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

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Answer 1 · 2018-03-08T13:18:19

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

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

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

$t=(-b+-sqrt(b^2-4ac))/(2a)=(9+-sqrt(81-72))/2=(9+-9)/2$

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 $x=+-3$

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

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

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