Question 707551: x^4-8x^2+16=0
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! x^4-8x^2+16=0
we can factor the above as:
(x^2-4)(x^2-4) = 0
(x^2-4)^2 = 0
.
But:
(x^2-4)
can be rewritten as
(x^2-2^2) "difference of squares" (special case)
can itself be factored as:
(x-2)(x+2)
.
That means:
(x^2-4)^2 = 0
is the same as:
[(x-2)(x+2)]^2 = 0
(x-2)(x+2)(x-2)(x+2) = 0
(x-2)^2(x+2)^2 = 0 (this is the factored form)
.
Solution:
x = {-2,2}
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