than 0 and 5 x plus a00:08:37
greater than 0 so now let's find out when00:08:42
what a x 1 is the root of the equation x00:08:44
100:08:45
it is equal to one-half is the root00:08:48
equations for a satisfying what00:08:52
requirements first x 1 1 2 he must00:08:58
lie in the segment from 1 2 to one he and00:09:00
it lies like that plus it should even satisfy00:09:02
with passion he must satisfy two00:09:04
inequalities 4x minus a greater than 0 and 5 x00:09:08
plus a is greater than 000:09:11
we substitute the second one under x1 and it turns out 400:09:15
multiply by one half00:09:16
minus a more than 0 5 by one second plus a00:09:21
more than 0 is obtained and less than 2 and00:09:26
more than -2 and a half conclusion, well00:09:31
less than two more than minus 2 and a half00:09:33
this means that a belongs to minus00:09:37
two and a half to two so here's the output in00:09:41
than with00:09:42
and with proper Andrey from minus two with00:09:44
half to 2 x 1 equals 1 2 is00:09:48
root of the equation on a given interval00:10:00
is the root of the equation and for what00:10:02
aek 2 is the root of the equation a and ko00:10:07
the second one which is equal to minus 2 a00:10:12
is the root of the equation for a00:10:15
satisfying the following requirements00:10:19
must lie on the segment from 1 2 to00:10:21
units00:10:22
it’s no longer obvious that he’s lying there00:10:25
so we must say a stroke for must00:10:27
lie00:10:28
for some from 1 2 to unity and he00:10:32
must satisfy o dose 4x minus a00:10:34
more than 0 and 5 x plus and more than 0 some00:10:37
it seems that it was possible not to write from them or00:10:39
baba because this root came from00:10:44
equations where these logarithms were or not00:10:48
Still, it’s better to check in short00:10:50
don't even think about it and just system00:10:52
write down to be on the safe side let's say00:10:55
this means we have such a system00:10:57
instead of exe we substitute -2 and here is 1 2 to00:11:04
units instead of exe to -2 and instead of exe to00:11:14
minus 2 a00:11:20
it turns out so let's go to minus 1 200:11:22
multiply it turns out a minus 1 4 minus00:11:30
a half00:11:32
the icons unfold because00:11:36
negatively multiplied further00:11:39
minus 8x minus and this is minus 9 and greater than 000:11:45
minus 10 plus00:11:47
and this is minus 9 and yes it’s not automatic00:11:52
holds o even a both restrictions in00:11:55
they turn into the same thing, that is00:11:57
it turns out it was, so let's write it down00:12:00
double inequality more common in00:12:02
in the usual form, that is, a should be from00:12:04
minus 1 2 inclusive up to minus 1 4 and00:12:06
there closely minus 1 multiply it turns out a00:12:10
must be less than less than zero not by00:12:14
multiply minus 1 and divide by -9 and00:12:18
it turns out a is less than zero a is less than zero00:12:22
in this case from minus 1 2 to admins 1 2 to00:12:26
minus 1 4 so well, the segment from minus 1 2 to00:12:33
minus there by 4 it is already less than zero then00:12:36
there is this segment left at baptism00:12:38
conclusion for a for a proper segment from00:12:44
minus 2 to minus 1 4 x second which00:12:51
was equal to -2 and is the root00:12:54
equations on a given interval00:13:02
root of the equation so let's now00:13:05
find out at what point they coincide 1 2 and00:13:08
minus 2 x 100:13:11
matches the old iq if anything00:13:18
occurs if x 1 which is equal to 1 200:13:25
equal to and which, that is, equal to minus 2 a00:13:28
but we find that we need to divide one half00:13:33
at -2 it turns out minus 1 4 output at a00:13:39
equal to minus 1 4 x 1 becomes equal00:13:46
and to the old one now let's apply a00:13:49
which influence whether x 1 or 2 are00:13:53
whether the equations are roots or not, that is, we00:13:55
put on the number line and all the important ones00:14:00
and the number line and we plot it like this x 100:14:05
is the root of a of minus two c00:14:07
half to 2 x 2 from minus 0 5 to minus00:14:13
0 25 and minus 0 25 must be turned on again00:14:17
and so two to the right of all here is chapane minus 200:14:22
with half to the left of all there is still minus 000:14:28
5 and minus zero 2500:14:36
we are interested in00:14:37
a less than minus 2 and a half a00:14:41
exactly, exactly minus 2 and a half a from00:14:45
minus two and a half to minus 0 5 a00:14:49
exactly -05 and from minus 0 5 to minus zero00:14:56
2500:14:57
a exactly minus zero 25 a from minus 0 25 to00:15:04
2 a is equal to 2 and a is greater than 2 so let's00:15:09
let's make small sticks00:15:26
and less than minus twenty so let's00:15:29
of x 1 is the root of minus two s00:15:32
half to two not inclusive then00:15:35
there is x 1 is the root here and it00:15:38
equals something I forgot 1 2 x 1 equals 100:15:43
2 here x 1 equals 1 2 here x 1 equals 1 200:15:48
here without 1 is equal to 1 2 here x 1 is equal to 1 200:15:54
here everything from minus two and a half to00:15:57
two where inclusive so and x 2 is00:16:01
root from -1 2 to minus 1 4 inclusive00:16:04
that is, x 2 is a root and equals00:16:07
minus 2 and here x 2 is the root hey00:16:10
rhine -2 here x 2 is the root equal00:16:13
minus 2 here if a is equal to minus 0 5 then x00:16:18
1 Ryan turns out to be one so here he is00:16:21
different all the time but here00:16:23
if a is equal to minus 0 25 then x 1 is equal to 1 200:16:26
and we have two00:16:29
matching roots, that is, only one00:16:31
root understand 2 matching that is00:16:34
just one root00:16:37
except if a is less than minus 2 s00:16:40
half of you don't see here at all00:16:41
it turns out there are no roots here00:16:43
if a is equal to minus 2 and a half also not00:16:46
turned out to be roots if00:16:50
and here it is from minus 2 and a half to00:16:52
minus 0 5 st we have one root if a00:16:55
-05 two different roots if from minus 0 500:16:57
to minus 22 different roots if minus 000:17:00
25 then 2 matching roots that is one00:17:02
root from -25 to 2 1 root00:17:05
if a is equal to 2t there are no roots if a is greater00:17:11
there are no two roots00:17:14
the question at which a will be only one00:17:18
root at what and we will have only one00:17:20
root on the segment so it suits us a00:17:23
these ones will only have x 1 as a root00:17:25
and equals minus 0 5 does not fit00:17:28
this is not suitable this is suitable for him00:17:30
there will be two matching ones, that is, one00:17:32
the root is also suitable, we write everything00:17:35
answer00:17:39
there will be one root00:17:41
from minus 2 and a half not inclusive00:17:44
up to -0.5 not inclusive and from minus 0.2500:17:55
inclusive up to because a equals minus00:18:00
0 25 gives two identical roots to two00:18:05
not inclusive, also with no two00:18:07
roots00:18:08
everything will come x will be exactly the same00:18:10
solution since we solved with the equation and00:18:13
the right part was transferred to the left, taken out00:18:15
common beyond parenthesis root root is zero00:18:18
or the difference of logarithms is 0 plus 100:18:21
plus x from zero to one00:18:23
solved these elementary equations00:18:25
got x 1 x 2 + z + x from zero to00:18:28
units Well, we transformed the ladies a little and now x00:18:32
1 is the root when when it00:18:34
satisfies all the limitations of the muse and00:18:36
is in the interval from 1 2 to units well00:18:38
well so in 1 5 2 to album units00:18:41
that's why they only demanded the image00:18:42
for him to do, here he does00:18:45
Odessa satisfies the drive of such a00:18:47
good x 2 satisfies there for the drive00:18:50
so cool now they match00:18:53
if a is equal to minus 1 4 understandable but00:18:56
number line shown00:18:57
important and showed at what and how much00:19:01
roots and circled tea at which only00:19:03
one root turned out to be such a task00:19:07
find all values for which00:19:09
equation has a single root at00:19:11
segment from 0 to 100:19:12
how to start a square is equal to a square00:19:16
how to start, well, it’s elementary00:19:19
right side to the left with minus the difference00:19:21
squares we paint to accept very00:19:26
easy to start obvious sometimes unclear00:19:30
where to start and here in my opinion it’s more obvious00:19:32
nowhere like ln x plus 2 a00:19:39
minus so ln x plus 2 squared00:19:50
difference of squares is equal to zero difference00:19:52
we write squares according to the formula00:19:54
draw a square just in case I00:19:56
I'll show you because there are people in whom00:20:01
for some reason he refuses00:20:04
brain work with simple formulas in00:20:08
more complex applications here00:20:11
ordinary difference of squares that is a00:20:12
square minus b square we see and can00:20:14
replace with the product of two brackets00:20:16
and minus b by plus would, that is, it turns out00:20:18
2 x plus00:20:21
ln x plus 2 minus this expression is00:20:27
there is minus 2x minus date bracket00:20:30
now the signs00:20:31
I reveal it turns out plus ln x + 200:20:35
multiply by the product a multiply by00:20:41
the sum of two terms, that is, 2 x plus00:20:44
ln x 2 + 2x minus00:20:51
ln x plus 2 so when we are the difference00:20:57
squares have been painted, still need to be hung dz00:21:01
yes it equals zero let's simplify 2 x00:21:07
minus 2 x destroyed it turns out l n plus 1 is 200:21:10
identical allene 2 ln x plus 200:21:16
you need to multiply by here Elena mutually00:21:19
destroyed remains 4 x equals zero00:21:25
This is the equation we got from00:21:28
initially you got this00:21:29
equation plus must be required so that00:21:31
od for fulfilled that is necessary00:21:34
demand that we also have x plus 2 a00:21:37
should have been greater was greater than 0 and00:21:42
it is necessary that x lies in the interval from zero to00:21:45
units, I’ll also push this straight into the system00:21:47
there was no need to write it down00:21:50
so on next when the work00:21:53
equals zero when at least one factor00:21:55
equals 0 authors exists, that is, we have00:22:00
or l.n.00:22:02
well two is not equal to zero 4 is not equal to zero00:22:04
that is, either x is equal to zero or ln is equal00:22:07
understand zero or ln x plus 2 and is equal to00:22:11
zero00:22:12
or x is equal to zero x plus 2 must be00:22:19
more than 000:22:20
x must be from 0 to one so ln00:22:26
equal to zero when when well ln means in00:22:30
the base of the little box and we build it into the first00:22:33
to the zero power and it turns out 1, that is00:22:35
pornographic one elementary00:22:38
everyone understands the logarithmic equation00:22:39
which one to decide of course yes that is x plus 200:22:43
and it is equal to one x is equal to zero x plus 200:22:51
and should be greater than 0, by the way, please note00:22:54
express 2 is equal to one, that is, oh yes for00:22:56
is executed at one of the roots well00:22:58
no matter, can we go there again, well, in general00:23:05
solve it will not be superfluous to express x in short00:23:12
that is, let me start with a zero x00:23:15
Let 1 be equal to zero and x 2 be equal to 200:23:19
we'll throw 1 -2 to the right, but we have two00:23:23
root00:23:25
dc and x must have a zero to00:23:30
units is good if each of the roots00:23:36
satisfies hell for00:23:37
and is from zero to one then we have00:23:39
there will be two roots and we need it to be00:23:43
the only root for the crack from zero to00:23:45
units means the first root must be00:23:47
suitable but the second one is not or vice versa 100:23:50
inappropriate and the second one is suitable or00:23:52
they must match, let's find out when00:23:55
which ones x 1 are suitable00:23:58
x 1 it is equal to zero for what a for a00:24:05
satisfying worst for and if x is from zero00:24:10
to one but x is equal to zero it is from zero to00:24:12
units include on the segment yes that is00:24:14
with a satisfying y it is even possible00:24:16
say when satisfying00:24:18
inequality x plus 2 greater than 0 x equals00:24:23
substitute zero for zero00:24:26
it turns out that two a is greater than 0, that is, a00:24:31
more than 0 output00:24:33
the conclusion is that when a is greater than 0x 100:24:39
equal to zero will be the root00:24:41
equations for other a x 1 will be00:24:44
violate decrees00:24:45
so that means at and more that is not00:24:47
will be curation aprio more00:24:49
0x 1 will be the root of the equation00:25:02
ikos 2 which is equal to 1 minus 2 and he00:25:11
is the root of the equation for what a00:25:13
with a satisfying means dc well00:25:23
although we have recently found out what ode00:25:26
forgets to do it, no problem00:25:28
let's figure it out again and he should be in00:25:31
segment from zero to one even if with00:25:33
zero it’s clear that he’s a unit in the squads00:25:35
already 0 then with the expression containing and not00:25:39
it’s clear we need to understand at what conditions it will be00:25:41
from zero to one so from this00:25:48
must be fulfilled therefore often we00:25:50
we substitute yes under x 1 -2 it turns out00:25:53
one minus 2 plus 2 a greater than zero to at00:25:57
anyway, that’s how we understood it and 1 -2 should00:26:00
be from zero to one first00:26:03
inequality gives us the answer and any in00:26:06
any00:26:07
one is more than Olya, and in double00:26:11
inequality how to solve we subtract00:26:12
one from all sides, that is, minus 100:26:15
minus 2 and zero and multiply by -1 second00:26:21
that is, it will be a 1 200:26:25
zero when on a negative number00:26:28
multiply the icons unfold that is00:26:31
output at a00:26:37
with proper from 0 inclusive to 1 200:26:47
x 2 will be the root of the equation and 200:26:51
equal to 1 minus 2 will be the root00:26:54
equations00:27:01
root of the equation00:27:12
yes even00:27:15
roots can coincide zero and 1 -2 can00:27:19
coincide00:27:21
let's find out at which ones and they coincide00:27:24
x 100:27:26
coincides, that is, there will be two roots but00:27:29
identical00:27:30
coincides with and to the old ones if anything00:27:33
happens if x 1 is equal to zero00:27:38
coincides means equal to the old 1 -200:27:41
and it turns out 2 and is equal to 1 and is equal to 1 200:27:46
the output for a is equal to 1 2 we will have00:27:56
single root matching x 1 kg00:27:58
the old ones are falling off the roots are falling off now00:28:04
let's see at what and how much00:28:07
roots and what kind of roots we will arrange00:28:09
large number line a00:28:19
at00:28:20
and greater than 0x 1 is essentially a root00:28:24
equation for a from 0 to 1 2 x 2 is00:28:27
root of the equation for a 1 200:28:29
they match, that is, you need to apply00:28:32
zero one second is just not enough00:28:35
usually more, that is, we apply a zero and00:28:39
1 2 and now we will watch everything00:28:46
at what and how many roots and what kind of roots00:28:49
and so we consider the situation when a00:28:56
less than zero when a is from 0 to00:28:59
when a is 0 when a is from 0 to 1 200:29:05
when a is exactly 2 and when a is greater than00:29:08
1 2 so if a is less than zero then x 100:29:13
is the root00:29:14
if a is less than 0 no expert is00:29:18
root when a is greater than 0, that is, x 1 is not00:29:20
aek 2 is the root00:29:22
x 2 is the root of 0 to 1 200:29:25
inclusive, that is, there are no roots here00:29:31
equals 0 is a root even more precisely when00:29:35
and equals 0 x 1 is the root no it00:29:38
is a root for a greater than 0 and x 200:29:41
is the root to x 2 is equal to 1 minus 2 a00:29:47
at from 0 to 1 2 inclusive00:29:50
if a is equal to zero then it turns out x 200:29:53
the only root of00:29:57
equals 0 circle and equals 0 does not fit00:29:59
if from 0 to 1 2 segment from 0 to 1 200:30:03
enter into greater than zero up to therefore x 100:30:08
equal to zero exists here, that is00:30:11
is the root of the equation and the interval00:30:15
from 0 to 1 2 are included in the segment from 0 to for00:30:17
the second one is of course therefore x 2 is equal to 1 here00:30:22
minus 2 and the roots are definitely not00:30:25
coincide because they do not coincide at 100:30:27
200:30:28
that's why we have two different roots here00:30:30
we feed two different things, I don’t need it, we don’t need it00:30:32
they are needed in such a way that there will be two00:30:35
times by roots if a is equal to 1 200:30:39
if a is equal to 2 then x 1 will be00:30:43
root of the equation for a greater than 0 and x 100:30:45
will be equal to zero x 2 will be equal to00:30:53
equals 1 200:30:54
it will be equal to 1 minus 2 but it too00:30:56
there is a square bracket here yes that's why he00:30:58
is also the root of equation 1 -2 a00:31:01
if a equals 1 2 it turns out 0 that is, y00:31:05
we have two roots but they coincide00:31:07
2 matching roots 2 matching roots00:31:20
so a equals 1 2 will be the answer00:31:22
further if a is more than 1 200:31:25
if a is greater than 0 then x 1 will be00:31:28
the root of the equation, that is, here x 1 is equal to00:31:29
zero and which00:31:33
but it is not the root of the equation for a00:31:35
more than her 2 so only one00:31:38
the root turns out to be Tokyo, everything suits us00:31:40
we write the answer and it suits us00:31:46
equal to zero and a from 1 2 to plus00:31:52
let's run through everything for infinity00:31:55
the solution initially has a square equal to00:31:58
how to solve a square, write down the difference00:32:01
squares, that is, we square to the right00:32:03
we transfer to the left00:32:04
difference of squares a minus b times plus b00:32:08
we simplify and the product is equal to00:32:10
zero plus 1 plus x must be from zero00:32:13
to one when the product is zero00:32:16
when at least one of the factors is 000:32:17
plus 1 plus x from zero to one in short00:32:20
we find the roots 0 and 1 -2 + a to z plus00:32:23
x from zero to one we need so that from00:32:25
two roots only one knife left00:32:28
need to find when there is only one00:32:30
root x 1 is a root for these00:32:33
and x 2 is a root for such00:32:36
and from 0 to 1 200:32:38
but they coincide with this and00:32:42
and we applied everything and looked at what00:32:45
A x 1 will exist for what a00:32:49
x 2 would exist and circle these00:32:51
situations when we either have one root00:32:53
or two roots but coinciding00:32:56
it turned out that with these00:32:58
and in the answer we have only one root00:33:04
the second solution is analytical00:33:07
the first one was graphic now00:33:09
analytically, well, let’s do that too00:33:11
on the right let's move it to the left to take it out00:33:14
en masse, that is, it turns out00:33:17
rewrite the left side minus 5x minus00:33:28
2 l n 2 x minus and all this is equal to zero00:33:35
put 5x minus 2 out of brackets and00:33:39
stays with us00:33:40
in brackets l n minus00:33:43
l.n.00:33:51
so when the product is equal to zero when00:33:54
at least one of the factors is zero and00:33:56
the second one exists, that is, it is necessary that00:34:01
each of them was equal to zero and00:34:04
to be carried out indicate roughly speaking00:34:06
that is, we say that 5x minus 200:34:09
must be equal to zero 2 bracket too00:34:12
must be zero00:34:22
in this case, the decrees x plus must be followed00:34:25
a must be greater than 0 2 x minus a00:34:28
must be greater than 0 and00:34:31
x must be between zero and00:34:33
units let's solve these two00:34:39
elementary equations it turns out00:34:45
two to the right 2 divided by 5 x equals 0.400:34:50
this is the first root00:34:51
second root how to find l n right and00:34:55
remove logarithms00:34:57
that is, x + a is equal to 2 x minus a and k00:35:05
pluses we rewrite more than 0 twos with00:35:08
minus greater than 0x from zero to one00:35:11
inclusive then we express x 2 now00:35:14
let's find it turns out x 1 equals 0 integers00:35:19
four tenths and xyg to the right and left and to00:35:27
2 equals 2a00:35:28
that is, we have two roots 1 2 and both of them00:35:31
should be deleted well no stop now00:35:36
I want to say that they both should00:35:37
satisfy dc or be in range from00:35:41
zero to one see if they are both00:35:43
satisfy there for00:35:44
and are in the interval from zero to one00:35:45
then we have about two roots00:35:51
it is necessary that exactly one of them00:35:53
satisfied dzy was in the segment00:35:55
parent is exactly 1 and the second is not or00:35:59
on the contrary 1 was 2 was not or vice versa 100:36:02
wasn't the second was or00:36:04
so that they coincide with each other, that is00:36:06
there seem to be three cases here, well, about00:36:10
coincidence and we'll talk at the end00:36:12
let's find out x 1 suitable for what00:36:15
and that is, when he satisfies there for00:36:18
and when it is in the range from zero to one00:36:19
and to the second at what a satisfying00:36:22
there for her from zero to one and when00:36:24
they match we'll see in the end00:36:26
shorter shorter00:36:30
x 1 is equal to 0.4 when at a00:36:38
satisfying it turns out about the dose00:36:43
that is, satisfying all this00:36:46
well x is from zero to one but not from00:36:48
zero to one, that is, roughly speaking, a00:36:50
must satisfy two inequalities x00:36:52
plus a is greater than 0 and 2 x minus a must00:36:57
be greater than 000:36:58
substitute x equals 0 4 tenths00:37:01
it turns out 0 4 + and more than 0 2 multiply00:37:07
by 0 4 minus and more than 0 that is, we00:37:11
Now we’ll find out at what a x 1 will be00:37:13
a good root which is a good root00:37:16
the equation will generally be more than -0400:37:18
and less than 08 output therefore when00:37:24
and with the proper from so it is 08 and so00:37:30
-04 to 0 8 x 1 will only be00:37:40
with these00:37:41
and x 1 will be the root of the equation00:37:45
in front of others and he is there, violating the recording there00:37:49
is not a root of the equation x 1 will be00:37:51
to be a root of the equation means when00:38:00
