# How do you solve the quadratic equation by completing the square: x^2-6x-8=0?

Jul 14, 2015

$x = 3 + \sqrt{17} , 3 - \sqrt{17}$

#### Explanation:

${x}^{2} - 6 x - 8 = 0$

In order to solve a quadratic equation by completing the square, we must force a perfect square trinomial on the left side. ${a}^{2} + 2 a b + {b}^{2} = {\left(a + b\right)}^{2}$

Add $8$ to both sides of the equation.

${x}^{2} - 6 x = 8$

Divide the coefficient of the $x$ term by $2$ and square the result. Add it to both sides of the equation.

${\left(\frac{- 6}{2}\right)}^{2} = {\left(- 3\right)}^{2} = 9$

${x}^{2} - 6 x + 9 = 8 + 9$ =

${x}^{2} - 6 x + 9 = 17$

We now have a perfect square trinomial on the left side, in which
$a = x ,$ and $b = - 3$.

Now we can solve for $x$.

${\left(x - 3\right)}^{2} = 17$

Take the square root of both sides.

$\left(x - 3\right) = \pm \sqrt{17}$

$x = 3 \pm \sqrt{17}$

$x = 3 + \sqrt{17}$

$x = 3 - \sqrt{17}$