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=√(5), x=-√(5)`

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=(-1 \pm √(1-4(-1)(20)))/(2(-1))\\&\begin{aligned}&=(-1 \pm √(1+80))/(-2) \\&=(-1 \pm √(81))/(-2) \\u &=(-1 \pm 9)/(-2)\end{aligned}\end{aligned}`

The variable u has two solutions,

`\begin{aligned}u &=(-1+9)/(-2) \\&=(8)/(-2) \\u &=-4\end{aligned}` and
`\begin{aligned}u &=(-1-9)/(-2) \\&=(-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) \\√(-4) &=\sqrt{x^(2)} \\\pm 2 i &=x\end{aligned}` and
`\begin{aligned}u &=x^(2) \\5 &=x^(2) \\√(5) &=\sqrt{x^(2)} \\\pm √(5) &=x\end{aligned}`

Thus, the roots are
`x=2 i, x=-2 i, x=√(5), x=-√(5)`