Could anyone else just see that x=0.5 because that is the only number for which x-1=-x, which must be true as 6 is an even power so the positive and negative version of x⁶ will be equal

@jonahansen

4 ай бұрын

Yes - and the two slim upward-concave curves meet at (1/2,1/64) when graphed.

@theunholybanana4745

4 ай бұрын

Yeah it was really obvious

@kai-speedrunningandstuff3292

4 ай бұрын

i did that

@Professor_Sargeant_JAMS

4 ай бұрын

To the original commenter, this inspired me to post a response video with another option.

@capaatmp1730

4 ай бұрын

Yes we can just guess this like that

|x|=|x-1| x=1-x x=0.5

@TFclife

4 ай бұрын

Hey got 0, just took the integral of both sides and realized that x^7 also = (x-1)^7, realized then made general rule so (x ^1)=( x-1)^1 Hence x=x-1 Multiplied both sides by x x²=x²- x 0= -x 0= x But if x=0 (X=x-1) Equation is contradictory. Ie:1= 0 Equation is not a subset of R

@generalflopper9697

4 ай бұрын

Wrong, x = -1 + x, not 1 - x

@zxl1440

4 ай бұрын

@@generalflopper9697 -1+x=x-1

@TFclife

4 ай бұрын

@@zxl1440 But that's not what you wrote, regardless look at my solution

@cucuo-wo5sv

4 ай бұрын

@@generalflopper9697dumbass

( 1 - 1/x) ^6 = 1 1 - 1/x = exp( 2 π i r /6) , r = 1, 2, 3, -1, -2, -3 1/x = 1 - exp ( i theta) x = 1/( 1 - cos ( theta) - i sin ( theta))

1/2 works as a solution. This also admits complex solutions (4 of them) with angles+-pi/3 and +-pi/6.

The reason that x has 5 values is not because “6-1=5”, but because if you expand the original equation on the right, both sides will have x^6. So in reality, the highest power of x is 5, which means there are 5 roots. Sorry if I have used the wrong terminologies, English is not my first language.

Since r^2 >= 0 for each r from R, you can easily take 3rd root. Then you get x^2=(x-1)^2 (here we can be sure both sides are positive) and the computation is much easier.

@leofigoboh1611

4 ай бұрын

But you lose many solutions that way

Alternate approach: square root both sides getting x³ = ±(x-1)³. The positive root becomes a quadratic, which is easy to deal with, and (2x-1) is a factor of the negative root, making another quadratic. It largely works out the same, just without the difference of squares step.

@Professor_Sargeant_JAMS

4 ай бұрын

Yet another approach is shown in a response video I posted.

Fantastic

Another way to do it is divide by x^6 (x=0 is not a root any way, so we can do it safely): [(x-1)/x ]^6 = 1 Then, we take 6-th root of 1 over the complex field and find 6 values exp(pi/3*i*n) for n=1..6: z1=1/2+i√3/2, z2=-1/2+i√3/2, z3=-1, z4=-1/2-i√3/2, z5=1/2-i√3/2, z6=1 Then, (x-1)/x =z x=1/(1-z). Substitute all of the branches z1 .. z6, exclude the values that are infeasible that don't yield any roots of x (such as z=1), and thus obtain the roots. I think it's a little bit faster.

take sixth root. x = +-(x - 1). since focusing on the positive part of the equation yields no solution, set x = - x + 1. this means 2x = 1, or that x = 1/2.

x=1/2; 10 secounds of thinking,.. For a given number and the result of subtraction to an even power to be the same result - it must be negative on one side and positive on the other, which will give a positive result in both cases, so it comes down to the equation: -x=x-1 or x =-(x+1), which is the same, because it comes down to -2x=-1 or 2x=1; x=1/2

@mahimabagri07

4 ай бұрын

Arre aryabhatha ke lost child

@hepdepaddel

4 ай бұрын

Yes, I found that too - and for the other 4 possible solutions you would have to spend a bit more time on applying real math to the problem. Or watch the video 😉

Instant answer X=1/2. 1/2 - 1 = -1/2 and the even exponent eliminates the negative sign.

