How to prove $\sqrt{\frac{x+1}{x^3-x^2}}=\frac{\frac1x+1}{\sqrt{x^2-1}}$

Question

I would like to prove the following equality $$\sqrt{\frac{x+1}{x^3-x^2}}=\frac{\frac1x+1}{\sqrt{x^2-1}}$$

It is tempting to square both sides of the equation and it is not hard to verify that they are equal. Then, I realized that by squaring the equation I may have altered the original equality and the proof is not really valid.

How do I then show properly that the equation holds?

Answer 1

In order to check equality of \begin{align*} \color{blue}{\sqrt{\frac{x+1}{x^3-x^2}}=\frac{\frac1x+1}{\sqrt{x^2-1}}}\tag{1} \end{align*} we must first determine the range of validity.

Left-hand side:

• We have to exclude $x\in\{0,1\}$ otherwise the denominator $x^3-x^2$ is zero. Since the radicand \begin{align*} \frac{x+1}{x^3-x^2}=\frac{x+1}{x^2\left(x-1\right)} \end{align*} has to be non-negative we consider \begin{align*} x-1>0,\quad &x+1\geq 0\quad\to\quad x>1\\ x-1<0,\quad &x+1\leq 0\quad\to\quad x\leq -1 \end{align*} and conclude the left-hand side is valid for $\color{blue}{x\in\mathbb{R}\setminus(-1,1]}$.

Right-hand side:

• We have to exclude $x\in\{-1,0,1\}$ otherwise a denominator is zero. Since the radicand $x^2-1$ has to be non-negative we consider \begin{align*} x^2-1>0\quad\to\quad x^2>1\\ \end{align*} and conclude the left-hand side is valid for $\color{blue}{x\in\mathbb{R}\setminus[-1,1]}$.

Region of validity:

We see the left-hand side of (1) is valid for $x=-1$, leading to zero, while the right-hand side is undefined in case $x=-1$. The region of validity is the intersection of the regions from both sides which is \begin{align*} \color{blue}{x\in\mathbb{R}\setminus[-1,1]}\tag{2} \end{align*}

Equality check:

We can now check for equality within the range of validity (2). We start with the right-hand side of (1) and obtain \begin{align*} \color{blue}{\frac{\frac{1}{x}+1}{x^2-1}}&=\frac{1}{x}\,\frac{x+1}{\sqrt{(x-1)(x+1)}}\\ &=\frac{1}{|x|}\,\frac{|x+1|}{\sqrt{(x-1)(x+1)}}\tag{3.1}\\ &=\frac{1}{|x|}\,\sqrt{\frac{(x+1)^2}{(x-1)(x+1)}}\tag{3.2}\\ &=\frac{1}{|x|}\,\sqrt{\frac{x+1}{x-1}}\\ &=\sqrt{\frac{x+1}{x^2(x-1)}}\\ &\,\,\color{blue}{=\sqrt{\frac{x+1}{x^3-x^2}}} \end{align*} and the claim (1) follows.

Comment:

• In (3.1) we use $\frac{x+1}{x}=\frac{|x+1|}{|x|}$ for $x\in\mathbb{R}\setminus[-1,1]$.

• In (3.2) we use the identity $\sqrt{x^2}=|x|$.

Answer 2

Divide by $x$ in the LHS: $$\sqrt{\frac{x+1}{x^3-x^2}}=\sqrt{\frac{1+\frac1x}{x^2-x}}$$Multiply by $\sqrt{1+\frac1x}$ on top and bottom: $$\frac{1+\frac1x}{\sqrt{\left(1+\frac1x\right)\left(x^2-x\right)}}=\frac{1+\frac1x}{\sqrt{x^2-1}}$$

Answer 3

since you're manipulating both sides and don't want to square you can multiply both sides by $\sqrt{ x^2 - 1 }$