How do you solve x^2-6x-4=0 by completing the square?

Jul 16, 2015

$x = 3 + \sqrt{13} , 3 - \sqrt{13}$

Explanation:

${x}^{x} - 6 x - 4 = 0$

Completing the square means that we will force a perfect square trinomial on the left side of the equation, then solve for $x$.

Add $4$ to both sides of the equation.

${x}^{2} - 6 x = 4$

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

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

${x}^{2} - 6 x + 9 = 4 + 9$ =

${x}^{2} - 6 x + 9 = 13$

We now have a perfect square trinomial in the form ${a}^{2} - 2 a b + {b}^{2} = {\left(a - b\right)}^{2}$, where $a = x \mathmr{and} b = 3$.

Substitute ${\left(x - 3\right)}^{2}$ for ${x}^{2} - 6 x + 9$.

${\left(x - 3\right)}^{2} = 13$

Take the square root of both sides.

$x - 3 = \pm \sqrt{13}$

Add $3$ to both sides of the equation.

$x = 3 \pm \sqrt{13}$

Solve for $x$.

$x = 3 + \sqrt{13}$

$x = 3 - \sqrt{13}$