# How do you solve sqrt(2x+1) - sqrt(x-5) = 3 ?

Mar 27, 2016

$x = 21 \pm 6 \sqrt{7}$

#### Explanation:

$1$. Start by moving $- \sqrt{x - 5}$ to the right side of the equation.

$\sqrt{2 x + 1} - \sqrt{x - 5} = 3$

$\sqrt{2 x + 1} = 3 + \sqrt{x - 5}$

$2$. Since both sides contain radical signs, square both sides.

${\left(\sqrt{2 x + 1}\right)}^{2} = {\left(3 + \sqrt{x - 5}\right)}^{2}$

$\left(\sqrt{2 x + 1}\right) \left(\sqrt{2 x + 1}\right) = \left(3 + \sqrt{x - 5}\right) \left(3 + \sqrt{x - 5}\right)$

$3$. Simplify.

2x+1=color(red)9+6sqrt(x-5)color(blue)+(color(blue)x color(purple)(-5))

$2 x$ $\textcolor{b l u e}{- x} + 1$ $\textcolor{red}{- 9}$ $\textcolor{p u r p \le}{+ 5} = 6 \sqrt{x - 5}$

$x - 3 = 6 \sqrt{x - 5}$

$4$. Since the radical sign still exists on the right side of the equation, square both sides again.

${\left(x - 3\right)}^{2} = {\left(6 \sqrt{x - 5}\right)}^{2}$

$\left(x - 3\right) \left(x - 3\right) = {\left(6 \sqrt{x - 5}\right)}^{2}$

$5$. Simplify.

${x}^{2} - 6 x + 9 = 36 \left(x - 5\right)$

${x}^{2} - 6 x + 9 = 36 x - 180$

$6$. Move all terms to the left side of the equation.

$\textcolor{v i o \le t}{1} {x}^{2}$ $\textcolor{t u r q u o i s e}{- 42} x$ $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{+ 189} = 0$

$7$. Use the quadratic formula to solve for $x$.

$\textcolor{v i o \le t}{a = 1} \textcolor{w h i t e}{X X X} \textcolor{t u r q u o i s e}{b = - 42} \textcolor{w h i t e}{X X X} \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{c = 189}$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{- \left(\textcolor{t u r q u o i s e}{- 42}\right) \pm \sqrt{{\left(\textcolor{t u r q u o i s e}{- 42}\right)}^{2} - 4 \left(\textcolor{v i o \le t}{1}\right) \left(\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{189}\right)}}{2 \left(\textcolor{v i o \le t}{1}\right)}$

$x = \frac{42 \pm \sqrt{1764 - 756}}{2}$

$x = \frac{42 \pm \sqrt{1008}}{2}$

$x = \frac{42 \pm 12 \sqrt{7}}{2}$

$x = \frac{2 \left(21 \pm 6 \sqrt{7}\right)}{2 \left(2\right)}$

$x = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \left(21 \pm 6 \sqrt{7}\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \left(1\right)}$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} x = 21 \pm 6 \sqrt{7} \textcolor{w h i t e}{\frac{a}{a}} |}}}$