# How do you solve  2x^2 + 12x − 14 = 0 by completing the square?

Feb 12, 2017

$x = - 7 \textcolor{w h i t e}{\text{XX")orcolor(white)("XX}} x = + 1$

#### Explanation:

$2 {x}^{2} + 12 x - 14 = 0$

$\rightarrow 2 {x}^{2} + 12 x = 14$

$\rightarrow {x}^{2} + 6 x = 7$

Now to "complete the square":
If ${x}^{2} + 6 x$ are the first two terms of the expansion of a squared binomial: ${\left(x + a\right)}^{2} = {x}^{2} + 2 a x + {a}^{2}$
then $2 a x$ must equal $6 x$;
that is $a = 3$ and ${a}^{2} = 9$

To "complete the square" we must add $\textcolor{m a \ge n t a}{9}$ to the expression,
but we can only legally do this if we add $\textcolor{m a \ge n t a}{9}$ to both sides of the equation:
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + 6 x \textcolor{m a \ge n t a}{+ 9} = 7 \textcolor{m a \ge n t a}{+ 9}$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow {\left(x + 3\right)}^{2} = 16$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow x + 3 = \pm 4$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow x = - 3 \pm 4$

$\textcolor{w h i t e}{\text{XXX}}$which can be written as: $x = - 7 \textcolor{w h i t e}{\text{X")orcolor(white)("X}} x = 1$