Question 32359
Solve:
{{{x^4-x^2-12 = 0}}} Rewrite this as a quadratic equation by temporarily substituting {{{y = x^2}}}.
{{{y^2-y-12 = 0}}} Solve by factoring.
{{{(y+3)(y-4) = 0}}} Apply the zero products principle.
{{{y+3 = 0}}} and/or {{{y-4 = 0}}}
If {{{y+3 = 0}}} then {{{y = -3}}}
If {{{y-4 = 0}}} then {{{y = 4}}}

But, {{{y = x^2}}}, so:

{{{x^2 = -3}}} Take the square root of both sides.
{{{x = sqrt(3)i}}} and {{{x = -sqrt(3)i}}}

{{{x^2 = 4}}} Take the square root of both sides.
{{{x = 2}}} and {{{x = -2}}}

The four roots are:
{{{x = sqrt(3)i}}}
{{{x = -sqrt(3)i}}}
{{{x = 2}}}
{{{x = -2}}}