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

Jan 28, 2016

 x = ± sqrt3 - 3

#### Explanation:

By adding 9 to both sides of the equation to obtain:

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

now ${\left(x + 3\right)}^{2} = 9 - 6 = 3$

${\left(x + 3\right)}^{2} \textcolor{b l a c k}{\text{ is a perfect square }}$

Taking the 'square root' of both sides :

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

hence x + 3 = ± $\sqrt{3}$

so x = ±$\sqrt{3} - 3$

there are two $x$ values

${x}_{1} = \frac{- 3 + \sqrt{15} i}{2}$

${x}_{2} = \frac{- 3 - \sqrt{15} i}{2}$

#### Explanation:

Completing the square method:
Do this only when the numerical coefficient of ${x}^{2}$ is $1$.
Start with the numerical coefficient of $x$ which is the number $3$.
Divide this number by 2 then square the result. That is

${\left(\frac{3}{2}\right)}^{2} = \frac{9}{4}$

Add $\frac{9}{4}$ to both sides of the equation

${x}^{2} + 3 x + \frac{9}{4} + 6 = 0 + \frac{9}{4}$

the first three terms now become one group which is a PST-Perfect Square Trinomial

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

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

${\left(x + \frac{3}{2}\right)}^{2} = \frac{9}{4} - 6$ after transposing the $6$ to the right side

${\left(x + \frac{3}{2}\right)}^{2} = \frac{9 - 24}{4}$

$\sqrt{{\left(x + \frac{3}{2}\right)}^{2}} = \pm \sqrt{\frac{9 - 24}{4}}$

$x + \frac{3}{2} = \pm \sqrt{\frac{- 15}{4}}$

$x + \frac{3}{2} = \pm \frac{\sqrt{- 15}}{\sqrt{4}}$

$x + \frac{3}{2} = \pm \frac{\sqrt{- 15}}{2}$

Finally, transpose the $\frac{3}{2}$ to the right side of the equation

$x = - \frac{3}{2} \pm \frac{\sqrt{- 15}}{2}$

take note: $\sqrt{- 15} = \sqrt{15} \cdot \sqrt{- 1} = \sqrt{15} i$

therefore

$x = - \frac{3}{2} \pm \frac{\sqrt{15} i}{2}$

there are two $x$ values

${x}_{1} = \frac{- 3 + \sqrt{15} i}{2}$

${x}_{2} = \frac{- 3 - \sqrt{15} i}{2}$