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

Aug 19, 2015

$x = 3$

#### Explanation:

Right from the start, you know that you can only use positive values of $x$, since taking the square root of a real positive number will always produce a positive value.

This means that you need to have $x \ge 0$.

With this in mind, start by squaring both sides of the equation

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

$2 x + 3 = {x}^{2}$

Rearrange this equation by moving all the terms on one side

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

You can find the solutions to this quadratic equation by using the quadratic formula, which for a general form quadratic

$\textcolor{b l u e}{a {x}^{2} + b x + c = 0}$

allows you to find the roots by using the formula

$\textcolor{b l u e}{{x}_{1 , 2} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}}$

In your case, you can write

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

${x}_{1 , 2} = \frac{2 \pm \sqrt{16}}{2}$

${x}_{1 , 2} = \frac{2 \pm 4}{2} = \left\{\begin{matrix}{x}_{1} = \frac{2 + 4}{2} = 3 \\ {x}_{2} = \frac{2 - 4}{2} = - 1\end{matrix}\right.$

SInce ${x}_{2} = - 1$ does not satisfy the condition $x \ge 0$, your original equation will have only one solution, $x = \textcolor{g r e e n}{3}$.

You can do a quick check to make sure that the calculations are correct

$\sqrt{2 \cdot 3 + 3} = 3$

$\sqrt{9} = 3$

$3 = 3 \text{ } \textcolor{g r e e n}{\sqrt{}}$