Question

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

Answer 1

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`