From the point of arriving at (a+b)(a-b) , we can equate each of the factors to 0 on the basis that at least one factor =0 and find the value of x , so in this case (x3 +(x-1)3)=0 would have given 1/2 Value.

take sixth root of both sides. since it is a non-principle even root, you have to account for it by making both sides have plus-or-minus signs. this means +-x = +-(x - 1). since setting both sides to have the same sign will yield no solution, we can set x = -(x - 1). this means that x = -x + 1, or that 2x = 1, or that x = 1/2. that does it at least for real solutions- obviously with complex ones you have to go into the nitty-gritty of it all, but it’s relatively simple at least to find the one real solution that this equation has.

Raising both sides to an even power. Hence the result will be positive, regardless of the sign of either side. So what value of X would satisfy abs(X) = abs(X-1) ? And what comes to mind is 0.5

Finding the real solution only is really easy if you do it right. When you raise a number to an even degree you will get the same thing as you would if you raise the negative version of that number to the same even degree, so all you have to do is figure out the solution to x-1=-x

X = .5, the exponent is even so the base can be negative or positive and they will both have a positive result, from there you just need two values that are the same absolute value and are both exactly 1 from each other, which is -.5 and +.5

@heco.

4 ай бұрын

-0.5 - 1 would be -1.5 which wouldn't work because (0.5)^2 is not equal to (-1.5)^6

@Kevinjoy2305

4 ай бұрын

​@@heco. How high r u

@kimdokjasdream9158

4 ай бұрын

There is 1 real and 4 imaginary solutions

@jshrekmaster

4 ай бұрын

​​@@heco.When you root the equation to the power of 6,the equation becomes +-x=+-(x-1). Theres like 4 combinations here and you can try evaluating each expression. But 2 of the combinations, +x=-(x-1) or -x=(x-1), has an answer. One of these combinations is the one they're referring to. It doesn't have to be the case that every combination must be true for it to be a valid answer, since they're all different expressions at the end of the day

@heco.

4 ай бұрын

@@Kevinjoy2305 higher than you could ever be.

It is 5th polynomial because if open the brackets you will get x^6-……. which will be cancelled with the lift hand x^6 So five roots are valid

@gregorypeace6162

6 ай бұрын

yes, it is fifth in fact.

@Astrobrant2

5 ай бұрын

Ah, I was about to ask why there were only five roots. Thanks! That explains it.

@hybridaccounts

4 ай бұрын

I don't agree. Infinite should also be a value so it remains a polynomial of degree 6

@OMGclueless

4 ай бұрын

At infinity the polynomial diverges, infinity is not a number you can compute a polynomial of, and it certainly is not equal to any other number.

Thin of x and x-1 as 2 complex numbers. Since they're equal in length and imaginary part, they're equal in real part, meaning that re(x) is 0.5. Let theta be the angle of x, and that of x-1 as pi-theta. Thus, the angle difference 6*(pi-2*theta) is an integer multiple of 2*pi, meaning that 12*theta is such. The theta can be within 6 values. Given the positive real part, the theta is 0, 1/3, 5/3 times of pi.The solution in rectangular form is 1/2, 1/2+sqrt(3)/2*i, 1/2_sqrt(3)/2*i

@sfx8979

4 ай бұрын

Plus when theta is 1/6*pi or 11/6*pi, the 2 conjugate roots are 1/2 + sqrt(3)/6*i and 1/2-sqrt(3)/6*i

To get the real answer you could just make it so ((x)^2)^3=((x-1)^2)^3 and we take the 3rd root and we will gey x^2=(x-1)^2 and the solution is x=0.5

As soon as I saw it, I thought it had to be 0.5. The fact that there's so many solutions is crazy when you consider complex numbers.

@michaellopez1583

4 ай бұрын

Same. Took about a second and a half. Then I was like, "What's going to take 12 minutes?" I should have imagined....

@NLGeebee

4 ай бұрын

Equations with x^6 mostly have 6 solutions. So with x=1/2 out of the way, there should be 5 complex solutions remaining. The only part that has no solution is 1=0. The fun part is that the (complex) solutions are not on a circle, but on a (vertical) line.

А не проще было сразу степени сократить. Получили бы Х=Х-1 |Х|=|Х-1| 2Х=|-1| 2Х=1 Х=1/2 Х=0.5 Ну или хотя-бы на ^3 сократил бы, чтобы не делать бессмысленных действий

@user-mt9rn6bp2i

9 ай бұрын

Иррациональные корни тоже считать надо.

