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solve. 10x^4-11x^2+3=0
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let x^2 = y, then your equation results to: \[10y ^{2} -11y+3 = 0\] is that familiar?
\[10y ^{2}-11y+3=0\] \[10y-5y-6y+3=0\] \[5y(\[x= \pm \sqrt{3/5} \]2y-1)-3(2y-1)=0\] \[(5y-3)(2y-1) =0\] then : \[5y-3=0 or 2y-1=0\] therefore y =3/5 or y=1/2 but remember y= x^2 this means : \[x ^{2} = 3/5\] and \[x ^{2} = 1/2\] implying : \[x= \pm \sqrt{3/5}\] and \[x= \pm \sqrt{1/2}\] question?
Thats what I got. thatnks :)
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