How do you solve 2x ^ { 2} + 12x + 9= 0?

Feb 2, 2017

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

Explanation:

Complete the square then use the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = 2 x + 6$ and $b = 3 \sqrt{2}$ as follows:

$0 = 2 \left(2 {x}^{2} + 12 x + 9\right)$

$\textcolor{w h i t e}{0} = 4 {x}^{2} + 24 x + 18$

$\textcolor{w h i t e}{0} = {\left(2 x\right)}^{2} + 2 \left(2 x\right) \left(6\right) + {\left(6\right)}^{2} - 18$

$\textcolor{w h i t e}{0} = {\left(2 x + 6\right)}^{2} - {\left(3 \sqrt{2}\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(2 x + 6\right) - 3 \sqrt{2}\right) \left(\left(2 x + 6\right) + 3 \sqrt{2}\right)$

$\textcolor{w h i t e}{0} = \left(2 x + 6 - 3 \sqrt{2}\right) \left(2 x + 6 + 3 \sqrt{2}\right)$

Hence:

$x = \frac{1}{2} \left(- 6 \pm 3 \sqrt{2}\right) = - 3 \pm \frac{3 \sqrt{2}}{2}$