Question 374955
This is really a disguised quadratic equation.
Let {{{u=x^2}}}, then {{{u^2=x^4}}}
{{{6u^2-13u+6=0}}}
{{{(3u-2)(2u-3)=0}}}
Two solutions in {{{u}}}:
{{{3u-2=0}}}
{{{3u=2}}}
{{{u=2/3}}}
{{{x^2=2/3}}}
{{{x=0 +- sqrt(2/3)}}}
{{{highlight(x=0 +- sqrt(6)/3)}}}
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{{{2u-3=0}}}
{{{2u=3}}}
{{{u=3/2}}}
{{{x^2=3/2}}}
{{{x=0 +- sqrt(3/2)}}}
{{{highlight(x=0 +- sqrt(6)/2)}}}