Question

solve for x, guys please be quick

Answer 1

Hence, here also applying the same formula

==> (x^4 - 4x^3 + 6x^2 - 4x + 1) + (x^4 - 20x^3 + 150x^2 - 500x + 625) = 82

==> 2x^4 - 24x^3 + 156x^2 - 504x + 544 = 0
==> x^4 - 12x^3 + 78x^2 - 252x + 272 = 0.

By long/synthetic division,
x^4 - 12x^3 + 78x^2 - 252x + 272 
= (x - 4) (x^3 - 8x^2 + 46x - 68)
= (x - 4)(x - 2)(x^2 - 6x + 34).

So, we have
(x - 4)(x - 2)(x^2 - 6x + 34) = 0
==> x = 4, 2, or x = 3 ± 5i

 

So, the real solutions are x=4 and x=2

Answer 2

We know that (a-b)^4 = a^4- 4a^3b + 6a^2b^2 - 4ab^3 + b^4   ---- (1)

On substituting the values in (1), we get

(x-1)^4 = (x^4 - 4x^3 + 6x^2 - 4x + 1).

(x-5)^4 = (x^4 - 20x^3 + 150x^2 - 500x + 625)

(x-1)^4 + (x-5)^4 = 82

2x^4-24x^3+156x^2-504x+626 = 82

2x^4 - 24x^3 + 156x^2 - 504x + 544 = 0

2(x^4 - 12x^3 + 78x^2 - 252x + 272) = 0

2(x^4 - 12x^3 + 78x^2 - 252x + 272) = 0

   (x-2)(x-4)(x^2-6x+34) = 0.

x^2 - 6x + 34 can be written as x^2 - 6x + 9 = -25

                                                    (x-3)^2 = -25

                                                    (x-3) = root -25

                                                    x = 3 + root 25.

x = 2,x = 4, x = 3+5i,x = 3-5i..


Hope this helps!