# How do you solve and check for extraneous solutions in 3sqrt(2x+4)=12?

Aug 10, 2015

#### Answer:

$x = 6$

#### Explanation:

First, isolate the radical on one side of the equation by dividing both sides by 

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} \cdot \sqrt{2 x + 4}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}} = \frac{12}{3}$

$\sqrt{2 x + 4} = 4$

Square both sides of the equation to get rid of the square root

${\left(\sqrt{2 x + 4}\right)}^{2} = {4}^{2}$

$2 x + 4 = 16$

Finally, isolate $x$ on one side of the equation by adding $- 4$ to both sides and dividing everything by $2$

$2 x + \textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} = 16 - 4$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} x}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} = \frac{12}{2}$

$x = \textcolor{g r e e n}{6}$