<s>[INST] Solve the following equation for $x$,$\left(\frac{\frac{1}{2}\cdot(n-x^2)}{x}\right)^2 =\frac{1}{2}\cdot(n-x^2)$ I am not great at transposition and wolfram alpha confused me so I would like to see the steps in solving for x. 
$$\left(\frac{\frac{1}{2}\cdot(n-x^2)}{x}\right)^2 =\frac{1}{2}\cdot(n-x^2)$$
Wolfram alpha gave me two incompatible answers
$$x=\pm\sqrt{n}, \ \ \ \ \ \ \ \  \sqrt{n}\not=0$$
$$x=\pm\frac{\sqrt{n}}{\sqrt{3}}, \ \ \ \ \ \ \ \  \sqrt{n}\not=0$$
The first answer is wrong the second is correct.
Can anyone explain why the first solution is wrong and how to derive the second?
 [/INST] Another way to look at the problem is to rewrite (assuming $x\neq 0$) $$\left(\frac{\frac{1}{2}\cdot(n-x^2)}{x}\right)^2 =\frac{1}{2}\cdot(n-x^2)\implies 4 x^2 \left(\frac{\left(n-x^2\right)^2}{4 x^2}-\frac{1}{2}
   \left(n-x^2\right)\right)=0$$ Expand and simplify to get $$3 x^4-4n x^2+n^2=0$$ Now, using $y=x^2$, the equation is just the quadratic in $y$ $$3y^2-4ny+n^2=0$$ the roots of which being $y=n$ and $y=\frac n 3$.
So $x=\pm \sqrt n$ and $x=\pm \sqrt{\frac n3}$ are the roots of the original equation.
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