@Demotivator.

9 ай бұрын

@@user-mt9rn6bp2i Я не математик, я инженер. Просто вспомнил как решать можно... Эти видосы в рекомендациях хорошая зарядка для мозгов..

@user-lf1bm8fm2g

7 ай бұрын

Зачем искать эти иррациональные корни, краткость сестра таланта.

@Kto-to-drugoy

7 ай бұрын

​​@@user-lf1bm8fm2g Затем, что решить уравнение означает найти все его корни (а не только действительные или рациональные) или убедиться, что их нет.

@freehck

5 ай бұрын

И любой нормальный человек так бы и сделал, поскольку обычно в задачах производится поиск именно вещественных корней. Автору стоило сразу сказать, что ищутся комплексные корни. В этом случае степени сокращать нельзя, поскольку на комплексной плоскости умножение -- это не только изменение модуля, но в том числе и поворот.

The real solution is obvious because of the +/- symmetry in the even power. The solution pair {2,3} and the solution pair {4,5} can be written in the same form, over a denominator of two, and should be written in the same form, because the powers of three are one half in the first pair and minus one half in the second, another symmetry which may lead to insight.

@Professor_Sargeant_JAMS

4 ай бұрын

There's even more symmetry here. See my response video with another option. For what hat other similar questions would the approach I showed work?

Wir haben hier zwei identische achsensymmetrische Graphen, der Scheitel der ersten Funktion liegt im Ursprung, der Scheitel der zweiten ist um eine Einheit nach rechts verschoben. Daher liegt der Schnittpunkt der beiden Graphen in der Mitte der beiden Scheitel, daher bei x=0.5 Eine Rechnung ist nicht nötig.

@user-lf1bm8fm2g

7 ай бұрын

Можно сразу в исходное выражение сразу пдставить Х=0,5 и не гонять это выражение так долго.

@Professor_Sargeant_JAMS

4 ай бұрын

See more about the symmetry in a response video I made.

If we substitute the solution results into the equation, is equality satisfied?

Solutions are x = 1/2, x = 1/2 ± (i sqrt(3)/2) and x = 1/2 ± (i sqrt(3)/6)

Good

@learncommunolizer

8 ай бұрын

Thanks

If it was mentioned that the solution cam be an integer then figuring out thay x is less than 1 and greater thab 0, that is 1/2 is a 5 second job, trick is to find out the rest 4 :)

@user-lf1bm8fm2g

7 ай бұрын

А зачем искать остальные 4 решения, так как комплексное число имеет не красивый вид.

@host_north

7 ай бұрын

@@user-lf1bm8fm2g но это не значит, что этих решений нет.

@MightyBiffer

5 ай бұрын

1/2 is not an integer

Before any calcul, x= 1/2 is solution others are imaginary You can't put negative number under root sign Use 1/2 exposant

Thought about it for between 5 and 10 seconds. X has to be one half.

@baliyahsonlittle2016

4 ай бұрын

Yeah the hard part is finding the imaginary solutions

Abi bu ne yaaa😂gözünü seveyim..neye yariyor bu😂eline sağlık...

No manipulation needed. X = 0.5 just by looking at it. (To the extent that there may be imaginary answers, that would require a bit more work.)

why cant you just use pascals triangle? Like put the equation as x^6 = (x-1)^6 and then use pascals triangle idk

@Professor_Sargeant_JAMS

4 ай бұрын

I used that in a response video. Check it out?

super

@learncommunolizer

9 ай бұрын

Thank you🙏❤️🙏

you overcomplicated it, here’s the short way remember a negative number to the power of an even exponent will end up positive, so when you do a root with a even index, then u need +- x^6 = (x-1)^6 +-x = +-(x-1) x = +-(x-1) x = -(x-1) x = -x+1 2x = 1 x = 1/2 and to check x^6=(x-1)^6 (1/2)^6=(-1/2)^6

I solved it in under a minute by trying :)

I haven't seen the video but just looking at the problem 1/2 looks as an obvious solution since being the exponent even we will end up with the same results on both sides, whatever that result is.

@hansvangiessen8395

7 ай бұрын

And the imaginary solutions?

