# How do you solve using completing the square method x^2+x-6=0?

Apr 1, 2016

Using the completing the squares method as requested.
$x = 2$ or $x = - 3$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + x - 6 = 0$

Adding $6$ to both sides to clear the constant from the left side
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + x = 6$

If ${x}^{2} + x$ are the first two terms of an expanded squared binomial
${\left(x + a\right)}^{2} = {x}^{2} + 2 a x + {a}^{2}$
then $2 a = 1$ and $a = \frac{1}{2}$

Therefore, in order to complete the square, we will need to add ${a}^{2} = {\left(\frac{1}{2}\right)}^{2}$ to both sides:
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + x \textcolor{red}{+ {\left(\frac{1}{2}\right)}^{2}} = 6 \textcolor{red}{+ {\left(\frac{1}{2}\right)}^{2}}$

Rewriting as a squared binomial and simplifying the right side
$\textcolor{w h i t e}{\text{XXX}} {\left(x + \frac{1}{2}\right)}^{2} = 6 \frac{1}{4} = \frac{25}{4}$

Taking the square root of both sides:
$\textcolor{w h i t e}{\text{XXX}} x + \frac{1}{2} = \pm \sqrt{\frac{25}{4}} = \pm \frac{5}{2}$

Subtracting $\frac{1}{2}$ from both sides
$\textcolor{w h i t e}{\text{XXX")x=5/2-1/2=2color(white)("XX")orcolor(white)("XX}} x = - \frac{5}{2} - \frac{1}{2} = - 3$