How do you find the real and imaginary solutions of $x^{4} - 8 x^{2} = - 16$ $x^4-8x^2=-16$ ?

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Answer 1 · 2016-11-28T00:12:50

$x = \pm 2$ $x=+-2$ , each with multiplicity $2$ $2$

Given:

$x^{4} - 8 x^{2} = - 16$ $x^4-8x^2=-16$

Add $16$ $16$ to both sides of the equation and transpose to get:

$0 = x^{4} - 8 x^{2} + 16 = {(x^{2} - 4)}^{2} = {(x - 2)}^{2} {(x + 2)}^{2}$ $0 = x^4-8x^2+16 = (x^2-4)^2 = (x-2)^2(x+2)^2$

So the solutions of the equation are $x = \pm 2$ $x = +-2$ , each with multiplicity $2$ $2$ .

How do you find the real and imaginary solutions of x^4-8x^2=-16x4−8x2=−16x^4-8x^2=-16?