# How do you solve using the completing the square method x^2 + 3x – 6 = 0?

Jun 26, 2018

$x = - \frac{3}{2} \pm \frac{\sqrt{33}}{2}$

#### Explanation:

${x}^{2} + 3 x = 6$

${x}^{2} + 2 \left(\frac{3}{2}\right) x \textcolor{red}{+ \frac{9}{4}} = 6 \textcolor{red}{+ \frac{9}{4}}$

${\left(x + \frac{3}{2}\right)}^{2} = \frac{33}{4}$

$\textcolor{b l u e}{\text{take the square root of both sides}}$

$\sqrt{{\left(x + \frac{3}{2}\right)}^{2}} = \pm \sqrt{\frac{33}{4}} \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$

$x + \frac{3}{2} = \pm \frac{\sqrt{33}}{2}$

$\text{subtract "3/2" from both sides}$

$x = - \frac{3}{2} \pm \frac{\sqrt{33}}{2} \leftarrow \textcolor{red}{\text{exact solutions}}$

Jun 26, 2018

color(brown)(x = (-1/2) (3 - sqrt33 ), -(1/2) ( 3 + sqrt33)

#### Explanation:

${x}^{2} + 3 x - 6 = 0$

${x}^{2} + 3 x + {\left(\frac{3}{2}\right)}^{2} - 6 = {\left(\frac{3}{2}\right)}^{2} , \text{adding " (3/2)^2 " to both sides}$

${x}^{2} + 2 \cdot \frac{3}{2} \cdot x + {\left(\frac{3}{2}\right)}^{2} = {\left(\frac{3}{2}\right)}^{2} + 6$

${\left(x + \frac{3}{2}\right)}^{2} = \frac{33}{4}$

${\left(x + \frac{3}{2}\right)}^{2} = {\left(\sqrt{\frac{33}{4}}\right)}^{2}$

$x + \frac{3}{2} = \pm \sqrt{\frac{33}{4}} , \text{ taking squre root on both sides}$

$x = - \frac{3}{2} \pm \frac{\sqrt{33}}{2}$

color(brown)(x = (-1/2) (3 - sqrt33 ), -(1/2) ( 3 + sqrt33)