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

Apr 29, 2016

$x = - 4 \pm \sqrt{14}$

#### Explanation:

We will use the difference of squares identity, which can be written:

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

with $a = \left(x + 4\right)$ and $b = \sqrt{14}$ as follows:

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

$= {\left(x + 4\right)}^{2} - 16 + 2$

$= {\left(x + 4\right)}^{2} - {\left(\sqrt{14}\right)}^{2}$

$= \left(\left(x + 4\right) - \sqrt{14}\right) \left(\left(x + 4\right) + \sqrt{14}\right)$

=(x+4-sqrt(14))((x+4+sqrt(14))

Hence:

$x = - 4 \pm \sqrt{14}$

May 2, 2016

$x = \sqrt{14} - 4$ OR $x = - \sqrt{14} - 4$
$x = - 0.258$ OR $x = - 7.742$ (3 dec places)

#### Explanation:

Completing the square is based on the consistency of the answers to the square of a binomial.

${\left(x - 3\right)}^{2} = {x}^{2} - 6 x + 9$
${\left(x - 5\right)}^{2} = {x}^{2} - 10 x + 25$
${\left(x + 6\right)}^{2} = {x}^{2} + 12 x + 36$
In all of the products above, $a {x}^{2} + b x + c$ we see the following:

$a = 1$
The first and last terms, $a \mathmr{and} c$ are perfect squares.'
There is a specific relationship between 'b' - the coefficient of the $x$ term and 'c'. Half of b, squared equals c.

Knowing this, it is always possible to add in a missing value for $c$ to have the square of a binomial, which can then be written as (x+?)^2

In ${x}^{2} + 8 x + 2 = 0$, 2 is obviously not the correct value of $c$.
It is therefore moved to the right hand side and the wanted value of $c$ is added to BOTH sides of the equation.

${x}^{2} + 8 x + \textcolor{red}{16}$ = -2 + color(red)(16) rArr [16 = (8÷2)^2]
${\left(x + 4\right)}^{2} = 14 \Rightarrow$ where 4 is either from b÷2 or sqrt16
$x + 4 = \pm \sqrt{14} \Rightarrow$ take the square root of both sides

This gives 2 possible answers for $x$.

$x = \sqrt{14} - 4$ OR $x = - \sqrt{14} - 4$