# How do you solve 2x^2 + 8x -3 =0 by completing the square?

Apr 12, 2017

#### Answer:

$\textcolor{g r e e n}{x = - 2 \pm \sqrt{\frac{11}{2}}}$

#### Explanation:

$2 {x}^{2} + 8 x - 3 = 0$

$\Rightarrow \textcolor{b l u e}{2} \left({x}^{2} + 4 x\right) = 3$

If $\textcolor{\mathmr{and} a n \ge}{4 x}$ is the middle term of a squared binomial with the form
$\textcolor{w h i t e}{\text{XXX}} {\left(x + a\right)}^{2} = \left({x}^{2} + \textcolor{\mathmr{and} a n \ge}{2 a x} + {a}^{2}\right)$
then
$\textcolor{w h i t e}{\text{XXX}} \textcolor{\mathmr{and} a n \ge}{a} = \textcolor{\mathmr{and} a n \ge}{2}$
and
$\textcolor{w h i t e}{\text{XXX}} \textcolor{g r e e n}{{a}^{2}} = \textcolor{g r e e n}{4}$ must be added to ${x}^{2} + 4 x$ to complete the square.

This will result in adding $\textcolor{b l u e}{2} \times \textcolor{g r e e n}{4}$ to the left side of the equation;
so we must add this to the right side as well to keep the equation valid.

$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{2} \left({x}^{2} + 4 x + \textcolor{g r e e n}{4}\right) = 3 + \textcolor{b l u e}{2} \times \textcolor{g r e e n}{4}$

Rewriting the parenthesized factor as a squared binomial and simplifying.

$\textcolor{w h i t e}{\text{XXX}} 2 {\left(x + 2\right)}^{2} = 11$

Now divide both sides by $2$, leaving only a squared binomial on the left:
$\textcolor{w h i t e}{\text{XXX}} {\left(x + 2\right)}^{2} = \frac{11}{2}$

Take the square root of both sides (don't forget that there will be both a positive and negative root on the right side):
$\textcolor{w h i t e}{\text{XXX}} x + 2 = \pm \sqrt{\frac{11}{2}}$

Subtract $2$ from both sides to isolate the variable $x$
$\textcolor{w h i t e}{\text{XXX}} x = - 2 \pm \sqrt{\frac{11}{2}}$

Apr 12, 2017

#### Answer:

$x \approx 0.345 \mathmr{and} x \approx - 4.345$

#### Explanation:

$2 {x}^{2} + 8 x - 3 = 0 \mathmr{and} 2 \left({x}^{2} + 4 x\right) - 3 = 0 \mathmr{and} 2 \left({x}^{2} + 4 x + 4\right) - 3 = 8$ [Adding 8 on both sides]

or $2 {\left(x + 2\right)}^{2} = 11 \mathmr{and} {\left(x + 2\right)}^{2} = \frac{11}{2} \mathmr{and} x + 2 = \pm \sqrt{\frac{11}{2}}$

or$x = - 2 \pm \sqrt{\frac{11}{2}} = - 2 \pm \frac{1}{2} \left(\sqrt{22}\right) \therefore x \approx 0.345 \mathmr{and} x \approx - 4.345$ [Ans]