# How do you solve x^2 - 8x - 20 = 0 by completing the square?

May 14, 2016

$x = 10$ or $x = - 2$

#### Explanation:

In addition to completing the square we can use the difference of squares identity:

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

with $a = \left(x - 4\right)$ and $b = 6$ as follows:

$0 = {x}^{2} - 8 x - 20$

$= {\left(x - 4\right)}^{2} - {4}^{2} - 20$

$= {\left(x - 4\right)}^{2} - 16 - 20$

$= {\left(x - 4\right)}^{2} - 36$

$= {\left(x - 4\right)}^{2} - {6}^{2}$

$= \left(\left(x - 4\right) - 6\right) \left(\left(x - 4\right) + 6\right)$

$= \left(x - 10\right) \left(x + 2\right)$

So $x = 10$ or $x = - 2$

May 14, 2016

$x = 10 , - 2$

#### Explanation:

color(red)("We know that " (a+b)^2=a^2 +2ab+b^2

color(red)("And "(a-b)^2=a^2 -2ab+b^2

${x}^{2} - 8 x - 20 = 0$

This can be written as ${x}^{2} + \left(2 \times x \times - 4\right) - 20 = 0$

Let $a = x$ and $b = - 4$

color(red)("Now, " (x-4)^2=x^2 -8x+16

${x}^{2} - 8 x - 20 = 0$

${x}^{2} - 8 x = 20$

${x}^{2} - 8 x \textcolor{red}{+ 16} = 20 \textcolor{red}{+ 16}$

${\left(x - 4\right)}^{2} = 36$

We know that $36 = {6}^{2} \text{ and } 36 = - {6}^{2}$

Find the square root of both the sides:

$x - 4 = \pm 6$

When, $x - 4 = 6$

$x - 4 \textcolor{red}{+ 4} = 6 \textcolor{red}{+ 4}$

$x = 10$

When, $x - 4 = - 6$

$x - 4 \textcolor{red}{+ 4} = - 6 \textcolor{red}{+ 4}$
$x = - 2$