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find the real or imaginary solutions of each equation by factoring x^4-8x^2=-16
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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.
is the solutions (2i,2i,-2i,-2i)
is that right
\[(2i)^4 -8 (2i)^2 = -16 \rightarrow 16*1 - 8 * 4 *(-1) = -16 \rightarrow 16 + 32 = -16 \rightarrow 48 = -16\] which is evidently false. Same goes for -2i
so wats the answer im completely confused
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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 \[x=-2, x=2, x=-2, x=2\]
i figured it our actually before u wrote back but thank u
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