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

##### 1 Answer

Nov 5, 2016

This quartic equation has roots

#### Explanation:

The difference of squares identity can be written:

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

We will use this several times.

Given:

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

This is a quartic in

#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#