Much simpler: let u=x-7. Then Solve (u+1)^4 + (u-1)^4 - 16 =0. Expand the two quartic term using the binomial theorem. Terms with an odd power of u will cancel. The remaining equation is 2u^4+12u^2 -14=0, divide by 2 ==> u^4+6u^2-7=0. From formula for quadratic equation u^2=1 or u^2=-7. $ solutions for u: -1, +1, -i√7 , +i√7, add 7 to obtain the four solutions for x. Using the right substitution u=x-7 instead of u=x-2 simplifies the computational steps enormously, since odd term of u cancel out, and all that remains to do is to solve a quadratic equation in u^2.

let u=x-7 x-8=u-1 x-6=u+1 will be easier to calculate

I like that you don’t skip any steps. Makes it easier to follow. Thanks 👍

2 ответа очевидны и без решения - это Х=6,и Х,= 8. Одна скобка обнуляется,, а вторая становится = 2, или -2, что в 4-ой степени будет 16

Nice. To solve without polynomial division, after reaching the cubic equation, you could rewrite the equation as u³-8-4(u²-3u+2)=0 which is equal to (u-2)(u²+2u+4)-4(u-2)(u-1)=0 (u-2)(u²-2u+8)=0 u = 2 => x = 8.

... Good day to you math friend, You are always so neat and well organized in your work, which makes watching a piece of cake and delightful ... I just have one short remark; the 2 non Real solutions are not Imaginary but Complex (conjugate) solutions ... standard form z = a + b * i , when a = 0, then z = b * i , and this is an Imaginary number but also still a Complex number, so the set of Imaginary numbers is a subset of the set of Complex numbers ... as always watched your clear presentation with much pleasure math friend ... best regards, Jan-W

7:32 i've lost, why did you make the substitution u=2 I doesn't have any sense to me... 😢

Методом подбора целых чисел находим, что x = 6; x = 8 Делим (x-6)⁴+(x-8)⁴-16 на (x-6) и на (x-8) Получаем квадратное уравнение 2x²-28x+112=0 Делим обе части уравнения на 2 и находим его корни x = 7 - isqrt(7); x = 7 + isqrt(7) Ответ: 6; 7-isqrt(7); 7+isqrt(7); 8

@user-lf1bm8fm2g

6 ай бұрын

@piece_o_shi зачем эти корни с комплексными числами.

@2106522

6 ай бұрын

​@@user-lf1bm8fm2g, потому что уравнение 4-й степени, а значит, у него 4 корня, и их все надо показать.

@user-lf1bm8fm2g

6 ай бұрын

@@2106522 а что эти комплексные корни дают?

you could solve that easier if you take: y=x-7

6, 8 (as real roots).

Simple solution 1. Rewriting given equation as a^4 + b^4 = 2^4 + 0 Hence a^4 = 0 and b^4 = 0 Or vice a versa Therefore, x-6 = 2 and x -8= 0 X= 6 or 8

@kimobrien.

4 ай бұрын

You didn't find the two complex solutions.

Just from looking at it. 6 and 8 are both valid solutions.. and to find another 2 solutions, we would have to calculate the reminig quadratic equation to see, if it has real solutions or only complex ones... i'm to lazy to calculate it in the fly so i'm guessing its presented in the video...

@kimobrien.

4 ай бұрын

You could have expanded out the whole polynomial and then divided by (X-6) and (X-8) to get a quadratic that would be solvable with the quadratic formula giving you complex roots.

Solution by insight 0+16=16 x=6, 16+0=16 x=8

X=8 , X=6

7x4x-1??? You forgot the brackets 7x4x(-1)

6

Х=8

@user-lf1bm8fm2g

5 ай бұрын

Х1=8; Х2=6

Х= 6 зачем все эти преобразования, поставте тупо Х=6.

@piece_o_shi...

6 ай бұрын

X = 6; 8

@user-nc2ts9wi2d

6 ай бұрын

И Х= 8

@piece_o_shi...

6 ай бұрын

@@user-nc2ts9wi2d x є R