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. 
 A: Substitute $u = x^2$ and the equation becomes $u^2 - 3u + 2$
A: Consider the product $(y-2)(y-1)$ for which the expansion is $y^2 - 3 y +2$. Now let $y = x^2$ to obtain $x^4 - 3 x^2 + 2 = (x^2 -2)(x^2 -1) = 0$. This leads to $x^2 = 2$ or $x^2 =1$. These two equations yields $x \in \{ - 1, 1, - \sqrt{2}, \sqrt{2} \}$. 
A: Let $x^2=t$
Then equation becomes $t^2-3t+2=0\Rightarrow(t-1)(t-2)=0\Rightarrow t=1 $ or $t=2$
$\Rightarrow x^2=1,2\Rightarrow x=\pm1,\pm \sqrt2$
A: Got it.  Thanks for the hints.  Here is what I did.
Let $u=x^2$. 
$x^4-3x^2+2 = u^2-3u+2 = (u-2)(u-1)=0$. Solve for u I get 
$u=2,u=1$ but $u=x^2=2$, so $x=\pm \sqrt 2$ and $x=\pm \sqrt 1$
A: 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 $\pm\sqrt{2}$.
