# How do you solve  2x^2-3x-14=0 by completing the square?

Dec 22, 2016

$x = \frac{7}{2} \text{ }$ or $\text{ } x = - 2$

#### Explanation:

The difference of squares identity can be written:

${A}^{2} - {B}^{2} = \left(A - B\right) \left(A + B\right)$

We will use this with $A = \left(4 x - 3\right)$ and $B = 11$ later.

$\textcolor{w h i t e}{}$
Given:

$2 {x}^{2} - 3 x - 14 = 0$

To avoid fractions as much as possible, let us premultiply this quadratic by $8 = 2 \cdot {2}^{2}$. One factor of $2$ makes the leading term a perfect square, then the other ${2}^{2}$ deals with the $2$ denominator in "$\frac{b}{2 a}$" which would otherwise cause us to have to work with $\frac{1}{2}$'s.

So:

$0 = 8 \left(2 {x}^{2} - 3 x - 14\right)$

$\textcolor{w h i t e}{0} = 16 {x}^{2} - 24 x - 112$

$\textcolor{w h i t e}{0} = {\left(4 x\right)}^{2} - 2 \left(4 x\right) \left(3\right) + 9 - 121$

$\textcolor{w h i t e}{0} = {\left(4 x - 3\right)}^{2} - {11}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(4 x - 3\right) - 11\right) \left(\left(4 x - 3\right) + 11\right)$

$\textcolor{w h i t e}{0} = \left(4 x - 14\right) \left(4 x + 8\right)$

$\textcolor{w h i t e}{0} = \left(2 \left(2 x - 7\right)\right) \left(4 \left(x + 2\right)\right)$

$\textcolor{w h i t e}{0} = 8 \left(2 x - 7\right) \left(x + 2\right)$

Hence:

$x = \frac{7}{2} \text{ }$ or $\text{ } x = - 2$