If $x$ is positive, then why does $\frac{1}{\sqrt{x+1} + \sqrt{x}} = \sqrt{x+1} - \sqrt{x}$?

Question

Given that $x$ is positive, $\frac{1}{\sqrt{x+1} + \sqrt{x}} = \sqrt{x+1} - \sqrt{x}$

I've been trying to convert the left side of the equation to the right side:

$$ \frac{1}{\sqrt{x+1} + \sqrt{x}}$$

But then how can I flip this round to be what I have on the right side?

I know that $\frac{1}{\sqrt{x}} = x^{-\frac{1}{2}}$, so I would think that this would give me

$$(x+1)^{-1/2} + x^{-1/2}$$

Which I thought would then convert to

$$-\sqrt{x+1} + (-\sqrt{x}) $$

so I'm not sure how the first part $\sqrt{x+1}$ got to be positive

Answer 1

Rationalize the denominator:

\begin{align} \frac{1}{\sqrt{x+1} + \sqrt{x}} \cdot \frac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x+1} - \sqrt{x}} &= \frac{\sqrt{x+1}-\sqrt{x}}{\underbrace{\left(\sqrt{x+1} + \sqrt{x}\right) \cdot \left(\sqrt{x+1} - \sqrt{x}\right)}_{\color{red}{\text{FOIL}}}} \\[0.3cm] &= \cdots \end{align}

Some notes:

If you want to show that two things are equal to each other, it's better to manipulate only one of them than it is to start by saying they're equal to each other and then manipulating both sides to verify, etc.

Your idea of converting to $(x+1)^{-1/2} + x^{-1/2}$ won't work because, in general, $\displaystyle \frac{1}{a+b} \ne \frac{1}{a} + \frac{1}{b}$.

Answer 2

Clear the denominator to get:

$$1 = (\sqrt{x+1} + \sqrt{x})(\sqrt{x+1}-\sqrt{x}) = (x+1)-x.$$

As Tilper points out, this is not a correct proof per se, but once you see this, you can reverse the ordering of your logic and write a proper proof.

In this case it would be more proper to write:

$$1 = (x+1)-x = (\sqrt{x+1} + \sqrt{x})(\sqrt{x+1}-\sqrt{x}). $$

Therefore since $\sqrt{x+1} + \sqrt{x} \neq 0$,

$$\frac{1}{\sqrt{x+1} + \sqrt{x}} = \sqrt{x+1} - \sqrt{x}.$$

Answer 3

You need to use a little trick with fractions. Recall that for any number $n$ that is not zero, $\frac{n}n=1$, and 1 times any number is itself.

So start with the left side: $\sqrt{1+x}-\sqrt{x}$. If we choose our $n$ to be $\sqrt{1+x}+\sqrt{x}$, then:

$$\sqrt{1+x}-\sqrt{x}=\frac{\sqrt{1+x}-\sqrt{x}}1\cdot\frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}+\sqrt{x}}$$

Then "FOIL" the top:

$$\frac{\sqrt{1+x}^2+\sqrt{x}\sqrt{1+x}-\sqrt{x}\sqrt{1+x}-\sqrt{x}^2}{\sqrt{1+x}+\sqrt{x}}=\frac{1+x-x}{\sqrt{1+x}+\sqrt{X}}=\frac1{\sqrt{1+x}+\sqrt{x}}$$

Answer 4

One can basically use $$ \frac{1}{a+b}=\frac{1}{a+b}\frac{a-b}{a-b}=\frac{a-b}{a^2-b^2} $$ valid for $a-b\ne0$. If $a=\sqrt{x+1}$ and $b=\sqrt{x}$, you have $$ a^2-b^2=(x+1)-x=1 $$ so $$ \frac{1}{\sqrt{x+1}+\sqrt{x}}= \frac{1}{a+b}=\frac{a-b}{a^2-b^2}= \frac{\sqrt{x+1}-\sqrt{x}}{1}= \sqrt{x+1}-\sqrt{x} $$

With some practice, you'll do the steps automatically: $$ \frac{1}{\sqrt{x+1}+\sqrt{x}}= \frac{1}{\sqrt{x+1}+\sqrt{x}}\, \frac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x+1}-\sqrt{x}}= \frac{\sqrt{x+1}-\sqrt{x}}{(\sqrt{x+1}\,)^2-(\sqrt{x})^2} $$