# How do you find the number of complex zeros for the function f(x)=x^4-9x^2+18?

Mar 8, 2018

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#### Explanation:

We can make the variable change $t = {x}^{2}$. With this change we obtain

${t}^{2} - 9 t + 18 = f \left(t\right)$. If we are lookin form zeros od $f$, then

${t}^{2} - 9 t + 18 = 0$. Apliying knowed formula

$t = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{9 \pm \sqrt{81 - 72}}{2} = \frac{9 \pm 9}{2}$

Both solutions for $t$ are: $t = 0$ and $t = 9$

Undoing the change of variable

$t = 0 = {x}^{2}$, so $x = 0$ (double root)
$t = 9 = {x}^{2}$, so $x = \pm 3$

Our initial equation has no complex roots (or zeros), but we can express them in complex form (with their imaginary parts zero)

$0 + 0 i$ double
$3 + 0 i$
$- 3 + 0 i$