Question 156148
Let {{{z=x^2}}}. So {{{z^2=(x^2)^2=x^4}}}. In short,  {{{z^2=x^4}}}


{{{15x^4-31x^2+10=0}}} Start with the given equation



{{{15z^2-31z+10=0}}} Plug in {{{z=x^2}}} and {{{z^2=x^4}}}




*[invoke quadratic_formula 15, -31, 10, "z"]




Remember, we let {{{z=x^2}}}. So this means that 



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



Take the square root of both sides for each case (remember the "plus/minus")



{{{x=sqrt(5/3)}}}, {{{x=-sqrt(5/3)}}}, {{{x=sqrt(2/5)}}}, or {{{x=-sqrt(2/5)}}}




Rationalize the denominator (if necessary)


{{{x=sqrt(15)/3}}}, {{{x=-sqrt(15)/3}}}, {{{x=sqrt(10)/5}}}, or {{{x=-sqrt(10)/5}}}



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Answer:


So the solutions are


{{{x=sqrt(15)/3}}}, {{{x=-sqrt(15)/3}}}, {{{x=sqrt(10)/5}}}, or {{{x=-sqrt(10)/5}}}