@MightyBiffer

5 ай бұрын

@@hansvangiessen8395 The first two imaginary solutions are somewhat obvious. There is a simple identity that is often taught when dealing with complex numbers. The third root of -1 is (1/2, Root(3)/2) and (1/2, -Root(3)/2) The third root of 1 is (-1/2, Root(3)/2) and (=1/2, -Root(3)/2) Squaring -1 and 1 yields 1, so the sixth power of these terms is 1. And they come in values that differ by 1.

Its nice and easy because: X⁶=(x-1)⁶ Thats why because: If x⁶=(x-1)⁶ then x=(x-1) so yeah The answer is x=(x-1) Same as the last one that is: x⁶=(x-1)⁶ So x=(x-1) Idk if its right but it needed to be like the same because: x⁶=(x-1)⁶ x⁵=(x-1)⁵ Then the others are: (x-1)⁴ and (x-1)³ and (x-1)² So x=(x-1)

You must use chain brackets or big brackets

This question like a IQ question rather than maths question!

1/2

❤😂

It must be 1/2. Easy to see if you draw a graph.

X=0,5 just popped in my mind

remove 6 from both sides x = 1-x combine like terms 2x=1 divide both sides by 2 x=0.5 or 1/2

@MightyBiffer

5 ай бұрын

First mistake: the equation was x^6 = (x-1)^6 so in your method that would reduce to x=x-1 not x=1-x. Second mistake: when removing the radical of an even power the rule is a^n = b^n rduces to |a| = |b|. So in this exqmple the reduction is |x| = |x-1| then solve for both x=x-1 and x=-(x-1) This first has no solution and the second x=-(x-1) can be solved for x = 1/2

@eyese

5 ай бұрын

@@MightyBiffer I don’t understand a thing you just said

@eyese

5 ай бұрын

@@MightyBiffer Can u put it simpler?

@MightyBiffer

5 ай бұрын

@@eyese Ok. In your coment yous said that you solved the equation X^6 = (X-1)^6 by removing the power of 6 from both sides. This left you with X = 1 - X which you then solved and got the correct answer. However you made two mistakes. One mistake is that you removed the power of 6 from the right hand of the equation and turned (X-1)^6 into 1-X. The actual phrase would have been X-1. If you had done this your equation would have been X=1-X which has no solution. The other mistake you made was just removing the power. When removing an even number power, you have to use the absolute function. For example, 1^6 = (-1)^6. this is because 1 raised to the sixth power is 1 and -1 raised to the sixth power is also 1. (-1 x -1 x -1 x -1 x -1 x -1 has an even number of negative signs so they all cancel each other out). So the proper reduction of the equation 1^6 = (-1)^6 would not be 1 = -1 but would be |1| = |-1|. At the same time reducing the equation with a higher-level power has the side effect of removing the number of possible solutions. In this example, the removal of the sixth power removed the ability for us to find the 4 additional complex solutions.

my brain after 2 seconds: 0.5

X=1/2 Only possible solution. Here’s why: 1) x=/=(x-1) 2) 6 is even power So => |x|=|x-1|, or |1-x|=|x| Then |2x|=1, x=|1/2| -1/2 is obviously wrong, thus the right answer is x=1/2

@probropalzlive6961

4 ай бұрын

Neglecting complex roots?

Ok! I’m waiting to see the 6th root! Something isn’t right unless you claim two real solutions on top of each other at x=1/2

@fredschneider7475

6 ай бұрын

The x^6's cancel out. So, he was solving a quintic equation, exactly 5 roots.

@H.S909

5 ай бұрын

Consider the trivial equation. x^6 = x^6 ...(1) Any complex number is a solution to this equation. Now the given equation in question is written as x^6 - x^6 + P(x) = 0, ...(2) where P(x) = (x - 1)^6 - x^6. But the author of this video has shown that there are only five roots, which is quite true. How do we account for the sixth powet if we want to? Well, we may put that we are here considering the intersection of the zero set of the equation (1) and of (2). How does it sound?

Мое решение x⁶=(x-1)⁶. т.к. 6 - четная степень, то по свойсту степеней |x| = |x-1| Здесь уже не сложно заметить (я в 8 классе, так что не могу пока что математически это доказать ) Что x = 0,5. Вот так вот. 😅

@user-eo3lr7vz6i

5 ай бұрын

Я закончила 8 класс более 20 лет назад и решила за несколько секунд. Ну а компле́ксные числа (i) - это уже извращение

@user-dx8sv4ph4t

5 ай бұрын

При извлечении корня теряются решения.

