find the real or imaginary solutions of each equation by factoring
x^4-8x^2=-16

Question

find the real or imaginary solutions of each equation by factoring
x^4-8x^2=-16

paounn · Answer

write the expression as $$x^4 -8x^2 +16 =0$$
Recognize in it the square of $$(x^2 -4)$$
Your equation becomes $$[(x+2)(x-2)]^2 = 0 $$, whoose soultions should be evident.

anonymous · Answer

is the solutions (2i,2i,-2i,-2i)

anonymous · Answer

is that right

paounn · Answer

$$(2i)^4 -8 (2i)^2 = -16 \rightarrow 16*1 - 8 * 4 *(-1) = -16 \rightarrow 16 + 32 = -16 \rightarrow 48 = -16$$
which is evidently false. Same goes for -2i

anonymous · Answer

so wats the answer im completely confused

paounn · Answer

From my first post, remembering that $$(a*b)^2 = a*b*a*b$$$$(x+2)(x-2)(x+2)(x-2)=0$$
so you get that $$x=-2, x=2, x=-2, x=2$$

anonymous · Answer

i figured it our actually before u wrote back but thank u