# The sum of a number and its reciprocal is equal to 4 times the reciprocal, What is the number?

Sep 11, 2017

There are two possibilities:

$x = - \sqrt{3}$ or $x = \sqrt{3}$

#### Explanation:

Suppose the number is $x$

Then the "reciprocal" of this number is $\frac{1}{x}$

We are told that:

"The sum of a number and its reciprocal is equal to 4 times the reciprocal"

Therefore:

$x + \frac{1}{x} = 4 \times \frac{1}{x}$

$\therefore x = 4 \times \frac{1}{x} - 1 \times \frac{1}{x}$

$\therefore x = 3 \times \frac{1}{x}$

$\therefore {x}^{2} = 3$

$\therefore x = \pm \sqrt{3}$

Sep 11, 2017

$n + \frac{1}{n} = 4 \cdot \frac{1}{n}$

on the left side, multiply n by $\frac{n}{n}$ (same as multiplying by 1) to get a common denominator so you can add the numerators...

${n}^{2} / n + \frac{1}{n} = \frac{4}{n}$

giving:

${n}^{2} + 1 = 4$

${n}^{2} = 3$

so...

$n = \pm \sqrt{3}$

...check your work, when you can:

$\sqrt{3} + \frac{1}{\sqrt{3}} = 4 \cdot \frac{1}{\sqrt{3}}$

$\sqrt{3} \cdot \frac{\sqrt{3}}{\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{4}{\sqrt{3}}$

$\frac{3}{\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{4}{\sqrt{3}}$

$\frac{4}{\sqrt{3}} = \frac{4}{\sqrt{3}}$

(the same sequence also applies for the $- \sqrt{3}$ case

Sep 11, 2017

I got two possible numbers: $\pm \sqrt{3}$

#### Explanation:

Let us call the unknown number $x$; we get:
$x + \frac{1}{x} = 4 \cdot \frac{1}{x}$
rearrange:
${x}^{2} + 1 = 4$
(with $x \ne 0$)
${x}^{2} = 3$
$x = \pm \sqrt{3}$