# How do you solve sqrt (4x + 8) - 1 = x - 2?

Jan 26, 2016

1) leave the square root alone on 1 side:
$\sqrt{4 x + 8} = x - 2 + 1 = x - 1$

2) square both sides:
$4 x + 8 = {\left(x - 1\right)}^{2}$

3) Expand the right side:${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$
$4 x - 8 = {x}^{2} - 2 x + 1$

4) Move everything to one side and you get:
${x}^{2} - 6 x + 9 = 0$

5) This. luckily factors out to:
${\left(x - 3\right)}^{2} = 0$
so $x = 3$

Jan 31, 2016

$x = 7 , - 1$

#### Explanation:

$\sqrt{4 x + 8} - 1 = x - 2$

$\rightarrow \sqrt{4 x + 8} = x - 2 + 1$

$\rightarrow \sqrt{4 x + 8} = x - 1$

Square both sides:

$\rightarrow {\left(\sqrt{4 x - 8}\right)}^{2} = {\left(x - 1\right)}^{2}$

Use the formula ${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$ for the equation on the R.h.s

$\rightarrow 4 x - 8 = {x}^{2} - 2 x + 1$

$\rightarrow 0 = {x}^{2} - 2 x + 1 - 4 x 8$

$\rightarrow 0 = {x}^{2} - 6 x - 7$

$\rightarrow x = 7 , - 1$