Question 189607
Solve for x:
{{{x^4-5x^2-24 = 0}}} 
You can write this as a quadratic equation with a temporary change of variable, thus...
Let {{{y = x^2}}} so that {{{x^4 = y^2}}}, then you substitute...
{{{y^2-5y-24 = 0}}} Factor this...
{{{(y-8)(y+3) = 0}}} Apply the zero product rule:
{{{y-8 = 0}}} or {{{y+3 = 0}}} so that...
{{{y = 8}}} or {{{y = -3}}} Now substitue back {{{y = x^2}}} so you have...
{{{x^2 = 8}}} or {{{x^2 = -3}}} and...
{{{x = sqrt(8)}}}, {{{x = -sqrt(8)}}} or {{{x = sqrt(-3)}}}, {{{x = -sqrt(-3)}}}
The roots are:
{{{x = sqrt(8)}}}
{{{x = -sqrt(8)}}}
{{{x = sqrt(3)i}}}
{{{x = -sqrt(3)i}}} where: {{{i = sqrt(-1)}}}