# How do you solve #2x^4-128=0#?

##### 1 Answer

Jan 10, 2017

The roots of this quartic equation are:

#+-2sqrt(2)" "# and#" "+-2sqrt(2)i#

#### Explanation:

The difference of squares identity can be written:

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

So we find:

#0 = 2x^4-128#

#color(white)(0) = 2(x^4-64)#

#color(white)(0) = 2((x^2)^2-8^2)#

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

#color(white)(0) = 2(x^2-(2sqrt(2))^2)(x^2-(2sqrt(2)i)^2)#

#color(white)(0) = 2(x-2sqrt(2))(x+2sqrt(2))(x-2sqrt(2)i)(x+2sqrt(2)i)#

So the roots of the quartic equation are:

#+-2sqrt(2)" "# and#" "+-2sqrt(2)i#