How do I solve $x^4-3x^2+2=0$? How do I solve $x^4-3x^2+2=0$ ?
I would appreciate some kind of hint here.  I have no clue how to start this problem. 

We can complete squares as follows:
\begin{align*}
x^4-3x^2+2&=0\\
x^4-3x^2&=-2\\
x^4-3x^2+\color{green}{\frac{9}{4}}&=-2+\color{green}{\frac{9}{4}}\\
\left(x^2-\frac{3}{2}\right)^2&=\frac{1}{4}\\
\left(x^2-\frac{3}{2}\right)^2-\frac{1}{4}&=0\\
\end{align*}
Now we can factor this difference of squares:
\begin{align*}
\left(x^2-\frac{3}{2}\right)^2-\frac{1}{4}&=0\\
\left(x^2-\frac{3}{2}+\frac{1}{2}\right)\left(x^2-\frac{3}{2}-\frac{1}{2}\right)&=0\\
\left(x^2-1\right)\left(x^2-2\right)&=0\\
(x+1)(x-1)\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)&=0\\
\end{align*}
Hence, the solutions are $\pm 1$ and $\pm\sqrt{2}$.
