Question 131891
{{{6x^4-19x^2+10=0}}} Start with the given equation



Let {{{u=x^2}}}


{{{6u^2-19u+10=0}}} Plug in {{{u=x^2}}}




{{{(2u-5)(3u-2)=0}}} Factor the left side (note: if you need help with factoring, check out this <a href=http://www.algebra.com/algebra/homework/playground/change-this-name4450.solver>solver</a>)




Now set each factor equal to zero:

{{{2u-5=0}}} or  {{{3u-2=0}}} 


{{{u=5/2}}} or  {{{u=2/3}}}    Now solve for u in each case




Now remember, we let {{{u=x^2}}}, so this means 



{{{x^2=5/2}}} or  {{{x^2=2/3}}} 



*[Tex \LARGE x=\pm\sqrt{\frac{5}{2}}] or  *[Tex \LARGE x=\pm\sqrt{\frac{2}{3}}]  Take the square root of both sides for each case



So our solutions are 



*[Tex \LARGE x=\sqrt{\frac{5}{2}}],*[Tex \LARGE x=-\sqrt{\frac{5}{2}}],*[Tex \LARGE x=\sqrt{\frac{2}{3}}]  or  *[Tex \LARGE x=-\sqrt{\frac{2}{3}}]