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

Aug 5, 2015

$x = 3 + \sqrt{6}$

#### Explanation:

Right from the start, you know that any solution you find must satisfy the conditions

• $2 x + 1 \ge 0 \implies x \ge - \frac{1}{2}$ $\to$ you need to take the square root of a positive number;

• $x - 2 \ge 0 \implies x \ge 2$ $\to$ the square root of a positive number is a positive number.

all in all, you need your valid solution(s) to satisfy the condition $x \ge 2$.

Since the radical term is already isolated on one side of the equation, proceed to square both sides to get

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

$2 x + 1 = {x}^{2} - 4 x + 4$

Next, move all your terms on one side of the equation and use the quadratic formula to determine the two solutions of this quadratic equation

${x}^{2} - 6 x + 3 = 0$

${x}_{1 , 2} = \frac{- \left(- 6\right) \pm \sqrt{{\left(- 6\right)}^{2} - 4 \cdot 1 \cdot 3}}{2 \cdot 1}$

${x}_{1 , 2} = \frac{6 \pm \sqrt{24}}{2} = \frac{6 \pm 2 \sqrt{6}}{2}$

This means that you have

${x}_{1} = \textcolor{g r e e n}{3 + \sqrt{6}}$ $\to$ valid solution because ${x}_{1} \ge 2$;

x_2 = cancel(color(red)(3-sqrt(6)) $\to$ extraneous solution because ${x}_{2} \cancel{\ge} 2$