and from minus 0 4 to 8 x 2 at what a00:38:05
x 2 it is equal to 2 and so it will be00:38:10
be the root of the equation for a00:38:12
satisfying dzi also he must00:38:18
get into the interval from zero to one here00:38:20
with him not everything is so obvious if 0400:38:22
clearly included in the interval from zero to00:38:23
units up to 2 but just take some00:38:26
but they enter, that is, this root must00:38:29
satisfy y dc that is x plus a00:38:31
greater than 0 must be 2 x minus a must00:38:33
be greater than 0 and it must fit into00:38:37
segment from zero to one00:38:41
it turns out that under x we substitute 2 a 2a + a00:38:46
greater than zero under x we substitute 2 that is00:38:50
two by two a00:38:53
minus greater than zero under x we substitute00:38:59
a210 in the end I couldn’t choose for me00:39:03
write x or two but let’s do it anyway00:39:06
We’ll probably leave x here and substitute it here00:39:15
2 and it turns out from zero to one should00:39:17
be solved to understand at what a x 200:39:22
will be a suitable root so it turns out 300:39:25
and more than 000:39:26
from which it follows that a is greater than 0 3 a00:39:30
more than 0 5 and more than 0 follows here on00:39:33
multiply one second, everything turns out and from00:39:37
0 to 1 2 it turns out that at00:39:43
but from zero, empty is still more important.00:39:50
for a from 0 to 1 2 square bracket here00:39:55
with these00:39:56
and x is second00:39:59
will be the root of the equation00:40:06
also write separately at what a roots00:40:12
match00:40:13
but we can see it a little later let's00:40:19
let's now say that x1 coincides with00:40:24
x sex 2 when if what if so x 100:40:32
we have 0.400:40:33
equal to the old 2 and that is, if a00:40:38
equals 0 4 divided by 2023 a00:40:42
equals 0 2 roots coincide00:40:47
Yes, after all, on the number line we would have it00:40:50
didn't see it so it's definitely here00:40:52
write x 1 matches to old00:40:54
because we have a equal to 0 2 nowhere yet00:40:56
this has not been noticed before00:40:58
information yes that's what x 1 matches00:41:00
with to old with this and this is important00:41:02
now let's go to the number line00:41:05
let's apply everything we know and see00:41:11
at what and how many roots00:41:12
number line and the first root00:41:16
is the root of minus 0 4 to 08 then00:41:19
there we apply minus 0 4 tenths we00:41:26
apply 0.8 zero one second and 02 like this00:41:33
that is, the next zero is 0 2 1 200:41:45
painted over again, just a dash like that00:41:49
0 0 5 1 2 0 5 and 08 it was still like that now00:41:57
we'll see at what and how long00:42:00
we'll get roots, well that's who I am00:42:02
I usually draw lines you can like how00:42:04
if you want to do it differently there, I’ll do it like this00:42:07
I do the usual purely for myself and so if00:42:13
let's first consider the situation if00:42:16
and less than -04 then what if a00:42:19
exactly, exactly -04 what will happen if a from00:42:24
-04 to zero then what if a is equal to 0 then what00:42:29
will be if a from 0 to 0 2 then what00:42:33
how many roots will occur if a is exactly 000:42:35
2 then how many roots if a is from 0 2 to 0 500:42:39
then how many roots if a is exactly 0 5 then00:42:42
how many roots if a is from 05 to 08 then00:42:46
how many roots if a is 08 then00:42:49
how many roots if a is greater than 08?00:42:51
how many roots let's give it all00:42:52
analyze00:42:53
so if a is less than -04 then the first00:42:58
root and is a suitable second00:43:02
the root too, that is, it’s simply not here00:43:06
roots00:43:09
if a is exactly 04 minutes00:43:12
exactly -04 then the first root is not00:43:15
the second one is also suitable if a between -0400:43:25
and zero then the first root is00:43:29
suitable and it is equal to 0 4 that is, here x00:43:33
1 is equal to 0.4 and x 2 x 2 appears only00:43:39
from 0 to 1 2 that is x 2 not yet means00:43:42
at and from -04 to zero we have one root like this00:43:46
and we are satisfied00:43:48
if a is equal to 0 to 1 the root exists and00:43:52
it is equal to 0.400:43:57
if a is 0 then 2 roots do not exist00:44:01
because it exists from 0 to 1 200:44:03
inclusive therefore a exactly zero me00:44:06
also satisfied with the answer and will move on00:44:08
if a is from 0 to 0 2 then the first root is there00:44:14
exists and is equal to 0.4 and the second root00:44:20
the second root for 0 to 0 2 exists00:44:25
from 0 to 0 2 is included in the segment but in00:44:27
interval from 0 to 1 2 and the root will be00:44:30
is equal to 2 x 2 will be equal to 2 and that is00:44:35
will change all the time if and equals00:44:37
0.1 then it will be 2 multiplied by 0 1 but obviously00:44:40
they won't match because they will match00:44:42
at 0 200:44:43
when equal to 0 you figured this out earlier00:44:46
we have two roots to 200:44:48
root00:44:49
I don’t need two minutes until 1, let’s continue00:44:51
see if a equals 0 200:44:54
if a is equal to 0 2 to 1 the root exists00:44:58
and it is equal to 0.400:45:02
if a is equal to 0 2 then the second root00:45:07
exists and it is equal to what it is equal to 200:45:10
multiply by and that is 2 multiplied by 0 200:45:13
that is, 0 4 we have, as it were, two roots like00:45:17
before but now they coincide, that is00:45:21
we can say that we have one that matches00:45:22
the root, you understand, that is, and me00:45:25
already satisfied with one matching root a00:45:29
equals 0 200:45:31
that's good it fits now if a00:45:35
between 0 2 and 0 5 1 the root exists and it00:45:41
is equal to 0.4 and the second root00:45:45
between 0205 also exists and it’s some kind of00:45:50
the other is just 2 and that means we have two roots00:45:52
I don't need 2 roots that's not what I need00:45:56
I'm looking for which will have one root like this00:46:00
if a is equal to 0 5 to 1 the root exists00:46:03
and it is equal to 0.400:46:07
if a is equal to 0 5 then the second root00:46:09
exists and it is equal to 2 a but if a00:46:14
equals 0 5 then 2005 is one and we have00:46:17
it turns out there are two different roots, I’m not the same00:46:22
and I'm looking for when there will be two different roots if a00:46:24
from 05 to 08 then the first root exists00:46:29
and it is equal to 0.4 and the second root is from 05 to00:46:35
08 doesn't exist because he00:46:37
exists up to 0 5 inclusive00:46:39
everything is missing, that is, we have one root00:46:42
means such a segment is such an interval00:46:44
I'm happy if a is exactly zero eight00:46:48
then 1 roots do not exist and the second one does not exist either00:46:52
that is, there are no roots at all if a is to the right00:46:58
than 08 it will also not give roots00:47:00
friend00:47:04
that's all that we circled goes in response00:47:07
there is an answer like this00:47:10
from minus 0.400:47:13
not inclusive up to zero inclusive00:47:16
then two separate tests00:47:18
tenths and from 05 to 08 not inclusive00:47:24
or another problem like this analytically00:47:29
decided that is, we threw the right one00:47:31
part to the left to take the total out00:47:33
bracket and say when the product00:47:35
equals zero, found two roots 042 and we have00:47:39
there is the same there is a segment from zero to00:47:41
units if x 1 satisfies00:47:44
Judas for her is in the segment then he00:47:46
suitable if x 2 satisfies there00:47:50
behind and is in a crack then he00:47:51
suitable so x 1 suitable for00:47:54
such a x 2 suitable for such a and00:47:59
we all do not coincide, they are equal to a equal to 0 200:48:03
we put all these on a number00:48:06
direct and look at what and how much00:48:08
roots and some roots and circle those00:48:13
situations where00:48:14
exactly one root turns out like this00:48:18
task find all values for each00:48:22
of which the equation has exactly one00:48:23
root on the interval from 0 to 3 meaning00:48:26
the solution is to find00:48:29
all the roots of the equation are there one two or00:48:31
three most likely three00:48:33
here one here two here legs 2 I don’t know00:48:38
Briefly speaking find all the roots, there are two or three of them and00:48:42
look at which ones and each of them00:48:45
is the root of the equation for 3 rows 3 then00:48:48
is at what and this root00:48:50
satisfies there for00:48:51
and is in the segment da trick end well00:48:54
then see at what and how much00:48:56
there will be some roots in the stern and00:48:57
in general, well, where to start is obvious before00:49:00
product equals zero when when00:49:01
each of the factors is equal to zero and also00:49:04
ode for fulfilled yes that is we write00:49:06
that l.n. 4x minus 1 equals zero or00:49:12
the second factor is zero00:49:24
plus under logarithmic should be00:49:26
greater than 0 plus radical must be00:49:31
greater than or equal to 0 plus x must be in00:49:40
segment from 0 to 3 approx. all this is included in the system00:49:43
you can push it so let's decide two00:49:47
equations can, in principle, be written directly into00:49:49
system to solve them and so go in parallel00:49:51
go to the end yes but you can just do them00:49:53
decide somewhere separately and then00:49:56
back to the system let's do this00:49:57
that is, we have equation 1 equation00:50:00
2 and we'll deal with them first00:50:02
equation 1 l n 4x minus 1 equals zero y00:50:07
natural logarithm in base e we00:50:10
and raise to the zero power we get00:50:11
logarithmic and classical00:50:13
solve the equation, that is, it turns out in00:50:17
0 equals one is 4x minus 1 one00:50:23
to the left that is 2 equals 4 x x equals 200:50:27
divide by 4 1 2 this is x 1 then the second00:50:33
equation solve root is zero when00:50:37
its radical is equal to zero, that is, x00:50:39
square minus 6 x plus 6x minus a00:50:45
square equals zero quadratic equations00:50:47
to equal zero when so well here you can00:50:53
try it somehow, of course, if it builds00:50:57
this equation is a circle obvious but00:50:59
if not build 100 if analytically00:51:03
you can decide by the discriminant00:51:05
find roots00:51:09
through fault, perhaps through fault, too00:51:12
it seems possible the sum is 6 and the product is 6 and00:51:15
minus and square, well you can guess00:51:17
probably but by discriminant probably00:51:19
it will be easier from the thickness of normal roots00:51:21
there is a group, I think it’s possible00:51:23
somehow a group can also be grouped00:51:24
can be grouped shorter00:51:26
either by discriminant y or by vieta but00:51:29
you need a lot of imagination or grouping00:51:32
let's group shine remnant you00:51:34
you can through the tap this is coefficient b00:51:37
this is the coefficient a this coefficient is c00:51:39
discriminant two roots is xcom 200:51:42
3 also 1 is already taken, well, grouping00:51:45
steeper than x square minus and square is00:51:48
difference of squares x minus a by x plus a00:51:51
and from minus 6x plus 6 a can be taken out00:51:56
minus 6 out of brackets will leave x minus y00:52:00
equals zero now x minus and we take it out00:52:03
parentheses and remains x plus and minus 6 when00:52:09
the product is equal to zero when at least00:52:11
one factor is zero, that is, x00:52:12
is equal to and this is x 2 is already and x 3 is equal to00:52:16
it turns out that well, here it is -6 on the right00:52:20
reroll and get 6 minus otherwise00:52:23
there are three roots of us 1 2 3 so let's go back to00:52:26
system let's go back to the system00:52:35
we have this kind of system to00:52:42
x 1 is equal to 1 2 x 2 is equal to and x 3 is equal to 600:52:50
minus a and we have three more conditions 4x00:52:57
minus 1 is greater than 0 well, we can solve this00:53:00
one to the right and divide x by 4 both sides00:53:03
more than by 4 so also x square minus00:53:09
6 x plus 6x minus and the square is greater or00:53:14
equals 0 and x must be between 0 and 3 in00:53:19
In principle, you can find the intersection here00:53:21
there are more than 1 4 of these fixies but00:53:24
less than or equal to three is obtained00:53:26
let's simplify it a little let it be00:53:28
a little simpler, that is, we will get something like this00:53:31
Well, there will be a little less systems00:53:34
although it was possible to substitute00:53:36
leave x 1 equal to 1 2 x 2 equal to a x 300:53:41
equals 6 mitus and00:53:45
x squared minus 6 x + 6 a minus a00:53:50
square must be greater than or equal to 0 and00:53:52
x must be that is, we find00:53:54
suppression more than 1 4 from 0 to 3 x00:53:57
must be from 1 4 to 3 inclusive all00:54:01
when x 1 is the root of the equation for00:54:04
cod when he satisfies dc and00:54:08
lies in the segment but she lies like that in00:54:10
segment therefore only when he takes away00:54:12
y even when x 2 is a root when00:54:15
he satisfies dc but in general he00:54:17
will hit the dose because he00:54:20
obtained from the radical equation equal to00:54:24
0 so he will definitely hit after us00:54:26
no big deal, let's check it again00:54:29
fashion when he lies in the segment in00:54:31
interval from 1 4 to 3 and when x 3 is00:54:34
be the root level00:54:35
when Zeno will be inside there again00:54:37
from here and when it lies in the segment in00:54:39
in the interval from 1 4 to 3 another 1 x 1 equals 100:54:46
2 when with a satisfying here dz00:54:54
greater than or equal to 0 well, that’s already x1 4 to00:54:58
3 but he is already from 1 4 to 3 1 200:55:00
between 1 4 three so just water for00:55:05
must be performed at the same time and jambs00:55:07
minus 6x plus 6x minus and the square is large00:55:12
equals 0, substitute the second one under x100:55:14
it turns out 1 4 minus 3 plus 6 a minus a00:55:18
square00:55:19
greater than or equal to 0, that is, minus a in00:55:22
squared plus 6x 1 4 minus 3 is minus00:55:26
11 quarters greater than or equal to 0 let's00:55:31
We multiply everything by -4, that is, we get 4 a00:55:38
square minus 24 + 11 less than or equal to 000:55:45
it looks kinda bad but ok00:55:49
discriminant 576 -16 by 11 176 good 2000:56:00
square equals 24 plus minus 20 divided00:56:06
by 8, that is, we are on the number line a00:56:10
We apply that forty four divided by 8 is00:56:14
five and a half and00:56:19
4 I'll take them, that is, 0 5 and for the second00:56:24
Well, let 05 be like this, let’s arrange it like this00:56:29
intervals plus minus plus signs00:56:31
we need and for which the value00:56:34
expressions less than or equal to 0, that is, we00:56:36
Minsk needs all the dots painted over00:56:38
these00:56:39
and x 1 is the root of please and the conclusion00:56:41
let's do therefore for and for00:56:45
proper interval from 0 5 to 5 s00:56:48
half x 1 is the root of the equation00:56:52
well x 1 is equal to 1 2 but let's put it this way00:56:56
is the root of the equation on the interval a00:57:02
at what a00:57:03
x 2 is the root of the equation let's00:57:06
let's find out x 2 which is equal to a00:57:14
is the root of the equation for a00:57:16
satisfying what requirements00:57:19
firstly it must be from 1 4 to 3 thousand00:57:25
x must be between 1 4 and 3 and it must00:57:29
satisfy u dose to be honest we00:57:31
we understand that he will definitely satisfy there00:57:32
for if you don't understand then turn it on00:57:35
this system won't get any worse00:57:39
it just won't affect anything00:57:42
will get well here very easily, that is, under00:57:44
x and we substitute a must be from 1 4 to00:57:46
3 and here a we substitute here it will be a00:57:50
square minus 6x plus 6x minus a square00:57:53
0 is greater than or equal to 0 for any a output00:58:00
at00:58:02
and with proper from 1 400:58:05
let's put it in decimals00:58:07
we write then when an belongs to 0.25 to00:58:11
3 inclusive x second which is equal to a00:58:15
will be the root of the equation00:58:26
so a x 3 which was 6 minus a00:58:30
is the root of the equation when x 3000:58:35
equals 6 minus a is the root00:58:39
equations for00:58:40
so with a here it was with a,00:58:42
satisfying for a00:58:44
, satisfying what requirements00:58:47
again x must be from 1 4 to 3 and must00:58:53
be performed ude for but again00:58:54
since 6 x is equal to 6 minus and we and00:58:57
got by solving the equation00:58:59
radical equals 0 then of course it is00:59:03
satisfies the decrees but if we don't00:59:05
understood then we write x square minus 6 x +00:59:10
6 a minus a squared is greater than or equal to 0 c00:59:13
system and so under x we substitute 6 minus00:59:15
and from 1 4 to 3 is inclusive00:59:20
so under x we substitute 6 minus 1 6 minus00:59:23
and now there will be a lot of tabs to solve at least00:59:25
obviously everything fits six by six00:59:27
minus plus 6x minus a square is large00:59:32
equals 0 subtract 6 from all three sides00:59:36
double inequality here it turns out00:59:39
minus a from 1 4 minus 6 minus 5 7 5 300:59:46
minus 6 minus 3 open the square00:59:51
difference 36 minus 12x + a squared minus00:59:57
3 6 + 6 + 6 a minus a square large01:00:02
equals 0 equals 0 turns out look 1201:00:06
12 and in return is destroyed by a square01:00:08
square is 36 minus 36, that is, on01:00:12
minus 1 multiply it turns out 5.75 3a01:00:18
but the icons need to be expanded, that is, because01:00:21
that we multiply negatively and there will be more01:00:23
or equal here there will be strictly more output01:00:28
at01:00:29
and with proper from 5-7 so from 301:00:33
it turns out yes01:00:36
yes 5.75 not inclusive x 3 which01:00:41
equals 6 minus a01:00:43
is the root of the equation on the interval well01:00:50
for crackling from 0 to 301:00:53
now let's see at what a01:00:55
the roots coincide here x 1 coincides with k01:00:58
old at what x 1 sex coincides01:01:04
old if what happens if x 101:01:10
equal to and which that is x 1 then we have 101:01:13
2 x 2 is equal and here is the output when a is equal to 0.501:01:23
x 101:01:25
to the rabbi they coincide with the old one01:01:26
further01:01:28
x 1 can coincide with ex third if01:01:35
what happens if x 1 is equal to x third01:01:40
that is, exactly five and a half conclusion01:01:45
at exactly five and a half x 1 is equal to x 301:01:51
match and x 2 can match sex with01:01:55
third if x 2 is equal to x third output01:02:10
when equal to 3 x 2 correct we will meet01:02:16
now let's see at what a01:02:17
how many more roots will we have in this time?01:02:19
roots we construct a large number line a and01:02:25
we apply everything to it that affects01:02:29
the number and type of roots means x 101:02:33
exists at a from 05 to five s01:02:36
half x 2 from 0 25 to 3 x 3 from 3 to 501:02:43
5 75 hundredths, that is, you need to apply the most01:02:45
small 025 then 05 then three five s01:02:48
half and 57501:02:49
so you need to lead to 025 101:02:55
then 05 then three it seems like yes so three01:03:03
five and a half 57501:03:08
five and a half and 575 we'll see what01:03:16
will be if a is less than zero 25 if a01:03:18
exactly zero 25 if a is from 0 25 to 0501:03:24
if exactly 0 5 if a from 0 5 to 301:03:31
if a is equal to 3 if a is from three to five c01:03:35
half if a is equal to five and a half if a is from01:03:39
five and a half to 5.75 if a is equal01:03:44
5.75 if a is more than 5.75 how much01:03:49
there will be roots what kind of roots so when01:03:52
exists x 1 from 0 5 to 5 and a half01:03:55
inclusive x 1 is equal to 1 2 that is, x is like that01:04:02
Let's01:04:03
small dashes01:04:16
ikos 1 again from 0.5 to 5 and a half01:04:20
inclusive i.e. x 1 is the root01:04:23
here and it is equal to 1 2 x 1 is the root01:04:27
when from 0 5 to 3 when a is equal to 3 x 101:04:32
is the root01:04:33
with a from three to five and a half x 101:04:36
is a root of a exactly five c01:04:38
half it and the root everything from 05 to five01:04:42
and half inclusive x 1 is01:04:45
root x 2 when is the root of 0 2001:04:48
25 acres not inclusive up to 301:04:50
inclusive i.e. x 2 is the root01:04:54
here here here and here from 0 25 not01:05:05
inclusive up to 3 inclusive and x 301:05:08
is the root from 3 inclusive to 501:05:10
of 5 not inclusive i.e. x 3 equals 601:05:13
minus and here here here01:05:22
and then at 5.75 it turns out there are no carnies01:05:30
so look if a is less than 025 then01:05:33
no roots01:05:36
if a is exactly 0 25 then there are no roots01:05:43
If01:05:44
and exactly 55 then there are no roots if a is greater01:05:51
than 75 75 then there are no roots so and now01:05:54
let's see if there's a couple coincidence x01:05:57
1 coincides with k old when a is equal to 0 501:06:00
that is, if a is equal to 0 5 then see x 101:06:04
the same as 2, that is, we actually have01:06:06
there will actually be one root here because01:06:10
the roots coincided you understand two roots but they01:06:13
matching so there will be only one01:06:15
the root means this and that suits me01:06:17
further when we still have roots there01:06:20
coincide at equal to five and a half 1 s01:06:22
the third coincides, that is, if a is exactly01:06:25
five and a half substitute then the first01:06:26
meet will coincide and there will be only one01:06:28
root or 2 matching ones can be written01:06:30
or just 1 that is exactly five s01:06:33
half will also respond with the second one01:06:36
the third coincides with the prior inside01:06:39
if a is equal to 3 toeic 2 is equal to 3 x 3 is equal01:06:42
6 minus 3 out of 301:06:43
and we have not 3 roots but two because 2 301:06:48
the same will be 1 2 and 3 you understand then01:06:50
there will actually be two roots01:06:53
we were looking for we were looking for what will happen01:06:58
exactly one root01:06:59
so if a is less than 025 there are no roots01:07:03
if a is exactly 0 25 there are no roots if a is from 001:07:06
25 to 0 5 1 root of 201:07:08
if a is equal to 0 5 then we have one root01:07:11
matching x 1 2 1 5 to 3 we have two01:07:15
