Solve the equation $\sqrt{x^2-1}=(x+5)\sqrt{\frac{x+1}{x-1}}$ Solve the equation $$\sqrt{x^2-1}=(x+5)\sqrt{\dfrac{x+1}{x-1}}.$$
I think that radical equations can be solved by determining the domain (range) of the variable and at the end the substitution won't be necessary which is suitable for roots which aren't very good-looking and nice to work with.
What are the steps to follow? We have $D_x:\begin{cases}x^2-1\ge0\\\dfrac{x+1}{x-1}\ge0\\x-1\ne0\end{cases} \iff x\in(-\infty;-1]\cup(1;+\infty).$ What next?

The equation's defined for $\;x>1\;\;\text{or}\;\;x\le-1\;$ , so now square the whole thing:
$$x^2-1=(x+5)^2\,\frac{x+1}{x-1}\iff x^3-x^2-x+1=x^3+11x^2+35x+25\iff$$
$$12x^2+36x+24=0\iff12(x+1)(x+2)=0\ldots$$
