# How do you solve for x in 1/x + x = 4?

Refer to explanation

#### Explanation:

First the equation holds for $R - \left\{0\right\}$.Multiply with x both parts of equation you get

x*(1/x+x)=4*x=>1+x^2=4x=>x^2-4x+1=0=>

This is a trinomial with roots given by

${x}_{1} {,}_{2} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

where $a = 1 , b = - 4 , c = 1$

So the roots are
${x}_{1} = 2 - \sqrt{3}$ and ${x}_{2} = 2 + \sqrt{3}$

Both roots are acceptable since they belong to $R - \left\{0\right\}$