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?

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
