Question

Can someone help me with this? Find the roots of -x^2-20=-x^4

Answer 1

The roots are $x=2 i, x=-2 i, x=\sqrt{5}, x=-\sqrt{5}$

Step-by-step explanation:

The equation is $-x^{2} -20=-x^{4}$

Switch sides, we get,

$-x^{4}=-x^{2} -20$

Adding both sides by 20, we get,

$-x^{4}+20=-x^{2}$

Adding both sides by $x^{2}$,

$-x^{4}+x^{2}+20=0$

To solve this equation, let us assume $u=x^{2}$ and $u^{2}=x^{4}$

Thus, rewriting this equation,

$-u^{2} +u+20=0$

Using quadratic formula, we get the value of u.

$\begin{aligned}&u=\frac{-1 \pm \sqrt{1-4(-1)(20)}}{2(-1)}\\&\begin{aligned}&=\frac{-1 \pm \sqrt{1+80}}{-2} \\&=\frac{-1 \pm \sqrt{81}}{-2} \\u &=\frac{-1 \pm 9}{-2}\end{aligned}\end{aligned}$

The variable u has two solutions,

$\begin{aligned}u &=\frac{-1+9}{-2} \\&=\frac{8}{-2} \\u &=-4\end{aligned}$ and $\begin{aligned}u &=\frac{-1-9}{-2} \\&=\frac{-10}{-2} \\u &=5\end{aligned}$

Since, $u=x^{2}$ and $u^{2}=x^{4}$

Thus, substituting the u-values, we get,

$\begin{aligned}u &=x^{2} \\-4 &=x^{2} \\\sqrt{-4} &=\sqrt{x^{2}} \\\pm 2 i &=x\end{aligned}$ and $\begin{aligned}u &=x^{2} \\5 &=x^{2} \\\sqrt{5} &=\sqrt{x^{2}} \\\pm \sqrt{5} &=x\end{aligned}$

Thus, the roots are  $x=2 i, x=-2 i, x=\sqrt{5}, x=-\sqrt{5}$