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

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

1
Mar 8, 2018

See details below

#### 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
$3 + 0 i$
$- 3 + 0 i$

• 13 minutes ago
• 14 minutes ago
• 19 minutes ago
• 21 minutes ago
• 9 minutes ago
• 10 minutes ago
• 10 minutes ago
• 11 minutes ago
• 11 minutes ago
• 12 minutes ago
• 13 minutes ago
• 14 minutes ago
• 19 minutes ago
• 21 minutes ago