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

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

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Answer 1 · 2018-03-08T13:18:19

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Explanation:

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

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

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

$t = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} = \frac{9 \pm \sqrt{81 - 72}}{2} = \frac{9 \pm 9}{2}$ $t=(-b+-sqrt(b^2-4ac))/(2a)=(9+-sqrt(81-72))/2=(9+-9)/2$

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

Undoing the change of variable

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

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

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

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

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Explanation:

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How do you find the number of complex zeros for the function f(x)=x4−9x2+18f(x)=x^4-9x^2+18?

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