# How do you solve the quadratic equation by completing the square: 5x^2 + x = −5?

Jul 28, 2015

${x}_{1} = - \frac{1}{10} - \frac{3 i \sqrt{11}}{10}$, ${x}_{2} = - \frac{1}{10} + \frac{3 i \sqrt{11}}{10}$

#### Explanation:

The first thing you need to do is get the equation to the form

${x}^{2} + \frac{b}{a} x = - \frac{c}{a}$

In your case, $a = 5$, so you get

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} \cdot {x}^{2}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{5}}}} + \frac{1}{5} x = \frac{- 5}{5}$

${x}^{2} + \frac{1}{5} x = - 1$

In order to solve this quadratic by completing the square, you need to write the left side of the equation as the square of a binomial by adding a term to both sides of the equation.

This term will be determined by the coefficient of the $x$-term. What you need to do is divide this coefficient by 2, then square the result.

$\frac{1}{5} \cdot \frac{1}{2} = \frac{1}{10}$, then

${\left(\frac{1}{10}\right)}^{2} = \frac{1}{100}$

The quadratic can thus be written as

${x}^{2} + \frac{1}{5} x + \frac{1}{100} = - 1 + \frac{1}{100}$

The left side of the equation can now be written as

${x}^{2} + \frac{1}{5} x + \frac{1}{100} = {x}^{+} 2 \cdot \left(\frac{1}{10}\right) x + {\left(\frac{1}{10}\right)}^{2} = {\left(x + \frac{1}{10}\right)}^{2}$

This means that you have

${\left(x + \frac{1}{10}\right)}^{2} = - \frac{99}{100}$

Notice that this equation has no real solutions, but it does have two distinct complex solutions.

Take the square root of both sides to get

$\sqrt{{\left(x + \frac{1}{1}\right)}^{2}} = \sqrt{- \frac{99}{100}}$

$x + \frac{1}{10} = \pm \sqrt{- \frac{99}{100}} = \pm \frac{3 \sqrt{- 11}}{10}$

Use the fact that the square root of a negative number can be written as

$\textcolor{b l u e}{\sqrt{- n} = \sqrt{\left(- 1\right) \cdot n} = \sqrt{\left(- 1\right)} \cdot \sqrt{n} = \sqrt{{i}^{2}} \cdot \sqrt{n} = i \sqrt{n}}$

In your case, the right side of the equaation is equivalent to

$\pm \frac{3 \sqrt{\left(- 1\right) \cdot 11}}{10} = \pm \frac{3 i \sqrt{11}}{10}$

This means that you have

${x}_{1 , 2} = - \frac{1}{10} \pm \frac{3 i \sqrt{11}}{10}$

The two solutions of the quadratic will thus be

${x}_{1} = \textcolor{g r e e n}{- \frac{1}{10} - \frac{3 i \sqrt{11}}{10}}$ and ${x}_{2} = \textcolor{g r e e n}{- \frac{1}{10} + \frac{3 i \sqrt{11}}{10}}$