Should there be 6 roots counting multiplicities? There are 5 distinct roots, but where is the duplicate root in his factoring? I only see 5 factors.

@KSL_Rhythm

5 ай бұрын

It looks like a sixth-order equation, but it's actually a fifth-order equation, so it's correct that five distinct roots exist.

The explanation about why we have 5 solutions is wrong. If the equation has x power of n, we expect n solutions NOT (n-1) solution. The reason we have only 5 solutions here is the equation is indeed x power of 5 because x^6 on LHS and X^6 on RHS are cancels out to each other. We can easily tell the coefficient of x^6 on RHS is 1 without expanding (x-1)^6.

@Hanible

8 ай бұрын

Nice! You're absolutely right

@Hanible

8 ай бұрын

By the way, weren't there an easier way to solve this by writing the complex e form from the start? Thanks in advance

@michellepopkov940

6 ай бұрын

I agree. I can only resolve this dilemma by positing that there are two coincident roots at 1/2.

Actually the equation is of fifth degree not sixth, therefore 5 solutions.

@is7728

9 ай бұрын

1/2 is a double root

@wolfbirk8295

9 ай бұрын

You are right because x^6 - x^6 = 0...

@salvatorecosta875

9 ай бұрын

@@is7728 1/2 is a simple root

@owengalendinoalfatah9533

7 ай бұрын

Why?

@user-cg5xv4zz2b

6 ай бұрын

6th. first degree x+c, second x^2 (...) +c, ... sixth x^6 .. +c ( and left side is sixth - since x^6. Am I wrong?

Can you give a practical use of this formula.?

@sytherplayz

5 ай бұрын

this isnt a formula, its a question. And Formulas in mathematics have various uses in many fields like in engineering and programming

x = infinity ?

Here is the chat GPT answer. Almost the same but not totally (why ?) : ChatGPT is talking : (after taking the sixth roots both side...) It appears that we obtained an inconsistent equation (1 = 0 ), which means there are no real solutions to the original equation X^6 = (X-1)^6 However, it's important to note that the equation X^6 = (X-1)^6 has complex solutions. These complex solutions involve roots of unity, and they are not real numbers. If you're interested in complex solutions, you can use the fact that (X-1)^6 = X^6 when X is a sixth root of unity. The complex sixth roots of unity are given by: Xk = e ^ 2πik/6 ​ where k = 0, 1, 2, 3, 4, 5 These complex solutions satisfy X^6 = (X-1)^6 . .

@shmuelzehavi4940

6 ай бұрын

No, you're wrong. x_k = e^(2πik/6) satisfies the equation: x^6 = (x-1)^6 only for: k=1 or k=5.

“By inspection” it is 1/2. This solution is a waste of dry erase marker.

Hello you have to give the domain. If the domain is R, there is only one solution.

There are supposed to be 6 solutions right?

why not +, - infinity?

0.5

X=±(X-1)

Logically i came to the answer in 30 seconds

@hansvangiessen8395

7 ай бұрын

Also imaginary answers???

Vietnamese student : shift solve🗿

@Professor_Sargeant_JAMS

4 ай бұрын

Shift as in use substitution?

Way way too complicated. Here is how to destroy this poor little thing. First of all we notice that X=0 is not solution of the equation and since x=0 x^6=0 we can divide everything by x^6. We obtain: (1-1/x)^6=1 Let's call z=1-1/x. Our equation is z^6=1. Let's solve this in C and write z=r.exp(i.theta). We can write our equation as follow: [r.exp(i.theta)]^6=exp(2.i.pi) or: r^6.exp(6.i.theta)=exp(2.i.pi) Which forces r=1 and theta=k.pi/3 where k is a relative integer. The possibles values are then 1, -1, j, -j, j² and -j² where j=exp(2.i.p/3). 1-1/x=1 has no solution. 1-1/x=-1 implies that x=1/2. 1-1/x=-j => 1/x=1+j but since j and j² are solutions of z²+z+1=0 that means that 1+j=-j²=exp(i.pi.3). Then x=exp(-i.pi/3)=-j 1-1/x=-j² => x=-j² for the same reasons. 1-1/x=j => x=1/(1-j) and this time there is no trick so let's get into it... 1/(1-j)=1/[3/2-i.sqrt(3)/2]=2/(3-i.sqrt(3)]=2(3+i.sqrt(3))/(9+3)=1/2+i.sqrt(3)/6. And finally for 1-1/x=j² we replace i.sqrt(3)/2 by its opposite and we find 1/2-i.sqrt(3)/6.

