Solve $\sqrt{\frac{4-x}x}+\sqrt{\frac{x-4}{x+1}}=2-\sqrt{x^2-12}$

Question

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:

$$2\cdot\sqrt{-\frac{(x-4)^2}{x^2+x}}+4\cdot\sqrt{x^2-12}=x^2-8-\frac{4-x}{x^2+x}$$

Answer 1

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.

Answer 2

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.

Answer 3

Once you identify $x=4$ as a real solution by inspection (mrtaurho's answer), you see that it is the only real solution. For all other $x$, the left side is not real. Check the signs of the radicands on the left side for all other possible cases $x>4, 0<x<4, -1<x<0, x<-1$.

There is one other possibility for a real solution. If one of the radicands on the left is positive and the other negative, you can separately try to fit the two terms on the left side to the real and imaginary parts of the right side. But before investing in heavy computations observe that if $-\sqrt{x^2-12}$ is imaginary at all then (according to the usual principal value definitions) it is a negative number times $i$, and any imaginary radicals on the left would be a positive number times $i$. You can't match the signs by this method.

So the only place to look for a real solution is $x=4$.