# How do you solve the quadratic equation by completing the square: x(6x-5) = 6?

Jul 22, 2015

The two solutions are $x = \frac{3}{2}$ and $x = - \frac{2}{3}$.

#### Explanation:

First use the distributive property to write $x \left(6 x - 5\right) = 6$ as $6 {x}^{2} - 5 x = 6$. Next factor the coefficient of ${x}^{2}$ out on the left side to write it as $6 \left({x}^{2} - \left(\frac{5}{6}\right) x\right) = 6$.

We can now complete the square: Take the coefficient of $x$ inside the parentheses, the $- \frac{5}{6}$, divide it by 2 to get $- \frac{5}{12}$, then square that number to get $\frac{25}{144}$.

Next, add the $\frac{25}{144}$ inside the parentheses on the left and compensate for that (balance the equation) by adding $6 \cdot \frac{25}{144} = \frac{25}{24}$ to the right hand side:

$6 \left({x}^{2} - \left(\frac{5}{6}\right) x + \frac{25}{144}\right) = 6 + \frac{25}{24} = \frac{144 + 25}{24} = \frac{169}{24}$.

The reason this is a good idea is that this technique has made the expression inside the parentheses on the left a perfect square. This last equation can be written as

$6 {\left(x - \frac{5}{12}\right)}^{2} = \frac{169}{24}$ (note also the appearance of the $- \frac{5}{12}$ again...this is no coincidence).

To finish, divide both sides by $6$ to get ${\left(x - \frac{5}{12}\right)}^{2} = \frac{169}{144}$ and then take the square root of both sides, allowing a $\setminus \pm$ sign on the right to get $x - \frac{5}{12} = \setminus \pm \frac{13}{12}$.

Now add $\frac{5}{12}$ to everything to get $x = \frac{5}{12} \setminus \pm \frac{13}{12}$. The two solutions are $x = \frac{18}{12} = \frac{3}{2}$ and $x = - \frac{8}{12} = - \frac{2}{3}$.