This is a quadratic equation in x^2.

Put u = x^2 ………….(1)

u^2 – 20u + 64 = 0

(u – 16)(u – 4) = 0

u = 16 or u = 4.

Hence from (1):

x = +/-sqrt(u) = +/-4 or +/- 2.

[x^2-4][x^2-16]. Now, both of these factors can be further factorised, as they are both the difference of two squares,

x^2-4 = [x+2][x-2], and x^2-16 = [x+4][x-4], so the complete factorisation is:

[x+2][x-2][x+4][x-4 = 0, giving as solutions,

x=-2, x=2, x=-4, and x=4. Hope this helps, Twiggy.

find 2 numbers that multiply to 64 and add to -20 which is -16 and -4

then you find what x should be to make the equation true wich is x^2

the equation is now

(x^2-16)(x^2-4)=0

to finnish, you take one equation at a time so

x^2-16 turns to x^2=16 which simplifies to x= + or – 4

do the same with x^2-4 making it x=+ or – 2

so x=+-4 or +-2

4 – 20, 2 + 64.

The answer is -16 and 66. In other words, -82.

(x-4)(x+4)(x-2)(x+2)=0

x=4 V x= -4 V x=2 V x= -2

It is the best answer.

The previous answers are false