we have three different prior roots inside01:07:18
there are 2 roots left because there are two01:07:20
coinciding with a from three to five c01:07:22
half two different roots at exactly01:07:24
five and a half we have one root at a01:07:26
from five and a half to 575 we have one01:07:29
root at again 75 and no more roots01:07:32
I just had to enter five s here01:07:34
circled it with half01:07:35
they are there for five and a half hours of pipes and01:07:42
this will be the answer, that is, what we01:07:46
Circled the hour and write the answer like this01:07:52
from 0.25 not inclusive to 0.501:07:58
inclusive and from five and a half01:08:02
inclusive up to 5.75 not included or01:08:07
let's go through the entire solution find all01:08:12
for each of which the equation has01:08:13
exactly one root on the interval from 0 to 301:08:15
when the product is zero when although01:08:18
one of the factors is equal to zero plus and01:08:20
we even add plus and say that x is from 001:08:21
up to 3 solve each equation worked01:08:24
3 roots plus 1 plus x and there, well, here we are01:08:28
crossed x more than one-fourth of 001:08:30
to 3 it will be from 1 4 to 3 x 1 will be01:08:33
be a root if it satisfies y01:08:36
even if it is in between it01:08:39
on 4 this is 3 he and so in the interval from a to01:08:41
4 to 3 means he needs to hit even01:08:43
he satisfies there for such a01:08:46
x 2 is a root if it is in between01:08:50
from a on 4 to 3 January there for that is01:08:53
with these a01:08:54
x 3 drive and quietly is the root x 101:08:58
coincides with k old when a is equal to 0 5 x01:09:00
1 coincides with ex third about five01:09:02
I'll meet you and half 2 right inside01:09:03
then we look at what and how much01:09:06
we have roots and what kind of roots they are01:09:08
here it is in the end01:09:10
circle those a for which it turned out exactly01:09:13
one root01:09:16
find all values of a for each of01:09:19
of which the equation has exactly one01:09:20
root on the segment from 0 given from what01:09:22
start right side left with minus01:09:25
throw take out general find roots on01:09:28
write the entire review x from 0 to units a01:09:31
then solve the equation and find x1 x201:09:34
maybe x3 and see at what a01:09:39
x 1 will be the root of the equation at01:09:41
what a x 2 for which they coincide and01:09:44
then at the end analyze at what01:09:46
what kind of roots and how many roots01:09:49
it will be the end, so let's move this one01:09:51
which is from right to left with a minus01:10:08
-16 a minus x by ln x minus01:10:16
all this is equal to zero, now we add l.n.01:10:19
general01:10:20
outside the bracket, that is, l.n. 6 a cons and c01:10:28
big bracket remains the difference01:10:29
logarithms 2 x plus 2 minus 2 minus ln x01:10:39
minus y equals zero when the product01:10:43
equals zero when at least one of01:10:45
factors equal to 0 2 exists, that is01:10:46
we say that lnr 6 and minus x is equal to01:10:53
zero or the difference of logarithms in is equal to01:10:58
zero01:11:06
in this case it must be executed up to z and x01:11:10
must be from zero to one 36 a01:11:11
minus x greater than zero other under01:11:16
logarithmic 2 x plus 2 minus 2 too01:11:20
greater than zero 6 a minus x greater than zero x01:11:24
minus a is greater than 0 and x must be from zero01:11:29
up to one we get such a system01:11:34
Now we solve the equation both equations01:11:36
very light, well here we have it01:11:41
it means yes and we are at zero01:11:43
degree and raise we get under01:11:45
logarithmic, that is, it turns out that01:11:53
it turns out that 6 and minus x is equal to one01:12:00
here we move ln to the right and remove it01:12:03
deer01:12:04
that is, it turns out that 2x plus 2 minus 201:12:09
equals x minus a01:12:12
we also rewrite the doses01:12:24
x from zero to one now let's01:12:30
let's express xv to each comparison, that is, x 101:12:35
equal to what01:12:37
x from the right to one to the left, that is, and01:12:39
sides in places other than 6 and minus 101:12:42
it turns out this is x 1 x 2 level what and01:12:45
as left left would be x 2 x minus x x01:12:50
and 21 minus two to the right, that is01:12:54
it turns out x 2 is equal to what minus three plus01:12:59
two01:13:00
these are the two roots we have01:13:03
must be executed until z l36h minus x01:13:06
greater than 0 2 x plus 2 minus 2 greater than 0 as01:13:13
long ago it was possible to divide by 2 well01:13:15
it doesn’t matter x minus a is greater than 0 and x is from zero01:13:19
up to and including one01:13:21
Next we need it to work out01:13:25
one root on the interval from 0 to 101:13:28
let's find out at what a x 1 will be01:13:30
is the root of the equation for segments at01:13:33
which a x 2 will be a root on01:13:35
segment from 0 to it when they coincide01:13:38
and then analyze at what a01:13:40
how much for roots01:13:42
in short x 1 is equal to 6 a -1 for what a for01:13:51
A01:13:52
satisfying everything even roughly speaking01:13:57
and from zero to one it should be01:14:01
that is, what should be in the first 601:14:05
and minus x must be greater than 001:14:08
Actually, the first equation came from01:14:12
the fact that 6 and minus x is equal to one then01:14:14
there is 6 and minus x is greater than zero01:14:15
runs automatically but nothing01:14:17
terrible thing, we’ll write it down again so that01:14:20
don't get confused like that 2 minus 2 is greater than 001:14:25
must be satisfied x minus a greater than 001:14:28
must be satisfied and x from zero to01:14:30
units must be fulfilled01:14:34
let's solve this little system01:14:37
then we do the same for them great01:14:42
since we will solve this system well let's01:14:49
we will substitute 6a -1 s for x01:14:51
beginning, that is, under x I substitute 6 minus01:14:56
1 will be 6 and then minus x, that is, minus01:14:59
6a + 1 must be greater than 0 for any a01:15:04
it will be greater than zero01:15:05
further along the dex we substitute 6 a minus 101:15:07
that is two01:15:08
multiplied by 6x minus 1 plus 2 minus 201:15:16
bodex substitute 6 a minus 1 minus a01:15:20
more than zero under x6 and minus 1 turns out01:15:30
What's in 1 answer and we'll reveal any later01:15:40
brackets will be 12 a minus 2 plus 2 minus01:15:47
26 minus and this will be 5a minus 1 more01:15:56
zero here we add one to everything01:16:01
all three sides of the double inequality01:16:03
that is, it turns out 16 a 2 well, any01:16:11
You don’t have to write it, it doesn’t affect anything01:16:13
so here 14 and more than 45 and more01:16:23
one here by 1 6 multiply all three01:16:28
parts of the double inequality, that is, a from01:16:32
1 6 to 2 sixths to one third well further01:16:39
but it turns out to be more than 4 14, that is, than01:16:43
2 sevenths and more than 1 501:16:47
and a from one sixth house 1 301:16:52
let's find by intersection01:17:01
these 3 answers, that is, we are on01:17:04
number line a01:17:07
apply this way 2 sevenths 1 5 1 pcs and who01:17:12
least of all well 1 6 obviously least of all01:17:23
then comes probably 1 5 t co2 sevenths01:17:26
almost 2 sims, well 1 5 is two01:17:29
tenths therefore if we compare the two01:17:32
tenths and 2 sevenths 2 tens of course01:17:34
less than 1 5 we have further so but 201:17:39
sevenths 1 3 if we compare01:17:41
Well, there’s a common denominator to find like this01:17:44
2 sevenths 1 3 common denominator twenty01:17:47
one additional multiplier of 3 here 7601:17:52
20 first and 720 first 1 3 big by 301:18:01
more but this is a draft01:18:03
leave how we compare fractions is01:18:06
for parameters probably for 18 tasks not01:18:09
it is important to indicate such moments in01:18:12
solution you can indicate it doesn’t matter so01:18:14
more than 2 sevenths and more than 1 5 a01:18:21
in the segment from 1 6 to 1 3 where three01:18:26
shading there the answer is from two01:18:28
sevenths to 1 3 output at01:18:32
and with proper from two sevenths to 1 301:18:38
x 101:18:40
which is equal to what is 6 and minus 1 will be01:18:46
be the root of the equation for these a01:18:49
x 1 will be the root of the equation01:18:51
know01:18:59
let's find out at which ones and there will be 2 of them01:19:01
be the root of the equation01:19:04
x 2 it is equal to what it was x 2 it was equal to01:19:10
us minus 3 plus 2 minus 3 plus 2 like us01:19:20
made out we said x 1 will be01:19:22
root of the equation for a satisfying01:19:24
ode launch garden zero to one the same01:19:27
most x 2 will be the root01:19:29
equations for a01:19:32
satisfying a to z plus x from zero to01:19:35
units so here it is01:19:39
we have these waters here and here01:19:42
from 0 to 6 and minus x must be greater than 001:19:46
2 x plus 2 to minus 2 must be greater01:19:51
0 x minus a must be greater than 0 and x01:19:56
must be between zero and one01:19:59
substitute x now minus 3 plus 201:20:02
and besides at what aek 2 will appear01:20:04
the root of the equation is 601:20:08
multiply by a minus x, that is, minus here01:20:13
this expression is getting plus 3 minus 201:20:17
greater than 0 2 must be multiplied by x, that is, by01:20:21
minus 3 plus 2 plus 2 minus 2 greater than 001:20:28
instead of Mexico we substitute minus 3 plus01:20:30
2 minus and greater than 0 and x sata minus 3 plus01:20:39
2 and it should be in double inequality01:20:42
from zero to one we solve it turns out 6 a01:20:49
plus 3 and that's 9 and that's more than two here01:20:54
open the brackets minus 6 plus 401:21:02
plus 2 minus 2 more than 0 minus 3 minus a01:21:09
this is minus 4 to the right, we flip the sides01:21:13
me in some places that is 4 less than two here01:21:18
let's subtract -2 from all sides01:21:20
it turns out minus 2 minus 3 minus 101:21:28
divide both sides by 9 and it turns out01:21:31
more than 2 9 minus 6 plus 2 and this is y01:21:37
we move minus 4 to the right, it will be 401:21:40
and less than 24 minus 2 is 2 and less01:21:46
than 1 2 well because we will divide by 401:21:50
any part so here let's take a minus01:21:52
one third and multiply both parts01:21:54
it turns out a 2 3 1 3 icons01:22:01
unfold01:22:10
Well, if we divide 4 by 4 less than two01:22:15
then it will be a less than 1 2 and we are already01:22:18
we wrote it down, that is, let's find it01:22:20
intersections we need to find intersections01:22:30
and more than 2 ninths and more is less01:22:34
than she 2 and a from 1 3 to two thirds01:22:41
number line so who is the least 1 301:22:49
or 2 ninths 2 ninths is less than01:22:52
02 even 1 3 is 03301:22:55
so of course 2 ninths of all 1 301:23:01
further01:23:03
half further and two thirds to the right01:23:07
everyone there 06601:23:09
Approximately, we apply more than 2 ninths01:23:12
a is less than 1 2 and a in the interval from 1 3 to01:23:21
two thirds where there are three hatches there is the answer01:23:23
that is, the conclusion for a for the proper one from 101:23:30
3 to 1 2 x 2 which is equal to minus 3 plus01:23:38
2 will be the root of the equation01:23:54
let's find out at what conditions they01:23:56
match01:23:57
x 1 x 2 that is, 6 a -1 must be equated to01:24:00
minus 3 plus 2 x 101:24:04
coincides with the old ones, well, let's do this01:24:13
if 6 and minus 1 is equal to minus 3 plus 201:24:22
when does this happen let's01:24:24
to the left it turns out 9 and is equal to 3 that is, a01:24:29
equals 1 3 output when a equals 1 3 x 101:24:38
By the way, what is 6 times one?01:24:41
the third -1 that is x 1 is equal to one as x01:24:47
2 also if a is equal to 1 3 minus 3 multiplied01:24:50
by 3 plus 2101:24:51
so x 1 is equal to 1 as x 2 that is they01:24:55
coincide, yes we all know at what a01:24:59
x 1 is the root of the equation for what01:25:02
and x 2 is the root of the equation and at01:25:05
which and the roots coincide, that is, it will be01:25:08
two roots of the equation but identical01:25:09
therefore there is only one root of the equations in01:25:11
in the end, let's now give a larger number01:25:13
Let's construct a straight line on to understand when01:25:15
what and how many roots and what kind of roots01:25:19
large number line01:25:27
which ones affect the number of roots01:25:32
that is, x 1 is the root of these01:25:35
A01:25:36
apply 2 sevenths one third then01:25:39
one third one second that is two01:25:41
sevenths one third one second only01:25:45
that's just them and we apply 2 sevenths01:25:49
one third and one second01:26:03
we will consider the situation when a01:26:05
less than 2 sevenths when a is exactly 201:26:08
sevenths when and between 2 sevenths and 1 301:26:14
when a is exactly 1 3 when a is between 1 3 and 101:26:21
2 and when a is equal to 1 2 and when a is greater01:26:26
than 1 201:26:29
so we make such small sticks if01:26:41
and less than 2 sevenths does x 1 exist01:26:44
is it the root of the equation if a01:26:46
less than 2 sevenths x 1 is the root01:26:50
the equations are for these01:26:51
and therefore if less than it is not a root but01:26:55
x 2 drive such a is the root01:26:58
equation x is less than two sevenths it is not01:27:00
root that is, with less two sevenths01:27:02
there are no roots is when a is equal to 201:27:11
sevenths someone root a equals 201:27:13
sevenths 2 sevenths are not included here therefore01:27:17
x 1 is not 2 sevenths are not included here01:27:20
therefore their sides are not equal01:27:23
2 sevenths again no roots01:27:27
if we take a from two sevenths to 1 301:27:31
the interval from two sevenths to 1 3 is included01:27:33
here therefore x 1 is the root that is01:27:37
x 1 is the root and it is equal to 6 a minus01:27:40
1 is it a root x 2 is it not01:27:45
root starting from 1 3 right 1 3 tons 301:27:48
given 2 so here it is not yet a root01:27:51
is and we have only one root01:27:54
root I’m actually looking for something like this when I’m alone01:27:56
root if a equals 1 3 then x 101:28:02
is the root01:28:03
to 1 3 is included here means x 1 is01:28:07
root and it is equal to one01:28:10
x 1 it is equal to 6 and minus 101:28:14
well, when a is equal to 1 3 it turns out to be equal to01:28:16
unit 1 3 included here included01:28:21
therefore the old one is also a root01:28:23
and it is also equal to one x 2 it actually01:28:27
usually equal to minus 3 plus 201:28:28
but when a is equal to 1 3 it is equal to one then01:28:32
we have, like, two roots01:28:33
but these are two matching roots so01:28:37
we get only one root01:28:38
different, different roots one thing two01:28:41
coinciding roots therefore such and me01:28:44
they also suit me in return too01:28:46
I'll take one third further if a from 1 301:28:50
the given 2 is located or the interval from 1 301:28:54
to 1 2 here I don't have it here01:28:56
therefore x 1 no x 1 not the root is found01:29:00
is the interval from 1 3 given 2 inside here01:29:02
of course yes that's why01:29:05
here x 2 will be the drive of these a01:29:07
x 2 will be the root of the equation01:29:10
minus 3 plus 2 x 1 there won't be only one01:29:16
the root suits me, I'm looking for something like this01:29:18
if a is equal to 1 2 x 1 will be01:29:22
the root of the equation is that he will never again01:29:23
won't be the smoking equation he01:29:25
is the root of the equation only when01:29:27
these and she is the second one already more and that’s it01:29:29
the rest will be even more that is x 101:29:31
no longer root x 2 no longer either01:29:34
a root is not a root when equal to 1 201:29:37
because it is not inclusive up to 1 201:29:40
the root is xtar and01:29:43
and for a greater than 1 2 it is also not a root01:29:46
we have no roots here01:29:49
neither the first nor the second are roots and there are none01:29:53
corr not everything works out suits us but from01:29:58
two sevenths to 1 3 1 3 too and from 1 3 to01:30:02
1201:30:03
so why should we somehow01:30:06
separate one third if it01:30:07
inside the gap that suits us01:30:11
you see, that is, the answer is two-sevenths01:30:14
to 1 2 just one third why it01:30:19
highlight separately if it01:30:22
if this point also suits us therefore01:30:24
the answer is about two-sevenths01:30:29
up to 1 star and go through the entire solution01:30:36
and so initially what we did we01:30:41
the right side was moved to the left so that01:30:43
put general out of brackets put out01:30:46
the result is the product of two factors01:30:48
is equal to zero, we equate each to zero01:30:50
plus yes for hang plus x from zero to01:30:51
units simplify simplify simplify01:30:54
we get that we have x 1 there is x 2 and o01:30:56
d z and k sat zero to one01:30:58
so x 1 will be the root01:31:00
equation if it satisfies everyone01:31:02
restrictions in general to everyone and is from01:31:05
zero to one we found out what it is01:31:07
occurs at the following a a 2 7 to 1 301:31:10
x 2 will be the root of the equation if01:31:13
it satisfies all modules01:31:14
Well, everyone is limited by water and is located from01:31:17
zero to one we found out what it is01:31:18
occurs when a from one third is given 201:31:22
the roots coincide with each other when a is equal to01:31:25
1 3 we equated them and understood that it was equal01:31:27
dendrite they fall off equal to one01:31:30
then we built a number line01:31:35
and looked at which roots01:31:38
there is where there are no roots where there is only 101:31:41
root grandfather only 2 so they circled those a01:31:43
in which we only got one01:31:45
root home abundance01:31:47
situation when we have two roots but 201:31:49
we only have one matching root01:31:51
find the root of one01:31:58
all values for which the equation has01:32:00
the only solution on the interval from 0 to01:32:01
301:32:03
there are some similarities here to x01:32:06
square minus 7 could be replaced01:32:07
but there is still 2 x minus, why change it too01:32:10
that to no other letter like x01:32:11
square minus 7 on the mind and 2x minus it in01:32:14
school01:32:15
but somehow it’s not a very good idea yet01:32:18
you know what comes to mind just stupidly01:32:21
squared in envy and the sum that is01:32:24
square the sum open on the left and there in01:32:28
during expansion x square minus 7 in01:32:29
it will be square and we will destroy them01:32:32
we will destroy them then the square root is01:32:36
2x minus and this is to destroy give then01:32:39
there is clearly enough to reveal01:32:41
simplify the square of the sum to otherwise we get01:32:46
equation but we must remember that01:32:48
radical is greater than or equal to 0 and well01:32:50
that sharply we have 0 to 3, that is, roughly01:32:53
speaking of visa we have 2 x minus a01:32:55
greater than or equal to 0 and here we have the equation01:32:58
when we decide we need to have roots01:33:01
equations so that the roots of the equation01:33:06
satisfied and the DZs lay in the segment well01:33:10
as it should, not all the roots are necessary01:33:13
Let's say this equation has three roots01:33:15
before it is not necessary that all three roots and01:33:17
satisfied dc and lay on the segment01:33:20
if this happens we have three roots01:33:21
and we need to have only one01:33:24
root, that is, there must be some roots01:33:26
were suitable and some01:33:28
inappropriate in short so confused you01:33:32
let's first simplify the obvious01:33:34
the first action is x square minus 701:33:37
take a bracket here and just a square01:33:39
write down the amounts just in case01:33:43
square of the sum in some more such01:33:45
in non-standard situations, not all people01:33:47
see that's why I have this formula and 7th grade01:33:51
Let us remind you that is the square of the sum we01:33:54
we reveal as follows with us01:33:56
it turns out x squared minus 7 squared01:34:01
plus twice the product x square01:34:05
minus 7 and root of 2 x minus a plus01:34:10
the square of the second expression that is01:34:12
square kills root right side01:34:17
I'll rewrite x squared minus 7 squared plus 2x01:34:21
minus and you still need it, but it’s not equivalent01:34:24
transition because we still have a limitation01:34:26
climbs out of the water for that is, we say01:34:28
what else is 2x minus and it should be more01:34:31
or equal to 0, well, everything is in the wall01:34:33
let's take another segment so that x is from 0 to 301:34:35
also here in the system01:34:39
Let's shove this into a system x01:34:43
square minus 7 squared x square01:34:45
minus 7 squared cancels out01:34:47
and 2x minus a also remains what we have01:34:51
product equals 0 here is the equation01:34:54
I also need it to fit under DC01:34:57
roots, well, as they should, again he said that01:35:00
if it fits then it's a suitable root01:35:02
well, that is, it will remain the root of the equation with01:35:05
it doesn’t fit, it won’t stay, you need to understand01:35:08
how many roots do we have?01:35:10
that is, we need to solve this equation 101:35:12
if you call him a one-off, but in general she is01:35:17
it’s easy to solve, we don’t know, we need it01:35:19
Of course I can write it out separately somehow01:35:22
I would fall asleep, I decided to give a little more detail01:35:24
I’ll explain in more detail, that is, equation 101:35:26
just a day there can be solved in the fields01:35:28
or right here in the solution x square minus 701:35:32
multiplied by root of 2 x minus a equals01:35:35
zero when01:35:38
when the product is zero when although01:35:41
if one of the factors is zero 201:35:45
exists Well, we’ll take care of existence here01:35:49
2x minus a must be greater than or equal to01:35:51
0 I at least first roots to roots01:35:54
I will find that is, two is equal to zero to be not01:35:56
Maybe this two is clear, it’s a parenthesis01:35:59
is equal to zero for some exe koch then01:36:02
is x square minus 7 equals zero01:36:05
and also the root is equal to zero if01:36:08
radical equals zero of 2x minus a01:36:10
equals zero shorter than x squared equals 701:36:14