@percy9228

8 ай бұрын

thanks for your solution. used more advance stuff but you nailed it!

@user-lf1bm8fm2g

7 ай бұрын

Зачем эти комплексные числа , это уродует решение, однозначно Х= 0.5.

@taflo1981

6 ай бұрын

A similar solution would be as follows and uses mostly geometric arguments in the complex plane. First observe that x⁶=(x-1)⁶ means that in the complex plane, x has the same distance from 0 and from 1, so its real part is 1/2. If we write x in polar coordinates as x=r*exp(i*a) (with a strictly between -π/2 and π/2, as the real part of x is positive), then x-1=r*exp(i*(π-a)) by symmetry. Now x⁶=(x-1)⁶ is equivalent to 6a and 6(π-a) differing by a multiple of 2π, which in turn is equivalent to a being a multiple of π/6. So the only possible values for a are 0, ±π/6 and ±π/3. From this, we get the values for x via x=1/2+i*tan(a)/2.

please be more precise from the beginning. Are we looking for complex or real solutions ?

@Professor_Sargeant_JAMS

4 ай бұрын

"A variable by any other name would smell as sweet," to misquote Shakespeare.

if i want to build a home with flor 3*i m^2 and walls sqr3*i meters, how can i build it? and who can live in this home, ant, human or elephant?😵‍💫

@user-lf1bm8fm2g

7 ай бұрын

Согласен с вами не на 100 а даже на все 300%.

@Elias___

4 ай бұрын

Imaginary ant, ant i, and imaginary human, human i only complex entities are allowed sadly.

X= 0,5 est trivial

"х" равен "∞" (бесконечности)

Bro why not just 1/2

There should actually be (up to) 6 roots for a 6th order polynomial. During the mass expansion and cancellation, one of the terms ended up being the constant (1), which promptly got ignored in this video. Had this factor had an x term that would be your sixth root. Otherwise, this implies that one of the solutions found is in fact a double root. Likely the real one but i dunno.

@sergeybrener251

5 ай бұрын

that's 5th order, open up your eyes, dude.

@probropalzlive6961

4 ай бұрын

Constants aren't solutions?? x²-(x-1)²=0. How many solutions? Only 1, the solution is x=0.5 I will define a function P_n(x)=x¹⁰+nx+1. P_10(x)-P_9(x)+1=0. How many solutions? Just one. x=-1.

This question is wrong , x can be infinite and using modulus is innapropriate since its not mentioned to just figure out the magnitude of a vector . Its wrong and unauthentic and he is making a paradox by solving in his way which is usually a common mistake for some wrong equations .

@Professor_Sargeant_JAMS

4 ай бұрын

Sorry, no. Check each candidate solution. Which do you claim doesn't make the two sides equal?

X=1/2

Why not 6 root why 5 root? İs 1/2 a double root? 360/6=60=pi/3 The solutions are sin(teta)×r+cos(teta)×i

X = 1 works

Обычно, если не оговаривается явно, x в математических уравнениях вещественное число. Для уравнений с комплексными числами используют z.

@f0039

6 ай бұрын

Может у них условности такие. Я вот не понимаю, почему он использует простые дроби, да ещё и зачем-то их расчленяет, вместо того, чтоб под общим знаменателем представить.

@ttxttx1177

6 ай бұрын

А можно по-русски ответить? С аглицким не дружу, да и не математик. Вот подставил я в калькулятор пару ответов (кроме первого) - равенство не выходит. Это я накосячил или где-то все же что-то не так?

@vdarasun

6 ай бұрын

@@ttxttx1177 , а я не знаю калькулятора для использования с комплексными числами. Тут ручками на бумажке надо проверять, как я думаю.

@canr772

6 ай бұрын

x в математических уравнениях - неизвестное число, и никого ебать не должно действительное или мнимое кли комплексное, самое главное что это число.

