Why is $\frac{\sqrt{x+1}-1}{x}$ equal to $\frac{1}{\sqrt{x+1}+1}$?

Question

I'm working with the expression $$\frac{\sqrt{x+1} - 1}{x}.$$ According to Wolfram Alpha "alternate form" section (http://www.wolframalpha.com/input/?i=%28%28x%2B1%29%5E1%2F2-1%29%2Fx) it is equal to $$\frac{1}{\sqrt{x+1} + 1}.$$ How can I go from one expression to the other?

Answer 1

$$\frac{\sqrt{x+1}-1}{x}=\frac{\sqrt{x+1}-1}{x}\times\frac{\sqrt{x+1}+1}{\sqrt{x+1}+1}=\frac{x}{x\left(\sqrt{x+1}+1\right)}=\frac{1}{\sqrt{x+1}+1}$$

Answer 2

Multiply the top and bottom by the conjugate $(x+1)^{1/2}+1$

Answer 3

Hint $\ $ Rationalize the $\, \color{#c00}{numerator},\,$ in exactly the same way one rationalizes denonominators, i.e. multiply both by the "conjugate" $\ \overline{1-\sqrt{x+1}} \,=\, 1+\sqrt{x+1}\,$ of the numerator.

Remark $\ $ Though not used as frequently as rationalizing the denominator, this does prove useful in various contexts, e.g. specializing the quadratic formula when the leading coeff $\,a\to 0.$

$$\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\ =\ \dfrac{2c}{-b \pm \sqrt{b^2-4\:a\:c}}$$

As $\,a\to 0,\,$ the latter gives the root $\,x = -b/c\,$ of $\,bx+c\,\ (= ax^2\!+bx+c\,$ when $\,a=0).$