then x is equal to plus or minus the root of 701:36:18
and here 2x is equal and that is, x is equal to the author01:36:24
their01:36:26
we are solving an equation so we must not01:36:29
neo express aix we are looking to appoint01:36:32
excite dex first second let it be01:36:34
will meet them, that is, let's return to ours01:36:36
system01:36:38
back to the system01:36:46
we solved the equation again, it was possible01:36:49
we decided to do it all verbally and01:36:51
understood that we have01:36:52
and xyg or root of 701:36:56
or and xyg minus root of 701:36:59
or xyg their author here we have three roots01:37:04
has equations with 2x minus a01:37:07
must be greater than or equal to 0 and when01:37:09
in this case x must be from 0 to 301:37:11
we notice that minus the root of 7 is not from01:37:14
0 to 3 means he just disappears like us01:37:18
we’ll write this and we’ll write this and note that01:37:27
x 2 does not belong to the segment from 0 to 301:37:41
cod from 0 to 3 and it all disappears01:37:46
simplicity we get in the system we01:37:48
it turns out that x 1 is equal to the root of 7 x 301:37:53
this is the author of their only two roots01:37:55
there are still potential yes and 2x minus a01:37:59
must be greater than or equal to 0 and x01:38:03
should be from 0 to 3 well you can already do that01:38:05
don't write on 1 start u let it01:38:08
will be saved again not wrong01:38:11
say that x 1 must satisfy o01:38:14
yes for x 2 too because if they are both01:38:17
satisfy even then we will have two01:38:19
This equation has two roots01:38:22
but we need that there was one root01:38:24
that is, we need one to satisfy01:38:26
y yes for and the second one did not satisfy the dose01:38:28
or the second root and maybe01:38:29
inappropriate if it is not from 0 to 3 but it is01:38:31
I'm already telling you another way there01:38:33
solutions in short how we decide how we01:38:37
we decide01:38:38
that is, let's say let's say this x 101:38:46
is the root of the equation and it is equal to01:38:49
root 7 when it is the root01:38:52
equation equal to root 7301:38:53
but belongs to but satisfying this one01:38:57
according to this inequality satisfying here01:39:03
the fact that he is from zero to three is so01:39:04
it is clear the root of the seeds is from zero to three01:39:07
prowess satisfying inequality01:39:17
2x minus a is greater than or equal to 001:39:20
that is, now we will find a for which x01:39:24
1 will be the root of the equation for01:39:28
To do this, you need to substitute the root of01:39:30
7 it turns out to be 2 times the root of 701:39:32
minus and greater than or equal to 0 that is01:39:36
and less than or equal to 2 roots of 7 you01:39:38
what did we find again if and this01:39:41
if a is less than or equal to 2 roots of 701:39:44
the x 1 is the root of the equation01:39:47
if a equals I don’t know there is 20 if a-20 then01:39:53
x 1 will not be a root because x 1 will be01:39:55
violate the decrees you understand if a less01:39:58
or equal to 2 roots 7 then x 1 root a01:40:01
if any is very large01:40:03
then x is the first to violate there for and he he01:40:05
disappears01:40:06
from the number of roots, that is, roots remain01:40:08
not two but only one I hope you understand01:40:13
so then the next root x 3 to 2 there01:40:16
we already forgot about him by x third01:40:19
secondly, it is the root of the equation01:40:21
when again for a it is the root01:40:26
for a satisfying the inequality but also01:40:30
it must lie in the range from 0 to 301:40:32
so satisfying the inequality 2x minus01:40:42
and the big ones are equal to 0 and x from 0 to 3 that is01:40:52
he still has to be with you until three01:40:54
again you can ask why us01:40:55
they didn’t write here that the root of 7 must01:40:57
be from 0 to 3 that well, from 0 to 301:40:59
and the author of them can be from 0 to 3 maybe01:41:02
not from zero to three you need to understand01:41:04
at what and he satisfies there and01:41:07
from 0 to 3 that is here01:41:10
if here we just substituted the root01:41:13
out of 7 in Odessa, then in this case we01:41:16
that means we substitute in dc here too01:41:21
we substitute that is 201:41:23
the author of their minus is greater than or equal to 0 and01:41:27
their author must be from 0 to 3 that is01:41:32
we will understand at what a01:41:34
x the third root of the equation is a01:41:39
minus and greater than or equal to 0 that is01:41:43
a minus a is 0 and greater than or equal to 001:41:47
I understand that we will need 0 but01:41:49
Now I’ll explain why I wrote it like that01:41:51
let's multiply by 2, it turns out from 0 to01:41:54
6 0 multiplied by which will be greater or01:41:59
equals 0 to anything, that is01:42:02
solution 1 inequality will be a any01:42:05
solution 2 from 0 to 601:42:09
if we find intersections then it will remain01:42:12
in the general intersection from 0 to 6 again01:42:16
what does found from 0 to 6 mean?01:42:20
means that if you have from 0 to 601:42:23
for example a is equal to 5 then x 3 will be equal to01:42:27
5 second two and a half means he01:42:30
is included in the segment from zero to three it is two01:42:33
and a half and he doesn't break even if01:42:37
you have a equals 5 x 3 x is two s01:42:41
half two by two and a half minus 501:42:43
equals zero, that is, under the root is what01:42:45
maybe it's not even broken, that is01:42:48
this will also be the root of the equation, but also01:42:51
since this x 3 will be the root of the equation01:42:53
only if from 0 to 601:42:54
if a is equal to 8 then x 3 will not fit in01:42:58
segment from 0 to 3 it will be 4 don’t blame01:43:02
power here, well, Audi for him he always01:43:06
satisfies vice for any he x 301:43:09
satisfies the design with others01:43:12
except from 0 to 6 it just won’t fit into01:43:14
segment from 0 to 3 that's what happened yes01:43:17
what to do with these answers what to do with01:43:20
with these [ __ ], that is, we understood at what01:43:25
here at the same time01:43:26
x 1 is a root for these a x 301:43:30
is the root01:43:31
let's see at what and how much01:43:34
the roots will be washed and that is what we do01:43:36
large number line and we put it on01:43:44
all and all and which interest us01:43:51
that is, we apply a zero, we apply a six01:43:57
so 2 roots of 701:43:59
this is 2 multiplied by 2 something, yes that is01:44:03
this is less than 6 2 roots of 7 somewhere here01:44:06
and the six is to the right of everyone I still want01:44:15
understand how the roots will coincide at what01:44:20
and the roots will coincide, that’s what he doesn’t want yet01:44:22
understand that is, at some a roots01:44:24
will coincide with us01:44:27
at what and the roots coincide, let's do it01:44:33
you know, for some and we will have x 101:44:35
kor kor root for some a x 2 will be01:44:38
root01:44:39
and at some point the roots coincide, let’s01:44:42
write let's find a let's find a for which x01:44:54
101:44:55
or we can directly understand this right now01:44:57
that is, it is not necessary to describe what I01:44:59
I’m writing, but I’m just hoping, just in case, so we’ll find it01:45:01
and at which x 1 coincides with to old01:45:07
matches means we have two identical01:45:10
root there is only one root and it will be01:45:13
also approach the answer so coincides with x01:45:16
2 expert matches sex 201:45:17
that is, we need to equate x 1 with x3s x 3 x01:45:22
1 exchange and x 3 turns out a equals 201:45:27
root of 701:45:28
if a is equal to 2 roots of 7 yes we have two01:45:31
root we have x 1 root of 2 but these are roots01:45:34
they will be the same, you understand01:45:38
so that means if I take I take something01:45:45
so often give us small sticks here01:45:48
let's do this for ourselves if I take a01:45:56
less than zero if I take something like this01:45:59
here is less than zero then x 1 is the first x01:46:01
exists if I take a less than zero and01:46:04
I will collect a minus 10 x 1 exists at a01:46:08
less than or equal to 2 roots of 701:46:09
that is, x 1 here will be x 1 equal to the root01:46:14
out of 7 it will be at a less than zero to a x 201:46:19
will x 2 exists only when a is from 001:46:23
until 601:46:25
you know that's why x 201:46:27
will not be here and who is not here is good01:46:30
if a is equal to 0 then we have everything01:46:33
the only root is this01:46:35
this suits us well, but less than zero suits us01:46:37
suits and less than zero suits us01:46:41
that means yes, good, but if a is equal to 0, then what?01:46:45
if a is 0 then x 101:46:49
exists because if a is less or01:46:52
equals than 2 roots 7 for example a equals 001:46:54
then x 1 is equal to the root of 7 to x 101:46:56
exists it turns out if a is equal to 0 then01:46:59
x 1 is equal to the root of 701:47:01
and if a is equal to 0 0 is included in the segment from 001:47:07
until 601:47:08
if a is equal to 0 then we will substitute01:47:10
under a zero and x 3 will be equal to 0 and 0 and x01:47:14
3 equals 0, that is, we have two different roots01:47:18
when a is equal to 0 we have two different codes01:47:20
I don't need it, I'm over it01:47:23
the only root on the segment remember01:47:25
conditions require only one root01:47:27
only decision01:47:28
so now we take a from zero to two01:47:33
roots 701:47:38
take from zero to two roots 7 well01:47:40
for example, for example, some one01:47:43
let's take one and if a is equal to 1 then x01:47:48
1 exists and it is equal to the root of 701:47:51
that is, x 1 is equal to the root of 701:47:54
if a is equal to 1 then x 2 exists because01:47:59
that it exists from 0 to 6 and if a01:48:02
equals 1 x 3 turns out equal to well there 1 201:48:07
well, or their author’s main thing is that there won’t be x 301:48:12
The root of 7 is also equal here, not like01:48:17
I said and thought, well, that is, x 3 not01:48:21
will be equal to the root of 701:48:23
because a must then be equal to 201:48:26
root of 7 on this interval but cannot01:48:31
equal to 2 roots of 701:48:32
that is, x 3 will be equal to a second but if01:48:36
a equals 1 to 1 2 if a equals 2 then 201:48:38
second but just01:48:39
x 3 will be equal to a second and x 1 x 301:48:44
different roots are different roots now yes01:48:47
because they are the same only at 201:48:49
roots 7 so we have two different roots here01:48:52
a total of two roots is not for me01:48:54
I need an answer, I'm looking forward to this if01:48:57
equals 2 roots of 701:49:00
if a is equal to 2 roots of 701:49:02
the x 1 exists because he01:49:06
exists when a is less than or equal to 201:49:08
the root of 7 it exists and is equal to the root01:49:10
of 7 that is, if a is equal to 2 roots of 701:49:13
So 1 is equal to the root of 7 x 3 x 301:49:19
there are from 0 to 6 and that means we01:49:23
substitute 2 roots from the family01:49:25
under a 2 roots of 7 divided by 201:49:29
it turns out x 3 is equal to the root of 7 that is01:49:32
yes we have two roots but here they were different01:49:35
the roots here were different01:49:36
and now we get the same roots01:49:41
identical roots, that is, different01:49:44
the roots are one thing, you know, we have01:49:46
coinciding roots that's why it's like that for me01:49:49
fits01:49:50
if this one didn’t suit me01:49:53
gave two different roots, something like this and to me01:49:56
fits she gives one two identical01:49:58
root that is one root root 71 y01:50:01
we are the only root I hope you are01:50:03
understood so if we take something in between01:50:06
2 roots of 7 and 6 type 5.9 for example yes01:50:13
if you take 5.9 then x 1 is no longer all x 101:50:17
exists to the left than two roots 7 to the right01:50:19
it doesn't exist and x 3 exists here that is01:50:24
here x 3 only exists and it is equal01:50:26
the author of them, well, drive these and from two roots 701:50:30
up to 6 for example 5.9 second like this01:50:35
the only root exists here01:50:36
the only root that suits me is01:50:39
there is and from two roots 7 to 6 is me01:50:44
very satisfied so if a is equal to 6 then x01:50:49
1 no longer exists x 101:50:51
for such a it no longer exists x 101:50:55
there is directly less than or equal to two01:50:57
root 7a x 2 exists at 0 to 601:51:02
means x 3 x 2 x 3 x 3 means equal to what01:51:06
if a is equal to 6 then on the second 6 x 3 is equal01:51:09
three is the only root for a equals 6 y01:51:13
there are only a few of us again, the only root01:51:15
but if a is more than 6 then it’s just roots01:51:19
no and01:51:21
x 1 no x 3 no that means the only one01:51:25
the root we got was purchased here01:51:26
and we write this answer, that is, the answer is01:51:30
This one suits us, but it's a minus01:51:34
infinity to 0 and then from two01:51:38
roots 7 inclusive01:51:41
up to 6 inclusive the end let's go over01:51:46
throughout the solution we see the equation and01:51:49
we notice that if the square of the sum is revealed01:51:51
that is, the most obvious transformation01:51:53
do that there is a lot of chop cut by01:51:56
mutually destroyed01:51:57
we did it, we got this01:52:00
equation it has three roots01:52:01
root of 7 minus root of 7 and their author01:52:06
but minus the root of 7 is not included in the segment01:52:09
from 0 to 3 that is, we have 2 roots left01:52:11
out of 3 we have one missing01:52:15
dc is a segment from 0 to 3 but01:52:18
it is wrong to say that both roots must01:52:20
satisfy o dzy be in the interval from 001:52:24
to 3 because if this happens if01:52:27
both roots satisfy the language and in the segment01:52:29
until three then we will have two roots and we01:52:32
we need to have only one left01:52:33
root that is, we need one of01:52:35
they fell out or they matched here we are01:52:40
we say that x 1 will be the root01:52:43
equation only if it satisfies y01:52:45
dc and lies in the interval from 0 to 3 but this is also01:52:48
so it’s clear the roots of the seed but it lies in01:52:50
segment from 0 to 3, here we substituted01:52:53
means x is the root of 7 and we realized that when01:52:55
like this01:52:56
x 1 is the root of the equation ax301:53:00
is the root of the equation if it01:53:02
satisfies even01:53:04
and lies in the interval from 0 to 3 it turned out01:53:07
what satisfies there for him at all01:53:09
always and in the interval from 0 to 3 lies at01:53:12
from 0 to 601:53:15
then we are on the number line on thoughts01:53:17
zero two roots 7 and 6 and started01:53:20
scan if a is less than zero then x 101:53:23
is and which is not if a is equal to 0 you have01:53:29
we have two different roots, I only need 101:53:31
the root to remain01:53:32
if a is from zero to two roots 7 then we have01:53:35
two different roots here at equal to 001:53:38
specific roots root of 7 and 0 and at01:53:41
from zero to two roots media different roots01:53:43
always different depending on different01:53:46
and if a is equal to 2 roots of 7 then we have01:53:50
roots 2 2 roots but they coincide, that is01:53:53
only one root is the root of 7 therefore01:53:55
it doesn't fit there naturally01:53:57
often people forget this and that’s what they get01:54:00
there are three out of four points and we01:54:02
let's get 4 4th floor let's not forget01:54:04
matching too he is also one01:54:07
the only root will ultimately give us01:54:08
if aa two roots 7 to 6 then x 1 is already01:54:13
no because x 1 drive exists01:54:14
such a a x 3 will be different all the time01:54:22
if a is equal to 6t x 1 a is long gone x 301:54:25
the only root is equal to three therefore01:54:27
we need you if a is greater than 6 then x 101:54:30
long gone x 3 won't happen either because01:54:33
that if a is greater than 6 then x 3 will not be01:54:35
fit into the segment from 0 to 3 that's all of us01:54:38
we write down the one that suited us like this01:54:40
task find all values for each01:54:44
of which the equation has exactly one01:54:46
root on the interval from 0 to 101:54:48
when the product is zero when although01:54:51
if one of the factors is equal to 0 201:54:53
exists that is, we must each01:54:54
set the multiplier to 0 and plus is indicated01:54:57
news and plus x from zero to one we will find01:55:01
roots will be x 1 x 2 x 3 most likely well01:55:05
here x 1 is equal to 0 25 it’s already clear x 2 and 301:55:09
and the third why I say because here01:55:11
x quadratic equation two roots rather01:55:13
in total there will be and so we have three roots and we01:55:15
let's find x 1 for which oh is the answer01:55:18
is the root at which ayxta swarm01:55:21
is the root01:55:23
for which a x 3 is a root then all01:55:26
and look at the number line at01:55:29
what kind of roots will the plan be like this01:55:31
so when the product is equal to zero01:55:35
ever the root is zero or01:55:40
l n equals zero01:55:50
either one or the other but the radical must01:55:53
be greater than or equal to 0 and right under01:55:57
logarithmic must be greater than zero01:55:58
strictly well01:56:05
x must be in the range from 0 to ones01:56:09
we get this system01:56:11
let's solve it and so the root is equal01:56:15
zero when the radical is equal to zero yes01:56:17
that means everything is clear there x is equal to 1 401:56:26
on its own it's x 101:56:28
what's next with the second equation01:56:32
here e to the base e to the 0th power is equal to01:56:34
this entire right side, well01:56:37
logarithmic, more precisely, so I'm at 001:56:39
one that is all all friend01:56:43
rhythmic equals one it turns out like this01:56:50
at the same time it is necessary that I will do 4x01:56:55
plus one is greater than 0 and actually01:56:57
you can solve it, that is, one to the right01:56:59
reroll 4x large equals 1 and to 401:57:03
divide it turns out x is greater than 1 401:57:05
should be x squared minus 2 x plus 201:57:13
minus and the square must be greater than 001:57:16
px must be between zero and one01:57:21
so with a quadratic equation but here01:57:24
throw a one to the left and it will be a square01:57:26
differences01:57:28
yes, let's look into this separately01:57:30
equation so as not to rewrite the extra one01:57:32
since everything else turns out that way01:57:38
it turns out let's convert the first one01:57:40
equation x squared minus 2 x plus 2 and01:57:45
minus one is plus 1 minus a square01:57:49
equals zero learn x minus 1 squared01:57:52
minus and squared is equal to zero difference01:57:56
it turned out squares, well, just reminding you01:58:02
not everyone sees in complex examples01:58:05
elementary formulas of different squares01:58:07
this formula turns into01:58:10
product of parentheses a minus and plus01:58:12
that is, x minus 1 minus and by x minus01:58:18
1 plus y equals 0, so when is the product01:58:24
equals zero when at least one factor01:58:25
is equal to zero, that is, x minus 1 minus a01:58:28
equal to zero01:58:29
x minus 1 plus a equals zero, that is, x01:58:33
it turns out 2 is equal to a + 1x it turns out 301:58:38
equals minus x plus 1 so we have x 101:58:41
x 2 x 3 back to the system01:58:52
we have x 1 equals 1 4 we have x 201:58:59
equal plus one us is x 3 equal01:59:05
minus a plus 13 roots we need01:59:10
so that we also have x more and01:59:15
equal to 4 x square minus 2 x plus01:59:21
2 minus and the square must be greater than 0 and01:59:24
x must be between zero and one01:59:27
Well, that’s the kind of system we have, so let’s do it01:59:31
we'll find out x 1 for which a is a root01:59:34
equation x 2 for which a is01:59:37
root of wounds x 3 for which a is01:59:39
let's start with the root of the equation at x 1 so x 101:59:44
it is equal to 1 4 for what a for a01:59:50
satisfying the following requirements01:59:53
firstly x must be greater than or equal to01:59:56
than 1 4 but he already satisfies this02:00:00
requirements02:00:01
x must be from zero to one it is already02:00:04
satisfies these requirements, the main thing is02:00:06
what he should do is this, well he should02:00:10
satisfy this inequality02:00:12
is understood that is, for a satisfying02:00:14
inequality x squared minus 2 x plus 202:00:19
minus and the square is greater than 0 let's find out02:00:22
at what and does it fit here so under x102:00:25
4 we substitute 1 4 we put this 1 1602:00:29
minus 2 times 1 4 plus 2 minus a square02:00:34
is greater than zero, that is, and the square is less than02:00:38
count 2 minus half02:00:42
one and a half + 1 16 so one and a half is for example02:00:50
24 1602:00:52
+ 1 1625 16 well of course someone02:00:59
the clock will write nonsense and less than02:01:01
plus or minus five quarters is complete nonsense02:01:03
you need to throw 2516 of them to the left and02:01:07
write the difference of squares that is02:01:11
a minus five quarters by a plus five02:01:17
fourth02:01:19
less than zero number line and plot02:01:23
minus 5 4 and 5 4 plus minus plus us02:01:30
need minuses conclusion for these02:01:35
and x 1 is the root of the equation derivation02:01:40
with a belonging from minus 5 4 to 5 402:01:49
x 102:01:50
equal to 1 4 is the root of the equation for02:01:53
segment02:02:00
good thing I'm fed up with the old x 2 is a + 102:02:04
do the same with him like this x 202:02:07
this is a + 1 it is the root of the equation02:02:12
with a satisfying which02:02:15
requirements it must be root larger02:02:19
or equal to than 1 4 he must be he02:02:28
must eliminate this inequality and02:02:29
it must be from zero to one then02:02:32
there are all three requirements here02:02:33
so that they execute like this x square minus 202:02:39
x plus 2 and for x 3 you will also need all this02:02:43
write more than 0x must be from02:02:46
zero to one is in principle possible, of course02:02:49
these ones cross these ones02:02:51
inequalities in principle but they are like that02:02:55
easy that everyone can decide like this02:02:58
it doesn't matter in general so let's mean02:03:01
decide x must be greater level than02:03:05
1 4 means a plus one should be02:03:09
