Question

How do you solve the quadratic with complex numbers given `x^4+16x^2-225=0`?

Answer 1

This quartic equation has roots `+-3` and `+-5i`

Explanation:

The difference of squares identity can be written:

`a^2-b^2 = (a-b)(a+b)`

We will use this several times.

`color(white)()`
Given:

`x^4+16x^2-225 = 0`

This is a quartic in `x^4`, but effectively a quadratic in `x^2`. So we can make a start by treating it as such and using whatever method we like to solve the quadratic:

`0 = x^4+16x^2-225`

`color(white)(0) = (x^2)^2+16(x^2)+64-289`

`color(white)(0) = (x^2+8)^2-17^2`

`color(white)(0) = ((x^2+8)-17)((x^2+8)+17)`

`color(white)(0) = (x^2-9)(x^2+25)`

`color(white)(0) = (x^2-3^2)(x^2-(5i)^2)`

`color(white)(0) = (x-3)(x+3)(x-5i)(x+5i)`

Hence roots:

`x = +-3` and `x = +-5i`