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:
$$2\cdot\sqrt{-\frac{(x-4)^2}{x^2+x}}+4\cdot\sqrt{x^2-12}=x^2-8-\frac{4-x}{x^2+x}$$
Best Answer
Left side exist only iff $${\frac{4-x}x}\geq 0\iff x\in (0,4]$$ and $${\frac{x-4}{x+1}}\geq 0\iff x\in (-\infty,-1)\cup [4,\infty)$$
So the only legitimate value for left side is $4$ which works.