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?
 A: Let $z = \frac{1}{2}(n - x^2)$. Then the equation becomes $\Big(\frac{z}{x}\Big)^2 = z$ or better $z\Big(\frac{z}{x^2} - 1\Big) = 0$. Solutions are $z = 0$ and $z = x^2$. If $z = 0$, then $x = \pm \sqrt{n}$. If $z = x^2$, then $x = \pm \sqrt{n/3}$.
A: Both of the solutions are possible since if you let $x = \pm\sqrt{n}$, then $x^2 = n$
Now, substiute in the equation you got
$$\left(\frac{\frac{1}{2}\cdot(n-n)}{\pm\sqrt{n}}\right)^2 =\frac{1}{2}\cdot(n-n)$$
$$\left(\frac{\frac{1}{2}\cdot(0)}{\pm\sqrt{n}}\right)^2 =\frac{1}{2}\cdot(0)$$
Which simply is $0=0$   
That is why the first solution is correct.  
Now let $x = \pm\frac{\sqrt{n}}{\sqrt{3}}$
$$\left(\frac{\frac{1}{2}\cdot(n-\frac{n}{3})}{\pm\frac{\sqrt{n}}{\sqrt{3}}}\right)^2 =\frac{1}{2}\cdot(n-\frac{n}{3})$$
$$\frac{\frac{1}{4}\cdot{(n-\frac{n}{3})}^2}{\frac{n}{3}}=\frac{1}{2}\cdot(n-\frac{n}{3})$$
$$\frac{3(n-\frac{n}{3})^2}{4n} = \frac{1}{2}n - \frac{n}{6}$$
$$\frac{3(n^2-\frac{2}{3}n^2+\frac{n^2}{9})}{4n} = \frac{1}{2}n - \frac{n}{6}$$
$Multiply\ both\ sides\ by\ 4n$
$$3n^2-2n^2+\frac{n^2}{3} = 2n^2 - \frac{2}{3}n^2$$ 
$$\frac{4}{3}n^2=\frac{4}{3}n^2$$
which implies that solution is also correct
A: 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.
