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

1 Answer
George C.
Nov 5, 2016

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#

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