more Iran than her 4 so now02:03:13
It’s going to take a long time to write, but let’s02:03:16
let's swap places doesn't matter so x square02:03:21
minus 2 bodex we substitute a + 1 it turns out a02:03:25
plus 1 squared minus 2 over x that is02:03:29
by plus 1 plus 2 minus and the square should02:03:34
be greater than 0 and a + 1 from zero to one02:03:39
inclusive02:03:43
so it turns out that a is greater than 102:03:53
4 minus 1 is minus three quarters and so02:04:00
here minus one from all three sides02:04:02
double inequality subtract that is from02:04:04
minus 1 to 0 it turns out, let's do it here02:04:08
simplify we reveal the square of the sum to us a02:04:13
square plus 2 plus 1 minus 2 and minus 202:04:20
plus 2 minus a square is greater than 0 a02:04:24
the square is destroyed 2a is also minus two02:04:28
plus two also one more than 0 always she02:04:31
by the way it seems that02:04:34
x 2 x 3 will satisfy this02:04:36
condition is always well because they are like02:04:40
the roots are out of the equation where is this thing02:04:44
was equal to one therefore we and02:04:47
it turned out that one is more than zero hour one02:04:49
more than 0 to 1 more than zero in any case02:04:53
any and all that remains is to find the intersections02:04:55
means a is greater than or equal to minus three02:04:57
fourth and it is from minus 1 to 0 that is02:04:59
What is the conclusion when and when proper from -302:05:08
fourth to 0x second which was a +02:05:15
1 is the root of the equation02:05:22
on a given segment let's pro x302:05:26
let's understand x thirds is minus a + 1 x 302:05:33
which is equal to minus a + 1 is02:05:38
root of the equation for a satisfying02:05:41
the same requirements as the old one, that is02:05:43
it is necessary that02:05:48
x was greater than or equal to 1 4 so that02:05:53
performed here x square minus 2 x02:05:58
plus 2 minus a square02:06:07
more than 0 to further so that x is from02:06:12
zero to one, such a system is needed02:06:16
now we solve we solve it turns out minus a02:06:21
+ 1 must be greater than or equal to it02:06:23
4 let's switch places02:06:27
that is, there will be 1 minus a under x02:06:30
substitute 1 minus and squared minus 202:06:34
by 1 minus and plus two minus square02:06:40
more than 0 and 1 minus and let's do it here02:06:44
no it doesn't matter02:06:46
I thought there was a difference no and so and so not02:06:49
it will be very convenient so we’ll decide one02:06:54
each02:06:55
and to the right, that is, a is less than or equal to02:06:59
one minus 1 4 than three quarters further02:07:02
open the square and get 1 -2 a +02:07:08
and the square is minus 2 plus 2x plus 202:07:14
so minus and square is greater than zero square02:07:18
and 2a were destroyed too, there were triplets too02:07:22
one is greater than zero for any and so and here02:07:25
subtract ones from all three sides then02:07:27
is -1 less than or equal to minus a02:07:30
less than or equal to zero by -1 multiplied02:07:35
that is, it turns out that a is less than or equal to02:07:38
than three quarters and a from 1 to 0 but02:07:44
the icons need to be expanded, that is, from 0 to02:07:47
1 but less than or equal to three fourths02:07:50
output at and at proper from zero to02:07:56
three quarters02:08:01
icos 3 is the root of the equation for02:08:05
segment02:08:16
we found for which a x 1 is a root02:08:19
on the segment from minus 5 4 to 5 fourths02:08:22
not inclusive at what a02:08:23
x 2 is a root on the segment from minus02:08:26
3 4 to zero and at what a x 3 is02:08:29
root of levels zero to three-quarters02:08:31
now let's look at all these and02:08:35
[music]02:08:36
Let's see where and how many roots there will be at02:08:39
different a02:08:40
number line and by the way, when do they02:08:44
coincide, this is also important to understand02:08:45
when they don't match they can match x02:08:49
first sex old x first sex third02:08:51
their 2 sex third x 1 matches although we02:09:00
we can see this directly, actually, come on02:09:02
let's check in advance x 1 coincides with iq02:09:05
old if anything but if x 1 is equal to and02:09:09
which that is if one fourth is02:09:11
and there is a plus 1 that is, and is equal to when02:09:15
minus three-fourths then they fall off conclusion02:09:22
when a is equal to minus 3 4 x 102:09:27
matches ex container02:09:34
x 1 coincides with the third when if what02:09:42
if x 1 is equal to x third then a02:09:48
equals 3 4 output when a equals 3 4 x 102:09:56
matches x3 teen02:10:02
and can match x 2 x 302:10:12
if a plus 1 is equal to 1 minus a that is 2a02:10:18
equal to zero02:10:19
that is, a is equal to zero when a is equal to 0x like this02:10:23
I'm right when I say that there is 1 minus and we have02:10:24
was a + 1 before02:10:28
output when a is equal to 0 x 2 x 3 are the same02:10:32
when a is equal to 0 x 2 the sex of the third coincides02:10:37
and so now the number line and we look02:10:47
at what and how many roots do we have?02:10:49
it turns out02:10:52
we are interested in02:10:54
and which from minus 5 4 to 5 fourths02:10:58
Firstly02:11:07
so from minus 5 4 to 5 fourths then02:11:11
we are interested in minus three-quarters zero02:11:16
zero and three quarters yes that is us02:11:23
interested in minus five quarters02:11:24
and five fourths then we are interested in02:11:33
minus three quarters and three quarters02:11:42
There's a zero between them, that's right, nothing is needed02:11:47
apply as from 1 means again from02:11:50
minus 5 4 to 5 4 a bit old minus 3 402:11:53
0 from 0 to 3 4 that is, we will consider02:11:57
situation when a is less than -5 fourths02:12:00
and let's see how much exicosis we have02:12:03
situation when a is exactly minus five02:12:05
let's see how many fourths we have02:12:06
evsikov when and from minus five fourths02:12:09
to minus 3 3 402:12:12
then a exactly - three three quarters when02:12:16
and from -3 quarters to zero when equal02:12:20
zero when from zero to three-quarters02:12:23
when is equal to three quarters when a is from02:12:28
three quarters and five quarters02:12:30
when it's exactly five quarters02:12:33
when a is more than 4 so if a02:12:38
less than -5 quarters give and stick02:12:41
we'll do more02:12:53
if a is less than -5 4 x 1 is02:12:57
the root x 1 is the root only of02:13:01
minus 5 4 to 5 4 that is, not yet02:13:04
is x 2 is the root of x 202:13:10
is a root for such02:13:12
and that is, it is not yet x 302:13:15
is a root in general for positive02:13:18
not yet, that is, not here02:13:20
the roots are just, well, minus five02:13:26
the fourth ones also have no roots there ikos 1 yet02:13:28
didn't appear but x 1 in short here we have02:13:33
here is from minus 5 4 to 5 402:13:35
in principle we can say that x 102:13:37
equals 1 4 here x 1 equals 1 4 here x 102:13:49
equals 1 4 here x 1 equals 1 4 here x 102:13:56
equals 1 4 here x 1 equals 1 4 here x 102:14:04
equals 1 4 here x 102:14:07
he's just from minus 5 4 to 5 fourths02:14:09
all the time there is like a root you understand like this02:14:13
that sex with old x 2 is the root of02:14:16
minus 3 4 0 from minus 3 4 include but to02:14:19
zero, that is, x 2 is the root here02:14:24
equals a + 1 to 3 4 inclusive that is02:14:27
x 2 is also a root here x 202:14:32
is also a root here x 2 is02:14:34
root also here and here a x 302:14:41
is the root oh no x 2 was02:14:47
root from minus 3 4 0 to 0 to 002:14:52
inclusive02:14:55
and x 3 is the root of zero to three02:14:59
fourths from 0 inclusive to 3 4 then02:15:01
there is x 3 equals 1 minus and x 3 equals 102:15:06
minus a02:15:07
x 3 equals 1 minus so let's do it02:15:13
let's see where our roots coincide at02:15:17
minus three fourths are matching02:15:21
roots because if a minus 3 4 who02:15:24
here there will be 1 4 that is, we have two here02:15:29
roots but 2 matching roots that is we02:15:37
circle the situation when we have x equal to 1 402:15:40
the only root for such and we02:15:43
circle minus three quarters because02:15:46
there are two of us matching aft, that is, one02:15:47
the root is the farthest our roots are02:15:51
coincided when prearo inside the fourth02:15:56
if you wipe 4 put it will be 1 4 then02:15:59
we have two matching roots again02:16:07
three quarters a equals 3 4 circle02:16:10
Next we circle from three quarters to 502:16:13
fourth because there is only one here02:16:15
root02:16:17
so it turns out there are no roots here02:16:22
roots and still feeding, they served and02:16:27
equals zero that is, if a equals zero then02:16:29
there are one fourth of us, then one02:16:33
and one, that is, we have x 2 x 302:16:36
matches x 1 other that is, we have like02:16:39
3 roots02:16:40
but these two coincide, that is, only two02:16:42
roots here you understand 3 roots but 202:16:45
coincide therefore we are twice aft02:16:47
and we don’t circle the zero and we don’t take the answer02:16:49
It’s clear that in response we only take what I02:16:54
I circled it now so the answer is from02:16:59
minus 5 4 to minus 3 4 inclusive and from02:17:07
three quarters inclusive up to 502:17:11
fourths not to be included or let's go over02:17:13
the entire solution product is zero02:17:17
when when at least one of the factors02:17:19
equal to zero that is, when 1 equals zero 202:17:22
I also love to curtain the garden plus x from02:17:24
zero to one well the first equation02:17:27
elementary x is 1 4 and second02:17:29
the equation02:17:30
well, also light and to the 0th degree equals under02:17:32
logarithmic we are a little bit with it02:17:36
we worked and realized that it gives us two02:17:38
root a + 1 m amine one minus on so y02:17:41
us02:17:42
the original equation gave 3 roots plus02:17:45
there is 1 plus x from 0 to one x 102:17:48
is the root of the equation if it is greater02:17:51
Iran than 1 4 but so much bigger than it02:17:53
4 if it is from zero to one, well, from02:17:55
zero to one means it just has to be02:17:57
this restriction is fulfilled and that's it02:18:00
executed under attack h x 2 is02:18:03
root if it is greater than or equal to it02:18:05
4 if there is more than 0 under the logarithmic02:18:08
if x is from zero to one with these02:18:10
aig 2 is the root02:18:12
x 3 should also be big before we02:18:15
by 4 lie in the interval from 0 to 102:18:17
it should be logarithmic02:18:18
greater than 0 with these a x 3 is02:18:21
the root of the equation x 1 coincides with k02:18:24
old means we need to equate them with02:18:26
so x 1 coincides with x third right here02:18:29
such x 2 coincides with them we will meet at a02:18:31
like this and we put it on the numerical02:18:37
direct and all and affecting the quantity02:18:40
roots and look at what and how many02:18:43
there will be roots02:18:44
and circle those situations when we have one02:18:46
the root turned out in the end02:18:47
for example, when two matching roots are02:18:50
after all, one curia here we have 3 roots but02:18:52
two of them coincided and only two roots02:18:54
We don't need such situations, we need them02:18:56
feed the situation when we have one root02:18:58
or 2 falling ones, that is, 1 or 202:19:02
falling here, not that there is one or here02:19:04
one just x 1 with such a02:19:07
it turns out that we will have one root02:19:11
find all values of a for each of02:19:14
of which the equation has exactly one02:19:15
root on the interval from 0 to 102:19:17
let's factorize the right02:19:20
radical and 0 it is clear how to decompose into02:19:24
the factors there are x minus 1 x plus a02:19:25
In short, in general, if the root is equal to the root then02:19:28
you can simply equate radicals02:19:31
question, why did the roots give us the conditions?02:19:34
Well, let us write more about the dose02:19:36
in addition, everything is so that we02:19:38
wrote the radical greater than or equal to 002:19:40
by the way, if the root is equal to the root and one02:19:42
radical we wrote that they are big02:19:44
equal to 0 2 automatic something equal to02:19:46
Karina, in short, in short, we need to simplify02:19:50
simplify the right-hand side by decomposing it into02:19:53
multipliers either by discriminant or with02:19:55
using describing and either using02:19:56
Vieta's theorems02:19:57
let me show you so let's do it like this02:20:02
let's break it down02:20:07
and then we’ll start solving by decomposing02:20:09
radical of the right side of the equation on02:20:16
multipliers02:20:30
Well, you and I remember the decomposition formula02:20:32
by factors here she is she is x minus x02:20:40
one for x minus x 2 that is by02:20:42
discriminant or fault then can be found02:20:44
roots and substitute into this formula and02:20:47
there will be factorization then02:20:48
there is a product of brackets and here02:20:50
we will add parentheses in that02:20:51
there to simplify so let's use02:20:56
I'll show you the discriminant first02:20:58
with the help, well, it’s enough to solve one02:21:01
way I'll just show you both02:21:04
using discriminant as using02:21:11
discriminant means we rewrite it02:21:14
3 square minus 3 plus 1 times x +02:21:21
poll roots we also substitute everything there02:21:22
under the form so the discriminant is equal to what b02:21:26
squared, that is, the coefficient ancestor c02:21:28
square02:21:29
the square will kill this minus, it turns out 3a02:21:32
plus 1 squared minus 4 times a02:21:36
on 3 and on c on a so it turns out on 9 a02:21:43
square plus 6x plus 1 minus 12 otherwise02:21:48
there will be 9a square minus 6x plus 1 then02:21:52
there is this 3a minus 1 squared02:21:58
find x first second by discriminant02:22:02
this turns out to be minus b, that is, 3a plus 102:22:06
plus minus the root of the expression in02:22:09
squared root of squared expression02:22:11
this is the module divided by 2a, that is, by 202:22:17
multiply by 3 by 6 modulus is the same02:22:22
the most that is 3a minus 1 plus minus that is02:22:24
this is plus minus 3a minus 1 module this is02:22:26
from a plus expression with a minus and then again02:22:28
plus minus in short we will have two cases02:22:31
when we add 3a to 3a plus one02:22:35
minus 1 and when we subtract 3a minus 102:22:37
group02:22:38
cooking that is, that is, x the first is equal to 302:22:43
plus 1 plus 3a minus 1 sixths x 2 is c02:22:49
minus or when we change signs02:22:54
minus 3 plus 1 sixths so here are ones02:22:58
mutually cancel 6 separate by 602:23:00
will be and here will be from rub 1 3 here02:23:11
we found the roots using the discriminant02:23:13
it was possible to find the roots using the theorem02:23:16
Vieta using Vieta's theorem02:23:25
Vieta's theorem states that02:23:27
the sum of the roots is equal to minus b divided by a02:23:31
that is, equal to minus this coefficient02:23:34
divide by three02:23:38
this is 3a plus one third a product02:23:45
roots equal to t and divided by a that is, from02:23:49
third it would seem to pick up the roots02:23:53
very difficult let's divide 3 term by term02:23:56
and at 3 it will be02:23:57
and one by three is one third02:24:00
and we think so and the sum of what numbers is a +02:24:05
1 3 if the product of a third well a02:24:09
and 1 3 actually that is, you can go to the roots02:24:12
it was easier due to that fault to guess to x 102:24:15
this is a and castoro this is 1 3 good02:24:18
factorization what are we talking about?02:24:19
says that we first write the coefficient02:24:23
and that is a three02:24:25
Next comes the bracket x minus 1 root and02:24:30
parenthesis x minus 2 root but three02:24:34
can be thrown into one of the brackets and into02:24:36
the second one is better x minus a02:24:39
multiply by 3 x minus 1 that's how we do it02:24:44
laid out02:24:45
this is broken down into a multiplier02:24:48
quadratic trinomial factored like this02:24:50
let's rewrite what we got02:24:54
Let's go back to the original equation02:24:56
there is how we transformed it in general02:24:59
everything could have been done in the fields like this02:25:01
Well, it’s probably better to show such things like this02:25:03
We'll get back to the official decision anyway.02:25:05
showed him one of the ways to vieta02:25:07
would show let's go back to the original02:25:10
equation so to the original equation02:25:21
means root x squared minus a02:25:27
the square is equal to the root of and change as we do02:25:31
there we simplified we simplified by bracket x02:25:34
minus a and by bracket 3x minus one02:25:39
root equals root02:25:41
root equals root let's left left02:25:44
Let’s also factorize it and it turns out x02:25:47
minus a by x + a is equal to the root of x minus02:25:54
and at 3x minus 1 when the root is equal to the root02:25:58
you can just stupidly equate radicals02:26:00
the advantages of writing about even memory, that is02:26:03
this equation is equivalent02:26:06
it turns out equivalently02:26:11
equality of radicals02:26:22
the only thing you need to write about is oh yes for02:26:25
this radical02:26:26
x square minus and square is greater or02:26:29
equals 0, should I write what is the second?02:26:31
radical is also greater than or equal to 0 no02:26:34
because we wrote that they are equal02:26:36
so if the first is greater than or equal to 0 202:26:38
automatically large equals 0 here's another x02:26:41
must be on the range from 0 to ones02:26:43
let's also write it into the system here02:26:44
this is how it turns out to be a system for how to solve02:26:49
equation well let's see what's on the right02:26:51
to the left with a minus and put a bracket behind02:26:52
the parenthesis turns out to be x minus and in short here02:26:59
minus and plains okay I will detail02:27:01
write x minus a by x plus a minus x02:27:08
minus and by 3x minus 1 is equal to zero02:27:13
here z and x from 0 to one should be02:27:19
Now we put the bracket outside the bracket styx02:27:23
cons we carry remains02:27:26
x plus and minus this is buying up, that is02:27:30
minus 3x + 1 x square minus x square02:27:35
greater than or equal to 0 xor 0 to one02:27:40
this is how it turns out to be a system that is before us02:27:43
when the product is zero when although02:27:46
if one of the set 0 2 exists then02:27:48
is we get the totality from us either02:27:51
1 bracket is equal to zero or 2 brackets02:27:54
is equal to zero and there is even x from02:28:02
zero to one02:28:06
so let's x 1 x to ruin02:28:09
let's express x 1 equals what the error to the right and to02:28:14
2 is equal to what, well, we need to solve the equation02:28:19
here we are and slice xx to the right02:28:22
there will be 2x equals a + 1 to if 3x and02:28:27
minus x reroll it will be on the right 2x02:28:30
on the left a + 1 then x equals a + 1 second02:28:34
understand yes a + 1 second x is equal to a + 102:28:38
second x is equal to a + 1 second but this is02:28:46
we solved this equation02:28:48
or let’s divide it term by term, it turns out02:28:50
half a + 1 2 might be easier this way02:28:55
will substitute further02:28:56
x square minus and square is greater or02:28:59
equal to 0 x must be between zero and one02:29:01
and so we have two roots if the first root02:29:08
satisfies there for02:29:09
and is from zero to one then it02:29:11
is the root of the equation notch canada02:29:14
one if x 202:29:16
satisfies this condition and is from zero02:29:18
up to one then it is a root02:29:21
equations for 3 scanner unit02:29:23
we can't have both roots at once02:29:27
were the cube roots of the equation for the segment02:29:30
which will move because we will have02:29:31
then two roots02:29:32
but we need to have only one root02:29:34
let's find out at what a x 1 will be02:29:37
suitable then we will find out at what aek 202:29:40
will be suitable then we will find out at what02:29:44
and they match02:29:45
and then we’ll write down at what a02:29:49
how many roots and some roots like this02:29:53
x 102:29:54
it is equal and so it is the root02:29:58
equations for a satisfying two02:30:03
conditions, firstly, this root must02:30:06
satisfy the water for the second time02:30:09
must lie in the range from 0 to ones02:30:12
substitute x 1 equals a under x02:30:16
we substitute a and we get a squared02:30:18
minus a squared b is equal to 0 and02:30:22
a must be from zero to one so 0 a02:30:27
the square is greater than or equal to 0 for any a002:30:31
greater than or equal to 0 for any and you know02:30:34
So all we have left is this02:30:37
double inequality that is, what is the conclusion02:30:40
output at02:30:41
and from zero to one inclusive02:30:45
x 1 which rabbi is the root02:30:49
equations on the interval and for others they02:30:54
are the roots of the equation well or yes for02:30:56
and violated either not sodium skill and lies02:30:58
so is the root of the equation a x 202:31:06
we have it equal to half a + 1 2 and so02:31:12
the root is the root of the equation at a02:31:15
satisfying review x square minus a02:31:21
square is greater than or equal to 0 and when is it02:31:24
lies in the interval from 0 to a minute, well02:31:26
so we substitute this thing for x02:31:29
on or this view is more convenient02:31:32
substitute what do you think, well, I think in /02:31:37
/ it’s more convenient to substitute a friend, it’s more convenient then02:31:40
it turns out that instead of x I substitute a +02:31:43
1 second squared minus a squared b02:31:48
equals 0 and a + 1 second must be from02:31:52
zero to one so we need to solve it02:31:58
solve by 4 can be multiplied and here by 202:32:02
it turns out a plus 1 squared minus 4 a02:32:09
square is greater than fat 0 by 4 multiplying02:32:12
because 2 squared 4 is this02:32:13
denominator 4 will be understand therefore I02:32:15
download 4 multiply top denominator me not02:32:17
it was like this here by 2, multiply it turns out02:32:19
a + 1 from 0 to 202:32:24
open the brackets a square plus 2 plus02:32:29
1 minus 4x square subtract here02:32:34
