# How do you solve the quadratic equation by completing the square: 3x^2-9x=-1?

Jul 28, 2015

${x}_{1 , 2} = \frac{3}{2} \pm \frac{\sqrt{69}}{6}$

#### Explanation:

Starting from the general form of the quadratic equation

$a {x}^{2} + b x + c = 0$

divide verything by $a$ to get it to the form

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

In your cae, $a = 3$, which means that you get

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} {x}^{2}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}} - \frac{9}{3} x = - \frac{1}{3}$

${x}^{2} - 3 x = - \frac{1}{3}$

To solve this quadratic by completing the square, you need to write the left side of the equation as the square of a binomial. This can be done by adding a term to both sides of the equation.

The coefficient of the $x$-term will help you determine what you need to add to both sides.

So, divide this coefficient by 2, then square the result to get

$\frac{- 3}{2} = - \frac{3}{2}$, then

${\left(- \frac{3}{2}\right)}^{2} = \frac{9}{4}$

Adding $\frac{9}{4}$ to both sides of the equation will produce

${x}^{2} - 3 x + \frac{9}{4} = - \frac{1}{3} + \frac{9}{4} = \frac{23}{12}$

The general formula for the square of a binomial looks like this

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

You can use this formula to rewrite the left side of the equation as

${x}^{2} - 3 x + \frac{9}{4} = {x}^{2} + 2 \cdot \left(- \frac{3}{2}\right) x + {\left(- \frac{3}{2}\right)}^{2}$

${x}^{2} - 3 x + \frac{9}{4} = {\left(x - \frac{3}{2}\right)}^{2}$

${\left(x - \frac{3}{2}\right)}^{2} = \frac{23}{12}$

Take the square root of both sides to get

$\sqrt{{\left(x - \frac{3}{2}\right)}^{2}} = \sqrt{\frac{23}{12}}$

$x - \frac{3}{2} = \pm \frac{\sqrt{23}}{\sqrt{12}} = \pm \frac{\sqrt{23} \cdot \sqrt{3}}{2 \cdot 3} = \pm \frac{\sqrt{69}}{6}$

${x}_{1} = \textcolor{g r e e n}{\frac{3}{2} + \frac{\sqrt{69}}{6}}$ and ${x}_{2} = \textcolor{g r e e n}{\frac{3}{2} - \frac{\sqrt{69}}{6}}$