TITLE: Solving $10x^4-13x^2+4=0$
QUESTION [6 upvotes]: I just came across a question in my paper that asks me to solve for $x$ in $10x^4-13x^2+4=0$
I've only learned how to factorize quadratics and the quadratic formula, but I'm not sure how to factorize quartics. Is there a way that I can apply my current knowledge to solve for this problem or if not, how would I be able to solve it?
For reference, the answer is $(5x^2-4)(2x^2-1)$ and therefore $x^2=0.8$ or $0.5$

REPLY [7 votes]: You have a quartic but it is simply a quadratic in disguise of a quartic so you may use the quadratic formula: $$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
For $$x^4 -13x^2+4=0$$
we have
$$x^2=\frac{13\pm \sqrt{13^2-4 \times 10 \times 4}}{2\times 10}=\frac{13\pm 3}{20}$$
So $$x^2=\frac12, \frac45 \implies x=\pm {\frac{1}{\sqrt2}},\space\pm {\frac{2}{\sqrt5}}$$