Shortcut to Radicals

Question

Can you help me how to solve this radical equation?

Solve for $x$: $\sqrt{\dfrac{5+x}{x-1}}=\sqrt{\dfrac{x-1}{5+x}}+\dfrac{3}{2}.$

Answer 1

The domain gives $-5<x<1$ and we have $$2(x+5)=2(x-1)+3\sqrt{(x+5)(x-1)}$$ or $$(x+5)(x-1)=16$$ or $$x^2+4x-21=0$$ or $$(x-3)(x+7)=0,$$ which gives $$x=3.$$

Answer 2

Raise this equation to power two. Etc. Be careful, analyze your process carefully and cleanly since squaring is not quite a reversible operation.

Answer 3

Let the term $\sqrt{\dfrac{5+x}{x-1}} =t$. Note that we have $x \in R \setminus (-5,1]$

Then $$t = \frac{1}{t}+\frac{3}{2} \\ 2t^2-3t-2=0\\ (t-2)\left(t+\frac{1}{2}\right) = 0$$

So you need to check if these values satisfy original equation.

As pointed in comments below, the surd cannot be negative. Thus we only have solution corresponding to $t=2$, and no solution corresponding to $t = -1/2$.

$$\sqrt{\dfrac{5+x}{x-1}} =2 \\ \dfrac{5+x}{x-1}=4 $$

Giving $x = 3$.