one from all three sides and double02:32:35
inequalities a from -1 to 102:32:39
so it turns out minus 3 and the square is plus 202:32:43
plus 1 is greater than or equal to 0 and from -1 to 102:32:51
we need to solve this thing and solve it in02:32:54
the system will probably have an inconvenience later02:32:57
or or ok what kind of roots are there?02:33:02
if according to the discriminant your roots are 402:33:05
+ 12 16 roots normal shorter02:33:08
what are the roots of this?02:33:12
inequalities so discriminant 4 + 12 1602:33:19
minus 2 plus minus 4 divided by minus 602:33:23
minus 2 minus 4 times minus 6 is 1 -2 + 402:33:31
this is 2 to minus 6 minus one third well I02:33:35
I don’t describe it again as a discriminatory02:33:37
they also think you understand how to do this02:33:40
minus plus minus we need pluses from02:33:43
colored dot and should be02:33:47
-1 to 102:33:51
find the intersection we need to find02:33:53
we can find the intersection by intersection02:33:56
Orally you can write it from -1 to 1 and02:34:03
while canceling from 1 1 1 3 to one well02:34:06
in short, let's show its intersections with02:34:11
we apply -1 we apply minus one third02:34:16
put one and so the first answer from02:34:19
-1 3 d and girls second answer from -1 to 102:34:24
where there are 2 hatches there is the answer, that is, the conclusion02:34:29
with and with proper from -1 3 to unity02:34:37
x 2 which is equal to what is 1 2 a + 1 202:34:49
is the root of the equation at a given02:34:51
segment for others and not for these or for02:34:55
even violates or does not lie on the segment02:34:56
so is the root of the equation on the interval02:35:05
well is the root of the equation okay so02:35:09
this means x 1 is the root of the equation at02:35:12
these a x 2 is the root of the equation at02:35:15
these and and when they coincide02:35:19
x 1 coincides with and to old when and when02:35:29
what situation so if it means x 1 is equal02:35:32
a a x 2 is equal to half a + 1 2 to x 102:35:40
equals 1 equals xt room so let's solve02:35:42
throw this equation 1 2 to the left02:35:44
there will be 1 2 and well, minus one second and this is02:35:48
will be 1 2 equals 1 2 equals one output02:35:54
when a is equal to 1 x 1 coincides with k old02:35:59
And02:36:04
x 1 equal to and which is equal to what02:36:08
if a is equal to 1 x 1 is equal to 1 week 2 is equal to02:36:12
1 2 + 1 2 is also 1 and so we are ready for02:36:19
do the last thing we do02:36:21
number line and construct a number line02:36:23
and apply to it everything that affects02:36:28
the number and quality of roots are shorter by02:36:32
number of roots02:36:33
Well, does x 1 x 2 exist?02:36:36
x 1 the root of the equation depends on a02:36:40
what is from zero to one yes that is us02:36:42
put zero one minus one third02:36:46
and one and one, that is, we apply02:36:50
minus one third zero one02:37:01
make three sticks02:37:14
Let's consider the situation when a is less than02:37:16
-1 3 when a is exactly minus one third02:37:20
when a is from -1 3 to zero when a is exactly02:37:26
zero when a from zero to one02:37:29
when a is equal to 1 when is greater than 1 so if02:37:34
and less than -1 3 how many solutions what for02:37:37
solution if a is less than -1 3 x 102:37:42
no it is not the root of the equation if02:37:45
and less than -1 3 x 2 no he is not02:37:47
root of the equation02:37:48
there are no roots if exactly minus one third02:37:55
then x 102:37:57
there is none x 1 is the root of only02:38:00
zero to one and x 2 is the root of02:38:03
at minus 1 3 is shorter that is x02:38:07
2 is the root of the equation x 2 in general02:38:10
was remember 1 2 a + 1 202:38:13
so with a equal to minus one third02:38:16
it is equal to 1 2 multiplied by02:38:21
minus one third + 1 2 that is, it is equal02:38:26
minus 1 6 + 3 sixths 2 sixths is 1 3 he02:38:33
equal to one third02:38:34
we have only one root, which means this02:38:36
I'm happy with it, she'll go in return02:38:38
further if a from -1 3 to zero x 1 is02:38:44
interval from -1 3 to zero02:38:46
they enter, well, they enter here02:38:50
no not included x 1 no and interval from -102:38:54
3 0 is included in this segment to include02:38:56
therefore for the second there is here there is x02:38:59
2 is just the only root of 202:39:01
therefore they are not satisfied with such people if a02:39:04
exactly zero then x 102:39:07
but is the root of the equation zero02:39:10
included in the interval from zero to one02:39:12
this means x 1 is the root of the equation02:39:16
x 1 is the root of the equation and it is equal to02:39:19
well, given that it is equal to 0 it is equal to 002:39:22
and x 2 is the root of the equation and 202:39:26
is the root of the equation for the drive of these a02:39:29
there's a zero here somewhere inside that means02:39:33
the second one is also the root of the equation it02:39:36
generally equal to usually 1 2 a + 1 2 means02:39:40
if a is equal to zero then it is equal to 1 2 for us02:39:44
it turned out that when a is equal to zero two02:39:46
We are looking for different roots and for which there will be02:39:49
one root02:39:50
it means I don’t need it anymore if from02:39:53
zero to one x 1 is the root of the equation a from zero02:39:58
to one are in the interval from zero02:40:00
to one means x 1 is the root02:40:02
equations02:40:03
x 1 are equal to a is by the way here 1 2 a +02:40:09
1202:40:10
so x 1 is the root of the equation x02:40:12
2 is the root of the equation on the interval02:40:14
from zero to one to02:40:16
the segment from zero to one is located on02:40:18
crack cancel from 1 3 from minus 3 to units02:40:22
that is, x 2 which is equal to one second a02:40:25
+ 1 202:40:27
here he is also the root and everything02:40:29
we have two roots02:40:30
we already had two roots02:40:34
this is zero 1 2 2 different roots and here02:40:36
maybe the same roots no roots maybe02:40:39
be the same only if a equals 102:40:41
a little further we will look like this if02:40:45
equals 1 x 1 is the root of the equation yes02:40:48
because or equals 1 and or equals 102:40:52
Means x 1 is the root of the equation he02:40:57
actually equal to a but given that a02:40:59
equals 102:41:00
it is equal to one x 2 is the root02:41:02
equation and 2 is the root of the equation02:41:05
given and therefore if a is equal to 1 . 202:41:08
is the root of the equation it is usually02:41:12
is equal to 1 2 a + 1 2 but if a is equal to 1 then02:41:15
it turns out to be equal to one yes we have02:41:17
again two roots for the third time in a row two02:41:20
root but in the data at the moment the roots02:41:23
we have two coinciding roots02:41:26
coincide, that is, there are one different roots02:41:29
thing02:41:30
and that’s why this suits me02:41:33
if a is more than one then ray a is more02:41:39
one is not here means x 1 not02:41:41
is more the root of the equation02:41:43
you can’t get more than one here either02:41:45
is therefore not located to the old one either02:41:47
is the root of the equation and here it’s simple02:41:49
there are no roots for a and there is more than one simply02:41:52
there are no roots and we have only one root02:41:58
purchased and02:41:59
and we write them in response from -1 3 to 0 1 302:42:07
inclusive 0 no and free-standing02:42:11
one now let's go over everything02:42:14
solution the root is equal to the root when02:42:17
when their radical questions are equal02:42:20
why write roots they would write02:42:22
radical is equal to under the knee so that02:42:25
Maldiz was hung just for this purpose and02:42:26
wrote roots but before radicals02:42:29
we equate these rights and rights02:42:32
and the radical was simplified, expanded either02:42:33
by disk Renata or by Vieta on02:42:35
multipliers using this formula02:42:36
laid out means we remove the roots and02:42:39
we say left radical is greater than or02:42:41
equals 0 and given that the radicals are equal02:42:43
then you know the right one is guaranteed too02:42:45
greater than or equal to 0, no need to write that02:42:47
should be right, that's what it should be from02:42:50
zero to one to how to solve this02:42:52
the equation just flip to the left02:42:54
you take out a parenthesis, you decide, you decide02:42:57
you get x 1 x 2 1 2 + 1 2 you get02:43:01
two roots let's find out at what a x 102:43:05
will be the root of the equation for02:43:07
segment that is x 1 should along02:43:09
satisfy the water behind him lie on02:43:11
segment it turned out that with such02:43:13
and x 1 is a root on the segment a for02:43:17
which a x 2 is a root on the segment02:43:19
he needs to satisfy the dose and lie down02:43:22
on the segment it turned out that x 2 is02:43:25
roots on segments for such a02:43:26
and for which a the roots must coincide02:43:31
equate them and it turns out that for a02:43:34
equals 1 they coincide and now we build02:43:37
number line and where we show when02:43:39
what and how many roots and what kind02:43:42
specifically the roots, I found out that there is one root02:43:46
acquired but only maybe with02:43:49
there are either two roots or none02:43:50
roots02:43:51
we all wrote down and at which it will be02:43:54
only one root, that's the problem02:43:57
find all values for each of02:44:00
of which the equation has exactly one02:44:01
root on the segment from 4 to 802:44:03
a fraction is equal to zero when the numerator is equal02:44:05
zero and the denominator is not zero but still02:44:08
radical must be greater than or equal to02:44:10
0 but since the denominator since since02:44:14
the root of the denominator is the radical02:44:15
cannot be equal to zero, that is02:44:18
the radical must be strictly greater02:44:19
zero numerator equals zero radical02:44:22
strictly more than me that is given02:44:24
equation it is equivalent to the system where02:44:28
the numerator is zero and the radical02:44:30
strictly greater than 0 we write the product of two02:44:35
brackets02:44:36
bullets are equal to zero and the radicand is strictly02:44:39
greater than zero so 10 x minus x square02:44:44
minus and the square is strictly greater than 0 and also y02:44:50
us x must be between four02:44:52
until eight until let's log in too02:44:54
let's turn on cocido, that is, x must be from02:44:56
four to eight, well, it’s up to us to decide02:45:02
equation product equals zero when02:45:05
when at least one of the factors is 002:45:07
in the second there is, that is, simply02:45:10
each to zero we equate x minus a02:45:11
minus 7 equals zero or x plus and minus 202:45:17
equals zero and must be fulfilled02:45:21
roughly speaking about the dose 10 x minus x square02:45:23
minus and the square is greater than 0 and excel to us02:45:27
needed only in the range from 4 to 8 so02:45:29
let's express x from the equations that is02:45:34
it turns out x 1 equals plus 7x second02:45:39
equals 2 minus and it should02:45:44
performed at a dose of 10 x minus x square02:45:46
minus and the square is greater than 0 and x must be02:45:50
from four to eight such a system02:45:54
must be fulfilled02:45:55
as you see if x 1 and satisfies02:46:00
there for02:46:01
and is included in the segment from 4 to 8 then it02:46:04
is the root of the equation02:46:05
and there is also a second one that will satisfy this02:46:08
for is the root of the equation we have two02:46:10
root but we need to have one root02:46:12
let's find out at what a x 1 is02:46:16
root then we will find out at what a02:46:18
x 2 is a root then we find out when02:46:21
what and they coincide and two roots02:46:24
turn into one single root02:46:25
and then we'll look at the end before answering02:46:28
at what and how many roots and what are they?02:46:31
roots and then we will know at what a02:46:33
there will be exactly one root02:46:35
so x 1 is the root of the equation here x02:46:39
1 it is equal to a plus 7 when it appears02:46:45
the root of the equation when at it is02:46:52
the root of the equation for a satisfies02:46:58
inequality is a to z and x must be from02:47:01
four to eight, that is, it should02:47:03
satisfy the system02:47:05
10 x minus x square minus a square02:47:10
greater than 0 and x must be between four and02:47:12
eight we will now find out at what a x 102:47:16
is the root of the equation then the same02:47:18
most done for which then when02:47:20
they match and then write the answer02:47:22
that means we have to substitute it right now02:47:25
Substitute a plus 7 for x, that is02:47:30
it turns out to be 10 times a plus 7 minus02:47:36
x squared is minus a plus 7 squared02:47:40
minus x squared greater than zero02:47:42
bodex we substitute a plus 702:47:49
and decide, well, how to decide here, let’s go to02:47:57
let's simplify a little, at least the second one is easy02:48:00
solve there and from what to what -7 subtract02:48:03
-7 out of three sides02:48:05
inequality here one here minus 3 and from -302:48:08
to 1 a 1 how to solve we open the brackets02:48:11
then ten a plus 70 minus square02:48:15
sums that is minus and squared minus02:48:17
14 a02:48:18
and -49 minus square is greater than zero02:48:23
so let's solve the first one separately02:48:26
inequality because you still have a long time02:48:27
I don’t want to write the second one to rewrite 5002:48:29
once we call the inequality one given02:48:32
so minus x square minus a square is02:48:34
-2 and it’s a square, so we’re the first to solve -202:48:39
and the square is further than 10x minus 14 and this is02:48:43
will be minus 4a02:48:46
70 minus 49 plus 21 must be greater than 002:48:50
multiply by minus 1 and it turns out to be 2 in02:48:56
squared plus 4 minus 21 when we multiply02:49:01
on negative icon unfolds02:49:03
less than 0 so discriminant b squared02:49:07
minus 4 by 2 and by -21 that is plus 16802:49:14
it turns out02:49:20
so 106 from 8 184 184 to 184 is02:49:28
184 is 4 times 4602:49:36
so it turns out that a is equal to a first02:49:41
second minus b minus 4 plus minus root02:49:46
out of 402:49:47
46 is 2 roots of 46 divided by 2a then02:49:51
is on 4 means a 1 is equal to what a 102:49:57
equals minus 4 divided by 4 is minus one02:49:59
and two roots and 46 divided by 4 is 1 202:50:05
root of 4602:50:07
and the second root is -1 minus 1 2 roots02:50:12
from hands 6 but we were not solving an equation but02:50:16
inequality therefore on the number line a02:50:19
We apply from the left the one who is smaller -102:50:22
minus one second roots 46 and to the right is that02:50:26
who is bigger, that is, minus 1 plus 1 202:50:30
root of 46 and set the intervals02:50:34
plus and minus signs02:50:36
plus we need those and in which02:50:39
It will be less than 0, that is, minuses are needed02:50:42
so now we need to find the intersection with a from02:50:45
-3 to 102:50:46
let's find the intersections02:50:59
-1 minus so we have -3 we have02:51:06
one and who are these numbers approximately02:51:10
what are they equal to02:51:11
so minus 1 plus half roots 4602:51:15
root of 4602:51:16
this is 6 something, let it be 6.8 for example02:51:20
half of 68 is 3.4 minus 12.402:51:25
this is approximately there to the right of everyone and minus 102:51:35
minus half of 68 minus 1 minus 3 402:51:42
minus 44 he turns out to the left of everyone and so02:51:49
we have the answer from one to the other and from -302:51:54
to 1 where 2 hatches there the answer is02:51:57
what is the conclusion02:51:58
from -3 to 1 what what happens from -3 to02:52:01
one from -3 to 1 x 1 is the root02:52:06
equations02:52:07
you understand, that is, he satisfies02:52:10
friends and is in the range from 4 to 802:52:12
therefore, for a from -3 to 1 x 102:52:21
which is the same as there x 1 that was a plus02:52:27
7x plus 7 is the root of the equation for02:52:33
segment02:52:37
and the satisfying dose is shorter02:52:40
root of the equation02:52:41
let's find out x2 when is the root02:52:44
equations and the second is equal we had 202:52:46
minus and we do the same thing02:52:49
so x 2 are two minuses and he is02:52:55
root of the equation for a satisfying02:53:01
system, that is, he is for02:53:04
it is necessary that 10 x minus x be executed02:53:07
square minus and square greater than zero to02:53:13
and x must lie in the range from 4 to 802:53:16
let's find out x 2 for what a is02:53:19
root of the equation02:53:20
let's substitute it turns out let's on the left02:53:26
write that is 10 multiplied by x is 1002:53:30
by 2 minus a minus 2 minus a squared02:53:35
and minus x square is greater than 0x from 4 to 8 then02:53:41
there are two minuses there must be four or more02:53:43
up to eight02:53:47
the knot 1 5 long to decide let's do it02:53:50
let's simplify by 1 2 for now we'll solve like this 20 minus02:53:54
10 and minus 4 minus 4 plus 4 minus to02:54:02
plus a square turns out to be a minus square02:54:04
there we change all the signs to because minus02:54:05
before the brackets there is a minus and the square is greater02:54:08
zero here we subtract 2 from all sides02:54:12
double inequality, that is, there will be 202:54:14
minus a and six we give similar02:54:19
minus 2 square minus 10 plus four is02:54:24
minus 6a02:54:27
2416 must be greater than zero by -102:54:31
multiply, that is, it turns out a from minus 202:54:38
to minus 6 but the icons need to be expanded02:54:40
because they pressed negative here02:54:43
must be from -6 to -2 and02:54:47
and that means we need to solve this inequality02:54:50
let's call it number 2 and solve02:54:53
it separately also with discriminant02:54:54
now let's solve the second inequality by02:54:58
discriminant02:54:59
firstly, you can divide by -2, that is02:55:03
will be a squared plus 3a minus 802:55:07
less than zero discriminant 9 + 3241 a02:55:15
equals minus 3 plus minus root of 4102:55:19
secondly, that is, we are on the number line02:55:23
apply -3 minus root of 4102:55:29
second and -3 plus root of 41 second02:55:35
place intervals with plus and minus signs02:55:38
plus02:55:39
we need minus signs and we need to find them02:55:43
we find the intersection with a from -6 to -202:55:47
baptism02:55:55
so we put it on the number line02:55:59
minus 6 and minus 2 and these two numbers02:56:02
so they will take approximately what they are equal to02:56:05
root and 40 1 is about 6 and a bit02:56:08
well let it be 6.402:56:10
minus 3 plus 6 4 that would be 3-4 in half02:56:15
this is one and seven about 1 whole 70's then02:56:18
there is he more right than everyone else02:56:27
and here minus 3 minus 6 and 4 it will be02:56:32
minus 9 4 in half minus 4.7 that is, he02:56:37
not to the left of everyone -6 to the left of everyone then goes02:56:41
here it is and then it goes minus 202:56:49
filled in and so one of the answers from -6 to02:56:54
-2 and the other from fraction to fraction where 202:57:01
shading there answer 202:57:03
well, the short answer is the drive of these and where02:57:09
now the intersection x 2 is the root02:57:11
equations therefore for these02:57:17
and that is, and with proper from -3 minus02:57:21
root of 41 second to minus 2 here at02:57:28
such a x 2 is the root of the equation02:57:31
is the root of the equation02:57:40
so let's now learn about x 1 x 2 at02:57:44
which a coincide let's find out x 102:57:51
matches old iq02:57:57
if something like this happens x 1 it was y02:58:01
us a plus 7 x 2 is 2 minus a if a02:58:06
plus 7 equals 2 minus02:58:10
an error to the left turns out to be 2 and a seven02:58:13
to the right it turns out minus 5 equals minus02:58:16
two and a half that is, if a is equal02:58:18
output when a is equal to minus 2 and a half x02:58:25
1 is equal02:58:26
and to the old even what they are equal to we can02:58:29
understand minus 2 and a half substitute02:58:31
or they would have served it here at up to minus 2 s02:58:34
half plus 7 equals four s02:58:35
half that is02:58:38
there are 2 roots but they coincide and everything is like02:58:40
there are 1 roots 1 pieces four s02:58:42
half kick and match so we all02:58:49
are now ready to analyze at what a02:58:51
how many roots will there be02:58:52
and what is behind the root we do the final02:58:54
number line a and plot all a02:59:02
influencing the appearance of x 1 x 2 and02:59:07
coincidences there x first 2 that is us02:59:09
apply02:59:10
and these are from -3 to 102:59:15
these and minus 2 and a half02:59:21
so we apply minus 3 and one we apply minus02:59:25
3 and let one be here -3 for example here02:59:29
there will be one further, you need to apply another minus02:59:34
2 and this is the number that is approximately02:59:38
minus 4702:59:39
that is, it is to the left more to the left it turns out to be more to the left02:59:42
always Libya we all have this one / -302:59:49
minus the root of 41 second02:59:54
minus 2 is further to the right than -3 obviously02:59:58
so minus 2 and a half is still needed03:00:00
apply minus 2 and a half will be exactly03:00:02
between minus 3 minus 2 minus 2 s03:00:04
half of them are now small sticks03:00:07
we do03:00:20
so let's look at a first03:00:25
which is less than this fraction is a minus03:00:27
3 minus the root of 41 second let's look at03:00:31
which are smaller than which then03:00:32
Let's look at a equal to this fraction03:00:35
what kind of roots will there be and how much will it feed?03:00:37
then let's look at and which are from fractions to03:00:43
-3 then look at a is equal to -3 then03:00:47
Let's look at a from -3 to minus 2 s03:00:49
half then look at exactly minus03:00:52
2 and a half03:00:53
then let's look at a from minus two c03:00:55
half to -2 then look at a03:00:59
equals minus 203:01:03
then let's look at a from minus 2 to03:01:06
units03:01:08
then look at equal to 1 yanao more03:01:10
units so if a is less than / if a03:01:15
less than /dx 1 is the root03:01:18
there is no equation x 1 is the root03:01:21
equations for such and that is, a fraction03:01:24
fraction is not included it is minus 4 something03:01:26
