# How do you solve 8x=3x^2 - 1 using completing the square?

Jul 28, 2015

${x}_{1} = \frac{4}{3} + \frac{\sqrt{19}}{3}$, ${x}_{2} = \frac{4}{3} - \frac{\sqrt{19}}{3}$

#### Explanation:

$8 x = 3 {x}^{2} - 1$

$3 {x}^{2} - 8 x - 1 = 0$

To solve this equation by completing the square you first need to find a way to write the left side of the equation as a square of a binomial.

Move the term that does not contain $x$ or ${x}^{2}$ to the right side of the equation.

$3 {x}^{2} - 8 x = 1$

Now divide everything by 3 to get

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} \cdot {x}^{2}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}} - \frac{8}{3} x = \frac{1}{3}$

${x}^{2} - \frac{8}{3} x = \frac{1}{3}$

Next, divide the coefficient of the $x$-term by 2, square it, then add the result to both sides of the equation. In your case, you have

$\left(- \frac{8}{3}\right) \cdot \frac{1}{2} = - \frac{4}{3}$, then

${\left(- \frac{4}{3}\right)}^{2} = \frac{16}{9}$

${x}^{2} - \frac{8}{3} x + \frac{16}{9} = \frac{1}{3} + \frac{16}{9}$

The left side of the equation can be written as

$\left({x}^{2} - \frac{8}{3} x + \frac{16}{9}\right) = {\left(x - \frac{4}{3}\right)}^{2}$

This will get you

${\left(x - \frac{4}{3}\right)}^{2} = \frac{19}{9}$

Take the square roots of both sides to find the two solutions

$\sqrt{{\left(x - \frac{4}{3}\right)}^{2}} = \sqrt{\frac{19}{9}}$

$x - \frac{4}{3} = \pm \frac{\sqrt{19}}{3} \implies {x}_{1 , 2} = \frac{4}{3} \pm \frac{\sqrt{19}}{3}$

The two solutions will be

x_1 = color(green)(4/3 + sqrt(19)/3 and ${x}_{2} = \textcolor{g r e e n}{\frac{4}{3} - \frac{\sqrt{19}}{3}}$