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?

$$\sqrt{x^{2}-1}=\left(x+5\right)\frac{\sqrt{x+1}}{\sqrt{x-1}}$$
Both sides are only defined for $|x|>1$ or $x=-1$
Squaring both sides,
$$x^{2}-1=\left(x+5\right)^{2}\left(\frac{x+1}{x-1}\right)$$
We can factor the LHS using difference of squares,
$$\left(x+1\right)\left(x-1\right)=\left(x+5\right)^{2}\left(\frac{x+1}{x-1}\right)$$
Assuming $|x|>1$, we can divide both sides by $x+1$ which is nonzero (We can check $x=-1$ as a sperate case).
$$\left(x-1\right)=\left(x+5\right)^{2}\left(\frac{1}{x-1}\right)$$
Multiplying both sides by $x-1$,
$$\left(x-1\right)^{2}=\left(x+5\right)^{2}$$
$$\Rightarrow \pm\left(x-1\right)=\pm\left(x+5\right)$$
There won't be any solutions for $(x-1)$=$(x+5)$ or $-(x-1)$=$-(x+5)$, so $(x-1)$ and $(x+5)$ must be additive inverses.
$$(x-1)=-(x+5)$$
$$\Rightarrow x=-2$$
Both sides of the original equation are defined for $x=-2$, so it's a solution.
Earlier we only considered values of x that satisfied $|x|>1$. If $x=-1$, both sides of the equation are $0$, so $x=-1$ is also a solution.
