Question 192032
Solve for x:
{{{15x^4-13x^2+2 = 0}}} Temporarily, let {{{y = x^2}}} so that {{{y^2 = x^4}}}. Make these substitutions.
{{{15y^2-13y+2 = 0}}} Factor.
{{{(3y-2)(5y-1) = 0}}} Apply the zero product rule.
{{{3y-2 = 0}}} or {{{5y-1 = 0}}}
{{{3y-2 = 0}}} Add 2 to both sides.
{{{3y = 2}}} Divide both sides by 3.
1) {{{y = 2/3}}}
{{{5y-1 = 0}}} Add 1 to both sides.
{{{5y = 1}}} Divide both sides by 5.
2) {{{y = 1/5}}}
Now substitute back {{{y = x^2}}} in both solutions.
{{{x^2 = 2/3}}} and {{{x^2 = 1/5}}} Take the square root in both cases.
{{{x = 0+-sqrt(2/3)}}} and {{{x = 0+-sqrt(1/5)}}}
The four solutions are:
{{{x = sqrt(2/3)}}}
{{{x = -sqrt(2/3)}}}
{{{x = sqrt(1/5)}}}
{{{x = -sqrt(1/5)}}}