# How do you solve 6x + 4 = x^2  by completing the square?

##### 1 Answer
Jul 19, 2015

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

#### Explanation:

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

Completing the square means forcing a perfect square trinomial on the left side of the equation.

Bring all terms to the left side.

$- {x}^{2} + 6 x + 4 = 0$

Multiply boths sides times $- 1$.

$- 1 \left(- {x}^{2} + 6 x + 4 = 0\right)$ =

${x}^{2} - 6 x - 4 = 0$

Add $4$ to both sides of the equation.

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

Divide the x-term by 2 and square the result. Add 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 on the left side. ${a}^{2} - 2 a b + {b}^{2} = {\left(a - b\right)}^{2}$, where $a = x \mathmr{and} b = 3$.

Substitute ${\left(x - 3\right)}^{2}$ into the equation.

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

Take the square root of both sides and solve for $x$.

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

Add $3$ to both sides.

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

Solve for $x$.

$x = 3 + \sqrt{13}$

$x = 3 - \sqrt{13}$