@Professor_Sargeant_JAMS

4 ай бұрын

​@@canr772"A variable by any other name would smell as sweet," to misquote Shakespeare.

X=0.5

Ez x=-0,5

My questions,how x^6=x^6-1^6 x^6 - x^6=-1^6 0=1 So Laughs😂😂

shouldn't x=1/2 also be a solution?

@dr.blockcraft6633

4 ай бұрын

7:10

@sayeshwarbhatia2132

4 ай бұрын

oh sorry, i missed that. thanks for letting me know​@@dr.blockcraft6633

Это решается в уме однако

@AJFantocii

5 ай бұрын

В решении слышно много раз - майонез)) Первый ответ все сразу увидели, не считая. А остальные ответы это и есть майонез!)))

Why can't you just take the sixth root off the bat to produce x = x-1?

@vinceturner3863

5 ай бұрын

Thence 0 = -1?

@probropalzlive6961

4 ай бұрын

x=abs(x-1).

1/2 its fkin easy

X=| X-1|

x=0.5

The real solution: 0.5

to start.... x^6 begs there's 6 solutions. To simply take the 6th root of both sides would lose 5 of them... 6logx = 6log(x-1) log(x) = log(x-1)

@Professor_Sargeant_JAMS

4 ай бұрын

Domain issue problem.

А где шестой корень?

@joseluissuarezmartinez7471

7 ай бұрын

No hay sexta raíz. La ecuación es de quinto grado porque si elevas a la sexta el segundo miembro los x^6 desaparecen.

Here is my though - someone explain to me if I am wrong. This is an impossible solution. Here is why. take the 6th root of both sides and you get x = x-1 - not possible. ---- ok, you can stop replying to this post - I got it. I understand. and thank you for being so gererous to reply

@user-xl1ig1bn6i

7 ай бұрын

When confronted with such a situation you should you use a different alternative method. Remember the polynomial is of order six, so you should expect to find at most 6 solutions. Always avoid the temptation to take the 6th root on both sides

@bvwalker1

7 ай бұрын

Well, nothing wrong with taking 6th root of both sides if your goal is only to find the real roots.

@user-ct2gg3lm8m

7 ай бұрын

Степень четная, 0.5^6=(-0.5)^6, так что 0.5 это действительный корень. Если мы извлекаем корень, то так как степень чётная получается не X=X-1, а |X|=|X-1|

@UltraDanicraft

7 ай бұрын

Amigo al tomar raíz sexta, debes poner valor absoluto, de ahí se deduce que x=1-x y al final obtenemos la única raíz real, que es 1/2. Un saludo.

@michellepopkov940

7 ай бұрын

Yeah. No real solution.

A simple vista esta ecuación nunca cumplirá la igualdad, al menos para raíces reales. La resolución de este ejemlo es de solo milisegundos, o quien me dice lo contrario ??

@Professor_Sargeant_JAMS

4 ай бұрын

The creator of the video himself shows a valid real solution.

@ivocosmi5303

4 ай бұрын

​@@Professor_Sargeant_JAMSverdad, la ecuación cumple para X=1/2, parecía que no cumplía para ningún número real 😮😮 , gracias por la aclaración 👍👍

Решил за 1 секунду методом подбора: Ответы: 0,5 и - 0,5, если бы степень была нечётной то наверное я бы не решил

Почему ответов 5, если степень 6🤔

It's a faulty question

Я понимаю, что ответ 0,5 , но как решать не знаю, только в 7 классе брух

이거 풀어야 하나요..? 0.5라는 거 그냥 알 수 있는 거 아닌가요. x=0.5선대칭이잖아요

@Well_And

5 ай бұрын

5차 다항식은 해가 5개입니다. 즉 허근이 존재한다는 것이므로 허근도 구해줘야 해요

@Professor_Sargeant_JAMS

4 ай бұрын

See more about the symmetry in the response video I posted.

not correct . a™=b™ that mean a=b . x=x_1 that mean 0=1 error result .

0

This is a ridiculously long way show something that's obvious from the start. Absolute value of x and x minus one has to be the same. Duh

Bro took it too lonv

I did it with a different method kzread.info/dash/bejne/lHxpsKeTiaSdh5M.html

Next time can the background music, please...