How do you solve using the completing the square method 9x^2 - 6x + 1=0?

Jun 9, 2018

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

Explanation:

$9 {x}^{2} - 6 x + 1 = 0$

First, divide every term by 9

${x}^{2} - \frac{6}{9} x + \frac{1}{9} = 0$

Move your $\frac{1}{9}$ to the RHS

${x}^{2} - \frac{6}{9} x = - \frac{1}{9}$

Simplify $\frac{6}{9}$

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

Now, your general form of a quadratic equation is $a {x}^{2} + b x + c = 0$
To complete the square, your equation looks something like this $a {x}^{2} + b x + {\left(\frac{b}{2}\right)}^{2} = - c$

What this means is that you divide your coefficient of x by 2 and then square your entire number

${x}^{2} - \frac{2}{3} x + {\left(\frac{\frac{2}{3}}{2}\right)}^{2} = - \frac{1}{9} + {\left(\frac{\frac{2}{3}}{2}\right)}^{2}$

${x}^{2} - \frac{2}{3} x + {\left(\frac{1}{3}\right)}^{2} = - \frac{1}{9} + \frac{1}{9}$

${x}^{2} - \frac{2}{3} x + {\left(\frac{1}{3}\right)}^{2} = 0$

${\left(x + \frac{1}{3}\right)}^{2} = 0$

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