How do you find the zeros of $g (x) = x^{4} - 5 x^{2} - 36$ $g(x)=x^4-5x^2-36$ ?

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Answer 1 · 2016-11-01T02:01:14

$g (x) = x^{4} - 5 x^{2} - 36$ $g(x) = x^4 - 5x^2 - 36$ has 4 zeros, $x = - 3, 3, - 2 i and 2 i$ $x = -3, 3, -2i and 2i$

Given:

$g (x) = x^{4} - 5 x^{2} - 36$ $g(x) = x^4 - 5x^2 - 36$

Let $u = x^{2}$ $u = x^2$

$g (u) = u^{2} - 5 u - 36 = 0$ $g(u) = u^2 - 5u - 36 = 0$

$0 = u^{2} - 5 u - 36$ $0 = u^2 - 5u - 36$

$0 = (u - 9) (u + 4)$ $0 = (u -9)(u + 4)$

$u = 9 and u = - 4$ $u = 9 and u = -4$

$x = \pm 3 and x = \pm 2 i$ $x = +-3 and x = +-2i$

$g (x) = x^{4} - 5 x^{2} - 36$ $g(x) = x^4 - 5x^2 - 36$ has 4 zeros, $x = - 3, 3, - 2 i and 2 i$ $x = -3, 3, -2i and 2i$

How do you find the zeros of g(x)=x4−5x2−36g(x)=x^4-5x^2-36?