# How do you solve 1/x + 1 = x/2?

Feb 27, 2016

$x = 1 \pm \sqrt{3}$

#### Explanation:

$1$. In order to solve this equation, each fraction must have the same denominator. To do this, multiply every term by $\textcolor{b l u e}{2 x}$ since the equation consists of fractions with denominators $\textcolor{\mathmr{and} a n \ge}{x}$ and $\textcolor{\mathmr{and} a n \ge}{2}$. When you multiply the whole equation by $\textcolor{b l u e}{2 x}$, you can get rid of the denominators.

$\frac{1}{\textcolor{\mathmr{and} a n \ge}{x}} + 1 = \frac{x}{\textcolor{\mathmr{and} a n \ge}{2}}$

$\textcolor{b l u e}{2 x} \left(\frac{1}{x} + 1\right) = \textcolor{b l u e}{2 x} \left(\frac{x}{2}\right)$

$2$. Simplify.

$\frac{2 x}{x} + 2 x = \frac{2 {x}^{2}}{2}$

$\frac{2 \textcolor{red}{\cancel{\textcolor{b l a c k}{x}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{x}}}} + 2 x = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} {x}^{2}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}}$

$2 + 2 x = {x}^{2}$

$3$. Rewrite the equation in the form, $a {x}^{2} + b x + c = 0$.

${x}^{2} - 2 x - 2 = 0$

$4$. Solve for $x$ using the quadratic formula.

$\textcolor{p u r p \le}{a = 1} , \textcolor{t u r q u o i s e}{b = - 2} , \textcolor{b r o w n}{c = - 2}$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{- \left(- 2\right) \pm \sqrt{{\left(- 2\right)}^{2} - 4 \left(1\right) \left(- 2\right)}}{2 \left(1\right)}$

$x = \frac{2 \pm \sqrt{4 + 8}}{2}$

$x = \frac{2 \pm \sqrt{12}}{2}$

$x = \frac{2 \pm 2 \sqrt{3}}{2}$

$x = \frac{2 \left(1 \pm \sqrt{3}\right)}{2 \left(1\right)}$

$x = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \left(1 \pm \sqrt{3}\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \left(1\right)}$

$\textcolor{g r e e n}{x = 1 \pm \sqrt{3}}$

$\therefore$, $x = 1 \pm \sqrt{3}$.