Solve $\sqrt{\frac{4-x}x}+\sqrt{\frac{x-4}{x+1}}=2-\sqrt{x^2-12}$ The equation is
 $$\sqrt{\frac{4-x}x}+\sqrt{\frac{x-4}{x+1}}=2-\sqrt{x^2-12}$$
I tried squaring both left side and right side then bringing them to same numerator but got lost from there ... any ideas of how should this be solved? 
I got to:
$$2\cdot\sqrt{-\frac{(x-4)^2}{x^2+x}}+4\cdot\sqrt{x^2-12}=x^2-8-\frac{4-x}{x^2+x}$$

Since there are three square roots you will need to square the whole equation at least three times. This is not the best way since it produces false solutions every single time and also the complexity of the equation will increase every time.
Therefore observe the numerator within the square roots of LHS and the square roots on the right 
$$\sqrt{\frac{4-x}x}+\sqrt{\frac{x-4}{x+1}}=2-\sqrt{x^2-12}$$
Note that for $x=4$ the LHS will be zero hence both numerators will become zero and the denominator will be defined. So we get
$$0=2-\sqrt{4^2-12}\Rightarrow 0=2-\sqrt{16-12}\Rightarrow 0=0$$
And therefore $x_1=4$ is a solution. Also it is the only real solution (according to WolframAlpha). The other two are given by $x_{2/3}= -0.889727 \pm 0.079797 i$ but honestly speaking I guess you only can get them by squaring over and over again eliminating the false solutions in the end. 
