# How do you find the zeroes for f(x)=2x^4-2x^2-40?

May 29, 2015

Simplify by replacing ${x}^{2}$ with $y$;
solve as a standard quadratic for expressions in terms of $y$ that result in zeros;
replace $y$ in those expressions with ${x}^{2}$ and solve for values of $x$

If $y = {x}^{2}$
$2 {x}^{4} - 2 {x}^{2} - 40 = 0$
is equivalent to
$2 {y}^{2} - 2 y - 40 = 0$

$\left(2 y + 8\right) \left(y - 5\right) = 0$

For zeros either
Case 1: $\left(2 y + 8\right) = 0$
or
Case 2: $\left(y - 5\right) = 0$

Case 1: $\left(2 y + 8\right) = 0$
$\rightarrow y = - 4$
$\rightarrow {x}^{2} = - 4$
There are no Real solutions, but there are Complex solutions $x = \pm 2 i$

Case 2: $\left(y - 5\right) = 0$
$\rightarrow y = 5$
$\rightarrow {x}^{2} = 5$
$\rightarrow x = \pm \sqrt{5}$