Question 122023
Try this:
Replace {{{x^2}}} with y.
{{{x^4-3x^2+2 = 0}}} becomes:
{{{y^2-3y+2 = 0}}} Solve by factoring:
{{{(y-1)(y-2) = 0}}} Apply the zero product principle:
{{{y-1 = 0}}} or {{{y-2 = 0}}} so then...
{{{y = 1}}} or {{{y = 2}}}
Now replace the y with {{{x^2}}}
{{{x^2 = 1}}} or {{{x^2 = 2}}} Take the square root of these:
{{{x = 1}}}, {{{x = -1}}}, {{{x = sqrt(2)}}}, {{{x = -sqrt(2)}}}