is not included in this segment x 1 is not03:01:28
root x 2 is the root of the equation03:01:31
no fractions are not included because these03:01:36
the parenthesis does not include the fraction and03:01:37
which does not exist, that is, to the left than / no03:01:40
no roots x 1 not 203:01:44
if a is equal to / again then x 1 being03:01:49
is the root of the equation no x 1 not03:01:52
included03:01:53
well, more precisely, a fraction and does not include from -3 to03:01:55
1 is not included in this risk therefore x 1 no03:01:58
and x 2 no there is a parenthesis yes therefore03:02:04
there are no roots if we take some03:02:10
number between fractions minus 3y minus 403:02:13
for example -4 is included in the segment from -303:02:18
This means that x 1 is not the root of x 203:02:21
root that is, if we take03:02:23
interval from fraction to -303:02:27
then this interval is included in this03:02:31
interval x 2 means will be here is03:02:33
root opinion x 2 will be the root here03:02:37
equation it will be equal to 2 minus03:02:39
x 2 is equal to 2 minus, that is, when03:02:42
these and we have only one solution03:02:46
circle this about the one we were looking for good03:02:50
so if a is equal to minus 3 x 1 is03:02:54
the root of the equation until we see a quadratic03:02:56
the bracket x 1 is the root of the equation03:02:58
if a is equal to minus 3t x 1 is03:03:00
the root of the equation and it is equal to what it is03:03:04
equal to a plus 7, like double a plus 703:03:07
but if a is equal to minus 3 then it is equal to03:03:11
it turns out it is equal to minus 3 plus 7 then03:03:15
is it equal to 4 4 x 2 what is x 2 equal to03:03:22
so if a is equal to minus three minus three03:03:24
come here started 2 is the root03:03:27
equation and 2 is equal to something else two03:03:30
minus but something else and not 2 minutes myself and03:03:33
we know that it equals -3 to03:03:36
equals 5x 1 equals four x 2 503:03:40
they are different two roots but something happened03:03:43
root to03:03:44
we don’t need two roots, that is, we don’t need -303:03:47
suitable if we take a from -3 to03:03:50
minus two and a half from -3 to minus03:03:54
two and a half interval is included in03:03:55
this segment yes means x 1 will be03:03:58
be the root of the equation a plus 7x so03:04:04
and from -3 to minus two and a half are included03:04:07
in this period, yes, that means 203:04:10
will be the root of the equation and it03:04:12
will be equal to 2 minuses and they will be different03:04:15
we will have two roots, I don’t need two03:04:19
roots I need there was one root03:04:21
if a is equal to minus two and a half then03:04:24
minus 2 and a half and are in this03:04:27
segment yes means x 1 is a root03:04:29
equations x 103:04:32
it will be a plus 7 but since we know a03:04:37
it turns out four and a half, well if -203:04:42
spray on you will sleep well x 2 so03:04:46
minus 2 and a half and in well is in03:04:48
in this interval yes means and which03:04:50
is the root of the equation and it is equal to 203:04:53
minuses but considering that we know what03:04:55
the same and the same will happen with half then03:04:57
before we had two roots two03:04:58
there are two roots again, but these are two03:05:01
coinciding roots therefore a is exactly minus03:05:09
2 and a half in return will go with him03:05:12
there will be two matching roots further if03:05:16
and from minus two and a half to -2 from03:05:21
minus two and a half to -2 are included in03:05:23
segment in this yes therefore x 1 is03:05:26
the root of the equation and it is equal to a plus 7a x03:05:31
2 equals 2 minus03:05:32
so their sides are from minus two and a half03:05:36
up to -2 are included in this segment up to therefore x03:05:38
2 will be the root of the equation will03:05:41
2 minus 7 we will have two different roots03:05:44
those that don't match already they only match03:05:46
about two and a half further if a is equal to minus 2 minus 303:05:50
minus 2 x 1 is the root of the equation -203:05:53
are included here yes therefore x 1 is03:05:56
root of the equation x 103:05:58
it is equal to x + 7, well, considering that a03:06:02
is equal to minus 2 it turns out it is equal to 5 so03:06:06
and minus 2 come in here and see the square03:06:10
the bracket means the second one is also there and he03:06:14
equals 2 minus and there were 2 minuses here03:06:20
2 and minus and here it was like that here too03:06:22
two minuses there 2 minus the author's root03:06:27
two minuses so two minuses and considering03:06:30
what and this is minus two makes four03:06:33
remember we had a situation x 1 4 2 5 a03:06:36
the forest is the opposite x 1 5 2 4 but we have it again03:06:38
two roots that are not the same, they coincide03:06:41
only at -2 alloy03:06:42
so we don’t circle this, we need one03:06:45
take the root03:06:46
interval from minus 2 to 1 interval from03:06:49
minus 2 to 1 per turn well lies here on03:06:52
this segment yes therefore x 1 here03:06:54
exists and it is equal to a plus 7 from minus03:07:00
2 to 1 are03:07:03
here from minus 2 to 1 are included in03:07:06
segment in the interval from fraction to -2 no03:07:08
that's why he's not here and here we are03:07:11
found and at which there will be only one03:07:13
solution ks 1 if a is equal to 11 when a is equal03:07:19
1 x 1 is the root of the equation03:07:21
yes, and because it’s square03:07:23
the bracket when a is equal to 1 x 1 is a root03:07:27
equation it is equal to a plus 7 but given03:07:30
that a is equal to 1 it turns out it is equal to 8 one03:07:32
plus seven eight a x 2 no x 2 at all03:07:36
more to the right than minus 2 x 203:07:38
but it is not the root of the equation03:07:39
therefore a is equal to 1 and we are further suited to a03:07:43
more than one and more than one is not included03:07:46
during this period03:07:47
therefore x 2 x 1 will not be here and which03:07:51
there are simply no roots here either and what comes out03:07:55
it turns out that we circled03:07:57
and for which there will be exactly one solution03:07:59
on the segment, that is, the answer is from -303:08:07
minus the root from the hands of one second to -303:08:13
then free-standing minus 2 s03:08:15
half and from minus 2 to one03:08:22
square bracket everything so again than03:08:27
we did the fraction equal to zero when03:08:29
when the numerator is zero and the denominator03:08:30
is not equal to zero and the radical must also03:08:32
be greater than or equal to 0 but since it is in03:08:34
the denominator is strictly greater than zero03:08:35
radical, that is, initially and03:08:38
equation equivalent to the numerator system03:08:41
is equal to zero and the radical is greater than zero but03:08:43
x must also fall within the range from 4 to03:08:45
8 the equation can be easily solved we found03:08:47
x 1 x 2 we also have, roughly speaking, dc03:08:50
x 4 to 8 and we think so and at what03:08:53
and x 1 will be the root of the equation03:08:54
very easy when he satisfies there03:08:57
for and when he is in the segment 4 to 8 we03:08:59
we find these and here they are, the drive of such and03:09:02
x 1 root of the equation a x 2 for what a03:09:05
is the root of the equation very easily03:09:08
when she hits friends and but also lies03:09:09
in the segment 4 to 8 it turns out that at03:09:13
these03:09:14
a and k is the second root of the equation okay a03:09:16
for which a x 1 coincides with x2 priors03:09:19
to minus 2 and a half grade is now at03:09:21
number line and plot everything03:09:24
values and on which it depends03:09:26
number of roots and types of roots03:09:28
what are the roots equal to, so we look at03:09:31
like this03:09:32
and x 1 there is no x 1 there is not and which03:09:35
no no roots either03:09:36
but with this there are roots03:09:39
there are no roots, but with this a x 1 x 1 there are no03:09:44
x 2 already exists, so there’s only one here03:09:47
the root means these and they suit me well03:09:50
this is how we scan to the end and find03:09:52
for which a we will have either x 1 or x03:09:55
2 thousand will be exactly one root or something03:09:57
two roots but they coincide at03:09:59
this is such a task03:10:03
find all values for each of03:10:05
of which the equation has exactly one03:10:06
root on a segment from 0 given03:10:08
Well, obviously x-man can’t be thrown to the left03:10:12
will be x square minus x taken out for03:10:14
bracket there will be bracket x minus 1 and here03:10:16
I x1 remove this bracket03:10:18
and then there will be a product of two03:10:21
factors equal to zero shorter than x squared03:10:24
move x to the left with a minus and further03:10:28
rewrite03:10:35
now we put x outside the bracket further03:10:40
rewrite03:10:46
now we remove the parenthesis as general03:10:48
factor03:10:54
x plus root of 3 x minus a must be03:10:58
is equal to zero so when the product is equal03:11:01
zero when at least one of the factors03:11:02
is equal to zero and the second one exists, that is03:11:05
we say that x minus 1 is equal to zero or03:11:11
x plus the root is equal to zero and03:11:19
radical and must be greater or03:11:21
equals 0 and x must be from 103:11:25
the unit series x must be from zero to03:11:30
units so we get such a system, well03:11:35
I already solved the first equation03:11:37
the second one needs to be solved there the second one needs to be solved03:11:39
decide the second you have to sit and decide like this03:11:43
so I think maybe the second one is separate03:11:45
write it out and now resolve it with the old one03:11:48
figure it out and then return to the system03:11:49
let's deal with03:11:54
in the second equation we return to the system03:11:57
so let's call it number 103:12:02
equation and let's figure it out x03:12:06
plus root equals zero03:12:11
let's leave the root in one country 3x03:12:14
minus y equals minus x equals x03:12:17
to the right and on the other side the root is equal to03:12:20
minus x what are you, how to decide03:12:22
rational equations we are talking about03:12:26
that the root cannot equal03:12:27
negative number yes that is we03:12:28
we say that the right side is larger or03:12:30
equals 0 and then square both sides03:12:33
build like this but if minus x is greater or03:12:41
equal to 0 means x is less than or equal to 0 and x03:12:43
there can only be 0 to ones that is03:12:44
only 0 suits us so x is equal to 003:12:49
in general so how how how the nights are here03:12:53
transition is an equivalent transition such that03:12:56
the right side must be greater or03:12:58
equals 0 and then can be squared03:13:00
both parts03:13:01
if the right side is 1 deuce then03:13:04
the radical must be a quadruple03:13:05
understand that is, if the right side is not03:13:07
negative then 3x minus a must03:13:09
be turns out to be equal to the right side in03:13:13
square minus x square like this here03:13:16
equivalent transition03:13:19
but let's notice, or let's let's03:13:24
let's continue03:13:25
minus x is greater than the mountain but 0 it is minus 103:13:28
multiply multiply by minus 1 and it turns out03:13:33
x is less than or equal to 0x less time03:13:37
than zero square and these03:13:41
we rewrite this if x is less or03:13:44
is equal to than 0 but in the segment it is from zero to03:13:46
units03:13:47
then the output is just x equals 0 to us03:13:52
fits understand we say what it is03:13:57
equation well this is the equation03:14:02
this equation has a root at03:14:09
segment from zero to one03:14:21
root on the interval from zero to one03:14:29
only if x is equal to zero you know because03:14:36
that x must be less than or equal to 0 but03:14:39
we only have a segment from zero to one03:14:40
0 hits03:14:41
x is either negative or zero03:14:44
negative we don't need xs segment03:14:46
Odynets therefore only a zero can be x03:14:49
equals 0 is removed, that is, at this03:14:53
equation into this equation means you need03:14:56
Substitute a zero under x and we will understand at03:14:58
what will happen let's write03:15:01
let's find at what a x is equal to zero at03:15:13
what happens and this happens, that is, it is necessary03:15:15
substitute into the original equation under x03:15:18
substitute the zero plus the root of 303:15:22
multiply by zero03:15:24
minus is equal to zero, that is03:15:29
radical equals zero minus a equals03:15:32
zero must be equal to zero03:15:34
output at03:15:38
a is equal to 0 when a is equal to 0 x is equal to well03:15:44
such a conclusion03:15:47
let's go back to our system03:15:49
back to the system03:15:59
we had a system and so x minus 103:16:04
equals zero let's finally decide03:16:06
the equation x is equal to 1 that is x 1 is equal to03:16:09
one so what is x 2 0 equal to and here is x 203:16:14
is equal to zero only it is equal to zero at03:16:16
equals zero, this should still be done03:16:20
condition 3x minus a greater than i equals 0 for03:16:23
each of them is clearly 2 executed03:16:24
so and k must be from zero to one03:16:26
Well, it’s clear that they are both from one03:16:29
it turns out x is equal to 0 from x 2 x 2 will be y03:16:33
let's call us the old one like this03:16:39
was x 1 let's return to the system means x 103:16:44
equal to zero on to so that only for him y03:16:46
yes for made from yes so here you go x 103:16:51
it is equal to unity for what a and for a03:16:57
satisfying the following conditions03:16:59
firstly must be performed in03:17:01
secondly, x must be in the interval from 003:17:03
up to 1 glass03:17:04
from zero to one therefore for questioning03:17:06
so that ode for fulfills so that 3x minus a03:17:08
was greater than and equal to 0 if x is 103:17:12
it turns out you need to multiply 3 by x by 103:17:14
minus a is greater than equal to 0 it turns out to be a03:17:17
less than or equal to 3 output when a is less03:17:25
or equal to 3 x 1 equal to one then03:17:30
is x 2 is equal to zero at exactly zero and03:17:32
no other way03:17:33
and x 1 is equal to one if less than or equal to03:17:35
than three someone will say and you finally03:17:38
checked that x 203:17:40
hit even let's check x equals 003:17:44
equals 0 put a zero and a zero here03:17:47
will be greater than or equal to 0 will be equal to 003:17:49
you understand that he is in the segment03:17:52
from 0 to one but no one needs it03:17:54
check that is essentially03:17:58
we found out at what a x 1 is03:18:02
root and for what conditions is a x 2 a root?03:18:05
Let's write the equations so I didn't finish it03:18:07
a little bit when a is equal to 0 and 2 is equal to 003:18:09
is the root of the equation at a given03:18:12
segment and when a is less than or equal to 3 x 103:18:22
is the root of the equation at a given03:18:29
segment can they coincide zero and03:18:34
unit03:18:35
no they don't match now let's03:18:38
Let's see at what and how much we have03:18:41
roots and what are the roots of the number line a03:18:49
we are interested in a exactly zero and a less03:18:52
or equal to 3 that is, we apply03:18:53
just zero to three and consider03:18:59
situation a less than zero a exactly zero a from03:19:05
0 to 303:19:06
ara inside and more 3 2 sticks03:19:13
we'll build it like this03:19:18
so if a is less than zero03:19:23
x 1 is the root of the equation x 103:19:26
is a root if a is less than03:19:28
3 a less than zero enters the beam a less03:19:31
or equal to 3 that is x 103:19:34
here is equal to one for such a x 1 is equal to03:19:37
unit03:19:38
and x 2 x 2 is a root only for a03:19:41
equals 0 so there is only x 1 root03:19:45
if a is equal to 0 then the zero is located03:19:49
less than or equal to three therefore x 103:19:51
is the root of the equation to units a x03:19:54
2 is also a root of the equation because03:19:56
that it is the root of the equation at03:19:58
is equal to 0, that is, 2 is equal to zero and we have here03:20:02
two roots before there was one root03:20:06
then two roots so if from zero to three03:20:09
0 to 3 is included in this ray therefore x 103:20:13
here is the root of the equation it is equal to03:20:15
alone and with the old ones never again03:20:17
is he then he was here like that03:20:20
if a is equal to 3 x 1 is the root03:20:22
equation is equal to one if a is greater than03:20:25
3 then we simply have no roots at all and03:20:28
so I had to find and at which03:20:31
we have exactly one root03:20:33
exactly one root exactly one root03:20:36
when a is less than zero when from 0 to 303:20:40
inclusive therefore the answer is from03:20:47
minus infinity to 0 and from 0 to 303:20:52
let's go over the entire solution inclusively03:20:55
find the value of a for each of which03:20:58
the equation has exactly one root03:21:00
segment, but the transformation x is also obvious03:21:02
left then take out x then take out03:21:04
parenthesis then when the product is equal03:21:07
zero when each bracket is equal to 0 plus03:21:09
1 curtain plus x from 0 to ones equation with root03:21:13
we're sorting it out here, we're starting it03:21:15
it’s up to us to decide, squaring it03:21:17
not useful, yes, that is, we started03:21:18
decide understood that x must be less03:21:20
zero but if we have a segment from zero to03:21:23
units then x is less than or equal to 003:21:26
only 003:21:29
why did we understand that x is zero in03:21:33
in the second equation x only the role of zero03:21:35
maybe it's equal and we all wanted it03:21:38
find out at what point a x is equal to zero and03:21:40
put a zero under x here and realized that03:21:43
also a zero03:21:44
and only if they are both zeros then we have03:21:48
there will be a solution on a given interval here03:21:51
if they are both zeroes the first equation is we03:21:54
solved, then the first equation x is equal to 103:21:57
the second equation x is equal to zero at equal03:22:00
0 and that's clear x 203:22:02
when a equals 0 is the root of the equation03:22:05
aix are the first for which a is a root03:22:07
equations for blowing hot doses and then03:22:11
that it is in the interval from zero to one and03:22:12
so and so it’s clear he’s just a unit03:22:15
it turns out that 1 x 1 satisfies for a03:22:19
less than or equal to 3 so with a less than or03:22:21
equals 3 x 1 is the root of the equation03:22:23
aprio exactly 0 and 2 is the root03:22:26
we carried the equations onto the number line03:22:28
everything depends on what they have03:22:31
we looked at the roots of the equation when03:22:33
and less than zero we only have x 1 at03:22:35
and equal to zero we have two roots for us this is not03:22:37
it is necessary at a from zero to three03:22:39
only x 1 at a exactly 3 x 1 at a03:22:42
there are no more than 3 roots at all03:22:44
we circled circled and at which it will be03:22:47
exactly one root and I am the answer03:22:52
find all values of a for each of03:22:55
whose equation has a unique03:22:56
root on the interval from -1 to 1 well on03:22:59
square equals square let's write it down03:23:02
difference of squares that is, we will bring03:23:04
the right side to the left with a minus turns out03:23:08
x square plus square root minus03:23:16
we also move the right side with a minus03:23:22
to the left plus the difference of squares is zero03:23:28
further03:23:29
difference of squares formula03:23:32
when the abbreviated multiplication formula03:23:35
appear in some more complex03:23:37
tasks, not everyone sees them, so I’m on03:23:39
I'll remind you of this one more time just in case03:23:41
shape a square minus b square03:23:45
turns into a minus b to plus b03:23:48
it turns out we have one bracket this is x03:23:52
square plus root minus that's all03:23:58
is minus 2x minus 1 minus root 203:24:03
the bracket is inside the bracket plus03:24:06
the other inside with a bracket, that is, x03:24:08
square plus root and minus x + 2x +03:24:14
1 + root and minus x equals zero so03:24:20
when the product is zero when although03:24:22
if one of the factors is equal to zero and03:24:23
the second exists so we have roots here03:24:27
mutually destroyed03:24:29
yes it turns out that we have either 103:24:34
bracket equals zero or 2 equals zero + dc03:24:37
that is, the radical is greater than or equal to 0 and03:24:40
plus x excellent given that is, we have03:24:43
it turns out either x squared minus 2 x03:24:46
minus 1 equals zero03:24:50
or x square plus square root plus square root03:24:53
is this plus two roots or let's do this plus03:24:56
2x+103:24:58
plus two roots and minus x equals 0 plus03:25:05
we also need to have more radicals03:25:08
or equal to 0 and x must be in the interval from03:25:13
-1 to 103:25:16
this is the system let's solve both03:25:21
equations03:25:22
let's find the roots and then see at what03:25:27
and the root03:25:28
is the root of the equation on the interval then03:25:30
there is when he satisfies this for and03:25:31
is on segments for which a x 103:25:35
at what my 2 maybe there will be x03:25:37
3 at what a will it be the root of the equation03:25:39
there in front of everyone, let's see what roots they have03:25:42
we get where the first one is shorter03:25:45
the discriminant equation is easy03:25:47
decided yes no no I won’t be here03:25:49
describe so as not to complicate the recording03:25:54
solution, well, the discriminant is 4 + 48 then03:25:58
there's eight that's four by 2 x 1 2 that's03:26:05
minus b that is two plus minus 2 roots03:26:09
of 2 divide by two shorter by03:26:12
I can find the discriminant for everyone's roots03:26:14
I hope so, that is, x 1 will be equal to 1 there03:26:18
plus the root of 2 x 2 is equal to 1 minus03:26:21
root of 2 and here in the second equation03:26:30
By the way, you can notice the square of the sum03:26:31
see x square plus 2x + 1 is x plus03:26:34
1 squared plus 2 times the root03:26:39
of x minus x is equal to zero and minus x is greater03:26:45
or equal to 0 x from 0 from -1 to 103:26:49
further see 1 minus root of 203:26:53
