# How do you solve using the completing the square method x(6x - 5) = 6?

May 9, 2016

$x = + \frac{3}{2} \text{ and } - \frac{2}{3}$

#### Explanation:

Multiply out the racket giving:$\text{ } 6 {x}^{2} - 5 x = 6$

Subtract 6 from both sides

$\text{ } 6 {x}^{2} - 5 x - 6 = 0$

Write as$\text{ } 6 \left({x}^{2} - \frac{5}{6} x\right) - 6 = 0$

Take the square outside the bracket and add the correction constant k

$6 {\left(x - \frac{5}{6} x\right)}^{2} - 6 + k = 0$

Remove the $x$ from $- \frac{5}{6} x$

$6 {\left(x - \frac{5}{6}\right)}^{2} - 6 + k = 0$

Multiply the $- \frac{5}{6}$ by $\left(\frac{1}{2}\right)$

$6 {\left(x - \frac{5}{12}\right)}^{2} - 6 + k = 0$

'~~~~~~~~~~~~~ Comment ~~~~~~~~~~~~~~~~~~~~~~~~~
Consider the part of $\left(x - \frac{5}{12}\right) \left(x - \frac{5}{12}\right)$

The ${\left(- \frac{5}{12}\right)}^{2}$ is an introduced value that is not in the original equation so we remove it by subtraction. However, this introduced error is in fact $6 {\left(- \frac{5}{12}\right)}^{2}$ due to the 6 outside the brackets
$\implies k = \left(- 1\right) \times 6 \times {\left(- \frac{5}{12}\right)}^{2} = - 1 \frac{1}{24} = - \frac{25}{24}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

color(brown)(6(x-5/(12))^2-6+k=0)" "->" "color(blue)(6(x-5/(12))^2-6-25/24=0

$\implies 6 {\left(x - \frac{5}{12}\right)}^{2} - \frac{169}{24} = 0$

$6 {\left(x - \frac{5}{12}\right)}^{2} = \frac{169}{24}$

${\left(x - \frac{5}{12}\right)}^{2} = \frac{169}{144}$

Taking the square root of both sides

$x - \frac{5}{12} = \pm \frac{\sqrt{169}}{\sqrt{144}} = \pm \frac{13}{12}$

$x = \frac{5}{12} \pm \frac{13}{12}$

$x = + \frac{18}{12} \text{ and } - \frac{8}{12}$

$x = + \frac{3}{2} \text{ and } - \frac{2}{3}$