Solve the following biquadratic equations, indicating the number of obtained solutions:
1) $$x^4-2x^2=0$$
2) $$x^4+x^2-12=0$$
3) $$x^4-25=0$$
4) $$x^4-3x^2+2=0$$
Development:
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.
Solution:
1) We have $$3$$ solutions: $$0$$, $$\sqrt{2}$$, $$-\sqrt{2}$$. 2) We have $$2$$ solutions: $$\sqrt{3}$$ and $$-\sqrt{3}$$. 3) We have $$2$$ solutions: $$\sqrt{5}$$ and $$-\sqrt{5}$$. 4) It has no solution.