Solve the equation $\sqrt{x^2-1}=(x+5)\sqrt{\frac{x+1}{x-1}}$

Question

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?

Answer 1

The equation's defined for $\;x>1\;\;\text{or}\;\;x\le-1\;$ , so now square the whole thing:

$$x^2-1=(x+5)^2\,\frac{x+1}{x-1}\iff x^3-x^2-x+1=x^3+11x^2+35x+25\iff$$

$$12x^2+36x+24=0\iff12(x+1)(x+2)=0\ldots$$

Answer 2

More directly, factorize the equation as

$$\sqrt{\dfrac{x+1}{x-1}}( |x-1|-(x+5))=0$$

which leads to $x+1=0$ and $(1-x)-(x+5)=0$.

Answer 3

$x^2 - 1 = (x+1)(x-1)$.

So you have $\sqrt{(x+1)(x-1)} =(x+5) \sqrt\frac {x+1}{x-1}$.

Now my first thoughts are dividing both sides by $\sqrt{x+1}$ and multiplying both sides by $\sqrt{x-1}$ to get $x-1 = x+5$.

BUT BE CAREFUL!

First off. When we divide both sides by $\sqrt{x+1}$ we are assuming that $\sqrt{x+1}\ne 0$. We must consider the case where $\sqrt{x+1} = 0$ so $x = -1$ as a potential solution.

Indeed, if $x = -1$ we have $\sqrt{x^2 -1} =0$ and $(x+5) \sqrt{\frac {x+1}{x-1}} = 0$. So $x=-1$ is a solution.

And we could divide both sides by $\sqrt{x+1}$.... except....

$\frac {\sqrt{(x+1)(x-1)}}{\sqrt{x+1}} \ne \sqrt{x+1}$ and $\sqrt{\frac {x+1}{x-1}}{\sqrt{x+1}}\ne \frac 1{\sqrt{x-1}}$ unless both $x+1$ and $x-1$ are positive. If they are both negative the terms we are refering to simply are not designed.

So we could two cases where we consider both $x+1, x-1$ are positive and a case were they are both negative, in which case we divide both sides by $\sqrt{-(x+1)}$ and get $\sqrt{1-x} = (x+5)\frac 1{\sqrt {1-x}}$.

But more sophisticatedly we shouldn't divide and multiply in two steps. We should simply multiply both sides by $\sqrt{\frac{x-1}{x+1}}$.

In this case we still must make the case where $x+1 = 0$ for a potential solution.

So if $x = -1$ that is a solution. We did that above.

If $x \ne -1$ multiply bot sides by $\sqrt{\frac{x-1}{x+1}}$ to get:

$\sqrt{(x^2-1)\frac {x-1}{x+1}}=(x+5)\sqrt 1$ or

$\sqrt{(x-1)^2} = x+5$ or

$|x-1| = x+ 5$.

(Note my initial thought of $x-1$ being positive was erroneous and we get $|x-1| = x+5$ not $x-1=x+5$)

If $x -1 \ge 0$ we get $x-1=x+5$ which has no solutions.

If $x -1<0$ we get $1-x = x+5$ or $x=-2$.

So our two solutions are $x=-1$ and $x=-2$ both of which fit in the domain that you calculated.

Answer 4

$$\sqrt{x^{2}-1}=\left(x+5\right)\frac{\sqrt{x+1}}{\sqrt{x-1}}$$ 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.