not included included but one plus root03:26:57
of 2 are not included03:26:59
segment from -1 to 1 so it disappears03:27:01
it disappears03:27:03
all that remains is x 2 x 2 equals 1 minus03:27:08
root of 2 it is included for now x 1 has disappeared03:27:10
because it's not included here let's03:27:13
we write down that we notice that x 1 is not03:27:21
belongs to the segment03:27:24
therefore disappears and belongs to the segment03:27:30
-1 to 103:27:34
so I dropped it, but here’s the equation03:27:37
She's how you decide to growl unusual up03:27:40
the square of the sum plus two roots equals zero03:27:44
notice that this expression is03:27:45
non-negative, that is, greater than or equal to03:27:47
0 and this is also greater than or equal to 0 tell me03:27:50
to me when the sum of two non-negative03:27:51
is equal to zero only when each of them03:27:55
equal to zero03:27:56
that is, each of the terms must be03:27:58
equals zero, let's do this here too03:28:02
note note that the equation x plus 1 in03:28:15
square plus two roots has a solution03:28:21
has a unique solution03:28:30
only if only if each of03:28:37
terms is equal to zero03:28:50
each of them that's why it was x03:28:55
1 x 2 x 1 not included it is not included03:28:57
segment is the equation it has03:29:00
a solution is obtained only if so03:29:05
only if x plus 1 equals zero and 203:29:12
the root of a minus x is equal to zero, understand that03:29:16
is when each term is equal to zero03:29:19
in this case it must be fulfilled at dc and at03:29:23
in this case x must be from 0 -1 to 103:29:26
person 0 I want to say this to everyone03:29:28
such a system turns out like this x 2 expert03:29:32
let's solve this little system03:29:34
it turns out that x 2 is equal to 1 minus the root03:29:38
from 2 we rewrite well here03:29:42
he can give one to the right03:29:44
will take x 3 x 3 equals minus one so03:29:47
if x 3 is equal to minus one let's go here03:29:51
let's substitute but we solve this system03:29:53
in short and under x minus 1 we represent that03:29:57
there is 2 times root x minus03:30:00
minus 1 plus 1 equals zero for what a03:30:03
this equation will be equal to zero well when03:30:06
which and the left-hand side will be equal to zero at03:30:09
-1 understand that is, if under x03:30:13
substitute minus 12 will equal minus 103:30:19
yes, you understand, and minus x should03:30:24
be greater than or equal to 0 and at the same time x03:30:26
must be between -1 and 103:30:29
this is what should happen here03:30:33
there were originally 3 roots it seems03:30:35
equations x 1 x 2 x 3 x 1 everything has already disappeared03:30:38
there are two roots left for the equation x 2 x 303:30:43
let's find out at what time03:30:46
x 2 is the root of the equation x 2 is equal to03:30:51
1 minus root of 203:30:54
for a satisfying the following03:30:58
the requirement must first be fulfilled03:31:02
even with this Mexico secondly he03:31:07
must fall within the range from -1 to 1 but03:31:09
well it goes like this so why do I need this03:31:11
write everything again03:31:13
and minus x is greater, that is, it will simply succeed03:31:15
must be carried out with him and then he03:31:17
is appropriate and understand so03:31:20
minus under x I substitute one minus root03:31:22
of 2 minus 1 plus root of 203:31:25
it turns out that big ones are equal to one03:31:28
minus root of 2 output when a is greater than or equal to one03:31:34
minus the root of 2 x 2 is the root03:31:37
equations for others and is not03:31:43
root of the equation on a given interval03:31:49
good and hour x 3 x 303:31:53
which is equal to -1 it is generally only for a03:31:59
is equal to minus 1, but only when a is equal to03:32:02
minus 1 is the root of the equation03:32:05
Is it fulfilled by dc, of course yes then03:32:09
there is firstly here we have a minus x equals03:32:11
zero that fits the big ones equals 0 in03:32:13
for the second we can substitute -1 for yes and03:32:16
Substitute -1 under x and it will also be equal03:32:19
zero, that is, o d is satisfied for these03:32:21
conditions and -1 is naturally included in03:32:24
segment from what is applied to the given one, that is, x03:32:25
3 is equal to -1 at03:32:27
and equals minus 1 understand x 2 is03:32:32
the root of the equation drives these a a x 303:32:36
is the root of the equation drive this a03:32:38
that is, with a equal to minus 1, so let's03:32:43
we write therefore for a equal to minus03:32:46
1 x 3 is the root of the equation now03:32:52
let's see at what and how much03:32:54
roots what kind of roots are they03:33:02
so upset straight03:33:11
straight line and on it we put the values that03:33:15
affect the number of roots and the appearance03:33:18
the roots are shorter and we apply one minus03:33:21
root of 2 and minus one is a from03:33:23
which depends there x 1 x 2 are03:33:25
roots or not x 3 so that means we03:33:28
apply -1 and 1 minus the root of evil is03:33:31
to the right, that is, we apply -1 and 1 minus03:33:36
the root of 2 MPa is made a little to the right03:33:40
small dashes03:33:48
we will consider the situation when a03:33:52
less than -1 when a is exactly -1 when a is from -103:33:58
to 1 minus root of 2 when a is exactly 103:34:02
minus root of 203:34:03
and when a is more than one minus the root03:34:05
from 2 so if a is less than -103:34:10
x 2 is the root of the equation or not x03:34:14
2 is a root only if a is greater03:34:16
or equal to one minus the root of 2 then03:34:18
there is x 2 there is no trace of it but x 303:34:21
is the root of the equation no he03:34:23
is the root of the equation only for a03:34:24
equal to minus 1, that is, there is simply no03:34:27
roots03:34:31
if a is equal to minus 1 then x 3 is03:34:36
the root of the equation and it is equal to -1 a x 203:34:40
hasn't started yet x 2 is the root03:34:42
equations here so from means a03:34:46
equals minus 1 is part of the answer when03:34:48
and equal to minus 1 turned out to be the only one03:34:50
root minus 1 xor x equals minus one03:34:52
well if a from -1 to 1 minus the root of03:34:57
2 then it turns out x 2 there no he is not03:35:02
the root of the equation x 2 is the root03:35:04
early equation drive and quiet and which03:35:07
no, x 3 is the root03:35:09
equations for a equal to minus 1, that is03:35:11
there are no roots here just if a is exactly 103:35:18
minus root of 2 x 203:35:20
exists and it is equal to 1 minus the root of03:35:22
2 x 2 is equal to 1 minus the root of 2 at03:35:26
there are some like that and we have one now03:35:30
minus root of 203:35:31
and that means 2 1 minus q2 and x 3 exists03:35:35
only well is the root smoother only03:35:37
if a is equal to minus 1 then there is such a03:35:40
I'm happy if there's more than one03:35:43
minus the root of 2 x 2 is equal to 1 minus03:35:45
root of 203:35:46
and x 3 no, by the way, is only a root03:35:52
when a is equal to minus 1, that is, they are not like that03:35:54
we're happy too, write down the answer03:36:00
Suitable for us, but free-standing is a minus03:36:04
one and the ray from 1 minus the root of 2 to03:36:08
plus infinity let's briefly go over03:36:11
to the whole decision we have a situation from the beginning03:36:15
square equals square03:36:17
throw the difference from the right square to the left03:36:20
squares we get the difference of squares03:36:21
write it as a minus b by a plus b03:36:23
simplify we get two equations 1 easy03:36:26
discriminant for two roots found one of03:36:28
they immediately disappear because they are not included03:36:30
in this segment, that is, x 1 is missing03:36:32
left x 2 and this equation is like this03:36:36
unusual to analytical we need to solve03:36:39
notice the square of the sum plus two roots03:36:41
me negative plus not negative03:36:44
is equal to zero only if each of03:36:45
terms is equal to zero, so he said that03:36:47
every term is equal to zero then x03:36:50
is equal to -1 when a is equal to minus one comes out03:36:54
that initially the equation had to do with03:36:56
three roots and we need to find and at which it will be03:36:59
only one of the roots will remain03:37:01
segment from -1 to 1 x 1 dropped immediately from03:37:05
the remaining two roots of 2 are03:37:08
root of the given a03:37:09
and x 3 is a root only for a03:37:13
equal to minus 1 we are on the number line03:37:15
caused everything and affecting that03:37:18
is xyg also a root of the equation?03:37:20
but for such and not x 2 x 3 are not03:37:24
the root of the equation for a is equal to minus 1 x 303:37:28
is the root of the equation when equal to -103:37:31
aek parties is not therefore total03:37:35
one guy means it suits us if03:37:37
and from -1 to 1 root of 203:37:40
neither the first nor the second is shorter03:37:42
we have also found a situation where they03:37:44
only x 2 is the root of the equation03:37:46
given and all this was written down in response03:37:48
end03:37:51
find all values of a for each of03:37:53
of which the equation has exactly one03:37:55
root on the interval from 0 to 1 let's say that03:37:59
move from the right to the left with a minus then03:38:01
we’ll make the general ones and then decide when03:38:06
the product is equal to zero when at least03:38:07
one link that I am equal to zero plus even03:38:09
for the entire in x from zero to one03:38:11
classic that is root x + 2 multiplied03:38:17
by ln x minus y equals x minus 1 times03:38:27
on ln x minus and so it means on the right03:38:34
move parts to the left equals zero03:38:36
write now let's take it out03:38:38
ln for purchase remains in the second bracket03:38:45
root minus x minus 1 thousand minus x plus03:38:51
1 equals zero03:38:53
so it must be said that each of03:38:58
multipliers equal to zero plus at the dose site03:39:00
that is, we say that ln x minus a about03:39:05
zero03:39:06
root x plus 2 minus x + 1 equals zero03:39:14
plus the radical must be greater or03:39:17
equals 0 x plus 2 and must be greater03:39:20
or equal to 003:39:21
rhythmic forgery must be greater than 003:39:23
strictly x must be between zero and one03:39:28
We’ll also write it into the system, it was possible not to03:39:31
I'll write it down for us, so now let's do it03:39:35
let's solve two equations, solve them outside the system03:39:42
so as not to rewrite everything else03:39:44
we have equation 1 there is comparison 203:39:47
let's solve them one equation 1 l n03:39:50
this is when the algorithm is at its core03:39:52
there is e to the 0th power equal to x minus and then03:39:56
there is one equal to x minus and that is x03:40:00
is equal to a + 1 this will be x 103:40:04
so second second second03:40:10
by the way, if x minus are a is equal to03:40:13
one then x minus a is greater than 0 and03:40:15
automatically executed i.e. for x03:40:17
1 x minus and more than zero is not possible03:40:19
demand, well, you can demand an extra one03:40:21
since the second means let's use the equation03:40:25
leave the root on one side and the rest03:40:26
let's estimate in the other direction x plus 2 equals03:40:31
throw excel and one03:40:33
x minus 1 and so the root is not clear03:40:36
what are we talking about this we are talking about03:40:39
that the right side cannot be03:40:41
a negative root cannot equal03:40:43
negative, that is, we say that x03:40:46
minus 1 or zero or more more03:40:48
or equal to 0 if so if there is a two03:40:50
for example, then the radical 4, well, that is03:40:53
the radical is equal to the right side in03:40:55
square03:40:57
roughly speaking, we square both03:40:59
parts x plus 2 equals x minus 1 in03:41:04
square03:41:05
here is an equivalent transition but if x03:41:10
minus 1 is greater than or equal to 003:41:11
then x is greater than or equal to one so x03:41:16
greater than or equal to one x plus 203:41:19
equals x minus 1 squared so if x03:41:22
greater than or equal to one and we have03:41:25
segment from zero to one03:41:27
then the only x that is included in03:41:30
the segment from zero to one is x equals 103:41:33
so under x we need one now03:41:35
substitute and understand at what and with us03:41:37
such x will be that is, this equation03:41:39
It has03:41:40
there may be many solutions, of course it is possible03:41:42
but yes this is the level03:41:43
she has two solutions, yes most likely03:41:46
yes it is square but can have two03:41:47
solutions but well, both solutions must be03:41:51
greater than or equal to one but given03:41:55
that x is more than one and given03:41:58
that x is from zero to one inclusive03:41:59
only x equals 1 can be a solution03:42:01
let's write down the second equation03:42:06
considering that we are looking for a root on the segment03:42:14
from 0 to 1 root on the segment from 0 to 103:42:29
so given that we are looking for the root of03:42:31
sharply to one only x equals 103:42:34
maybe this root is only x03:42:37
it turns out 2 x 1 occupied x 2 equals 103:42:40
could this be the root03:42:42
here of all here x is greater than 1 there03:42:45
one two three four and so on only03:42:46
x alone equals 1 could be it03:42:49
root let's find out at what a03:42:57
x equals 1 will be this root, that is, we03:43:00
we substitute it here, that is, it is possible and yes03:43:04
cut it there when we erected the square03:43:06
we substitute the second into the equation03:43:09
one yes yes let's say we'll find it we'll find it03:43:17
at what and at what03:43:21
and x second equal to 1 will be03:43:26
will be the root of the equation03:43:38
equation root of x plus 2 equals x03:43:43
minus 1 thousand by the root of equation 2, well03:43:47
here, in short, we substitute x equals 1 and03:43:49
let's understand at what time this happens03:43:51
is will be the root of 1 plus 2 equals 103:43:56
minus 1 that is, the root is equal to zero means03:43:59
radicand equals zero03:44:01
means 2 and is equal to minus 1 is equal to03:44:05
minus one second output at a equals03:44:10
minus 1 2 x 2 equals 1 is the root03:44:15
equation by the way is it necessary03:44:18
check about d is running at dc03:44:20
so x that is 1 plus 2 times minus03:44:23
1 second will be 00 greater than or equal to 0 to03:44:27
here we no longer violate x minus 103:44:31
minus minus 1 2 more than 0 to x from zero to03:44:35
fighter units from zero to one to03:44:37
in short, when a is equal to minus 1 2 x 2 is equal to 103:44:42
is the root of the equation and all uds for03:44:50
there, well, more precisely, that’s all it means03:44:54
satisfies ude for so is the root03:44:56
equations03:44:58
let's go back to the system let's go back to03:45:01
system03:45:02
we get two equations03:45:10
the equation gave us Sochi roots x 1 equals03:45:13
a + 1 x the second turned out to be equal to one03:45:17
when a is equal to minus 1 2 and should03:45:25
be carried out here for03:45:27
x plus 2 a is greater than or equal to 0 x minus a03:45:32
greater than 0x must be from 0 to ones well03:45:37
project two we already understood everything when03:45:39
what and it is the root of the equation at03:45:41
others and not let's pro x 103:45:44
let's understand that is, we say x 1 which03:45:47
equals a + 103:45:50
is the root of the equation for a03:45:52
satisfying the following requirement03:45:57
firstly x plus 2 must be greater03:45:59
or equal to 0 in the second x minus a should03:46:06
be greater than 0 remember we said that03:46:07
this will definitely happen, let's say03:46:10
we didn't notice this, we'll get it again03:46:12
now and x must be from zero to03:46:14
units we get such a system we need it03:46:19
solve and we will understand at what a x 103:46:22
is the root of the equation so under x03:46:24
we substitute a + 1, that is, we get a +03:46:28
1 + 2 a greater than or equal to 0 a plus 103:46:36
minus and greater than 0x03:46:41
from zero to one so a + 2 is 3 a03:46:47
large is equal to minus 10 and greater than03:46:54
-1 a + 1 from zero to one that is03:47:03
it turns out that what and big is equal to03:47:05
minus 1 3 0 multiplied by what is more than03:47:09
negative number for anything here03:47:14
subtract one from all three sides03:47:16
double inequality, that is, a from minus03:47:18
one to zero we will find the suppression we will find03:47:24
crossing we don't go crossing03:47:33
the intersection of a is equal to minus03:47:35
one third03:47:36
and from the place but obviously from -1 3 to zero03:47:39
intersections here it turns out well let's do it03:47:47
painted, let's find the punishments03:47:49
intersect then the number line and draw03:47:51
-1 put a zero put minus one03:47:56
third we need to check everything than minus03:47:59
one third03:48:00
and everything is in the interval from minus 1 to 0 where 203:48:03
shading there is the answer output at and at03:48:09
proper segment from -1 3 to zero03:48:14
x 1 which is equal to a + 1 is the root03:48:18
equations but on a given interval03:48:20
is the root of the equation03:48:26
it remained to understand at what a x 103:48:31
matches with and to the old ones03:48:32
well, considering that x 2 is a root03:48:36
only when you see minus 1 2 for sure if03:48:38
they coincide only at minus 1 203:48:40
let's check this x 103:48:45
matches the old iq if anything03:48:49
occurs coincides with and to the old if03:48:52
they are equal, that is, x x 1 this + 1 of 203:48:56
this one plus one matches the old x03:49:07
2 this one turns out to be zero03:49:10
so that is, when a is equal to zero, the roots03:49:15
coincide nu prea equals zero 1 no root03:49:18
in general, well, this is very 2 roots, no at all, but03:49:23
the first one is, that is, they coincide but x 203:49:29
then there is no and which, when a is equal to 0, will not exist03:49:34
That's why03:49:36
it turns out they won't match, it doesn't matter03:49:38
means the number line and the number line03:49:45
A03:49:46
and we need to apply, which means influencing03:49:51
whether the root is x1 or x2 and so on03:49:54
put a equal to minus 1 2 and minus 1 3 and03:49:59
0 so minus 1 2 to minus 1 2 minus one03:50:04
second to the left of all minus one third a little03:50:10
more to the right 0 more to the right so let's 303:50:21
let's make three cells like this03:50:23
let's look at the situation when a03:50:31
less than -1 2 when a is exactly minus one03:50:34
second03:50:35
when and from minus 1 2 to minus 1 3 when03:50:42
and equals minus 1 303:50:44
when a is from -1 3 to zero when a is 003:50:50
and when a is greater than 0 and so x 1 is03:50:57
root of the equation for a from -1 303:51:00
inclusive up to 0 inclusive that is x03:51:02
1 is the root here and it is equal to a + 103:51:06
and given that a is equal to minus one third03:51:09
it is equal to minus 1 3 plus 1 that is two03:51:11
third x 1 is the root here it is03:51:15
equal to a + 1 always different value of a03:51:17
here x 1 is equal to plus 1 and given that a03:51:20
is equal to 0 it is equal to one it turns out x 103:51:25
in three places03:51:26
exists and x 2 is the root03:51:30
equations only for a equal to minus 1 203:51:33
x 2 is equal to one only in this case03:51:38
obtained when a is less than 1 2 minus03:51:42
up to 2 there are no roots at03:51:46
exactly minus 1 2 1 root03:51:48
with a from minus 1 2 to minus 1 3 no03:51:51
roots for a is equal to minus 1 31 roots for03:51:59
from -1 3 0 1 root when equal to 0 1 root03:52:03
and at the end there are no roots for a greater than 0 and03:52:09
Let's circle those a for which we got one03:52:11
root03:52:12
this is minus 1 2 minus 3 from -1 3 003:52:16
inclusive, that is, there is only one answer03:52:21
the root will be if a belongs to either03:52:25
minus 1 2 if a is equal to minus 1 2 or03:52:29
and lies in the interval from -1 3 to zero then03:52:34
there are two roots here, nowhere at all03:52:35
we didn't find it, let's go over it briefly03:52:38
the solution must be found and at which03:52:41
the equation has exactly one root we03:52:43
throw the right side to the left to take it out03:52:45
ln for buying we take out saying that everyone03:52:48
the multiplier is 0 plus 1 plus x of zero03:52:50
to one we find x 1 with iq old here in03:52:56
in general we begin to solve the equation we understand03:52:57
that x is greater than or equal to one so x03:53:00
must be in the range of zero to one and03:53:01
however, it is no longer equal to one03:53:02
it turns out only x is equal to 1 maybe03:53:05
be the root of the equation here03:53:06
we say that x equals 1 will be03:53:08
the root of the equation is specifically in this case03:53:10
and with others it won’t be all we return to03:53:14
we tell the system that we have x 1 x 203:53:17
existing only but is03:53:18
which are the roots of the equation only for03:53:20
this but plus 1 plus x should be fromDescription:
Если в задании 18 про параметр есть уравнение с ОДЗ, где требуется найти а, при которых будет РОВНО ОДИН КОРЕНЬ на каком-то отрезке, то скорее всего работает этот способ решения. Привет, меня зовут Евгений Пифагор, и я готовлю к ЕГЭ по математике более 10 лет. Да, можно решать ещё и графически или ещё какими-то способами, но способ из этого видео самый универсальный. Если не получается понять это видео, то возможно сначала стоит посмотреть видео про то, что такое параметр: https://www.youtube.com/watch?v=_Fn0d4gScDE 👍 ССЫЛКИ: Материалы к видео: https://vk.com/shkolapifagora?w=wall-40691695_54443 Подпишитесь: https://www.youtube.com/c/pifagor1?sub_confirmation=1 VK группа: https://vk.com/shkolapifagora Видеокурсы: https://vk.com/market-40691695 Insta: https://www.facebook.com/unsupportedbrowser 🔥 ТАЙМКОДЫ: 00:00 — Вступление 05:12 — Задача #1 19:06 — Задача #2 33:03 — Задача #3 48:19 — Задача #4 01:09:15 — Задача #5 01:31:55 — Задача #6 01:54:40 — Задача #7 02:19:10 — Задача #8 02:43:56 — Задача #9 03:10:02 — Задача #10 03:22:51 — Задача #11 03:37:50 — Задача #12 03:54:07 — Задача #13 04:15:31 — Задача
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