Problems from Biquadratic equations

Development:

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

$$x^4-2x^2=0\Rightarrow x^2\cdot (x^2-2)=0\Rightarrow \{\begin{array}{c}{x}^2=0\Rightarrow x=0\\ x^2-2=0\Rightarrow x^2=2\Rightarrow x=\pm \sqrt{2}\end{array}$$

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

$${\displaystyle t^2+t-12=0\Rightarrow t=\frac{-1\pm \sqrt{1^2-4\cdot 1\cdot (-12)}}{2}=\frac{-1\pm \sqrt{49}}{2}=\frac{-1\pm 7}{2}\{\begin{array}{c}{t}_1=3\\ t_2=4\end{array}}$$ 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: $$t^2-25=0\Rightarrow t^2=25\Rightarrow t=\pm \sqrt{25}=\pm 5$$

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

$${\displaystyle t^2-3t+2=0\Rightarrow t=\frac{-3\pm \sqrt{(-3{)}^2-4\cdot 1\cdot 2}}{2}=\frac{-3\pm \sqrt{9-8}}{2}=\frac{-3\pm 1}{2}\{\begin{array}{c}{t}_1=-1\\ t_2=-2\end{array}}$$ 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.

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