Biquadratic equations

1. Since we do not have independent term we can extract common factor:

$$x^4-2x^2=0 \Rightarrow x^2\cdot(x^2-2)=0 \Rightarrow \left\{\begin{matrix} x^2=0 \Rightarrow x=0 \\ x^2-2=0 \Rightarrow x^2=2 \Rightarrow x=\pm\sqrt{2}\end{matrix}\right.$$

We have three solutions $0$, $\sqrt{2}$, $-\sqrt{2}$.

2. We can make the change of variable $x^2=t$, so we have the equation

$$\displaystyle t^2+t-12=0 \Rightarrow t=\frac{-1 \pm \sqrt{1^2-4\cdot1\cdot(-12)}}{2}= \frac{-1 \pm \sqrt{49}}{2}= \frac{-1 \pm 7}{2} \left \{\begin{matrix} t_1=3 \\ t_2=4\end{matrix}\right.$$ Undoing the changes:

$$x^2=t \Rightarrow x=\pm\sqrt{t}$$

$$x=\pm\sqrt{3}$$

$x=\pm\sqrt{-4} \Rightarrow $ no solution.

Therefore we obtain $2$ solutions: $\sqrt{3}$ and $-\sqrt{3}$.

3. Applying the change of variable: $$t^2-25=0 \Rightarrow t^2=25 \Rightarrow t=\pm\sqrt{25} = \pm5$$

Undoing the changes:

$$x^2=t \Rightarrow x=\pm\sqrt{t}$$

$$x=\pm\sqrt{5}$$

$x=\pm\sqrt{-5} \Rightarrow $ no solution.

Therefore it has $2$ solutions: $\sqrt{5}$ and $-\sqrt{5}$.

4. We do the change of variable $x^2=t$, so we have the equation

$$\displaystyle t^2-3t+2=0 \Rightarrow t=\frac{-3 \pm \sqrt{(-3)^2-4\cdot1\cdot2}}{2}= \frac{-3 \pm \sqrt{9-8}}{2}= \frac{-3 \pm 1}{2} \left \{\begin{matrix} t_1=-1 \\ t_2=-2\end{matrix}\right.$$ Undoing the changes:

$$x^2=t \Rightarrow x=\pm\sqrt{t}$$

$x=\pm\sqrt{-1} \Rightarrow $ no solution.

$x=\pm\sqrt{-2} \Rightarrow $ no solution.

Therefore it does not have a solution.