# How do you solve sqrt (3x+10 ) =x and find any extraneous solutions?

Sep 2, 2016

Solution is $x = 5$ ($x = - 2$ is extraneous solution).

#### Explanation:

Here we can not have $3 x + 10 < 0$ i.e. $x < - 10 \setminus 3$ and any value of $x < - \frac{10}{3}$ is extraneous.

Squaring both sides in $\sqrt{3 x + 10} = x$, we get

$3 x + 10 = {x}^{2}$ or

${x}^{2} - 3 x - 10 = 0$ or

${x}^{2} - 5 x + 2 x - 10 = 0$ or

$x \left(x - 5\right) + 2 \left(x - 5\right)$ or

$\left(x + 2\right) \left(x - 5\right) = 0$ i.e.

either $x + 2 = 0$ i.e. $x = - 2$

or $x - 5 = 0$ i.e. $x = 5$

As none of these id less than $- \frac{10}{3}$,

Solution is $x = - 2$ or $x = 5$, but $x = - 2$ id extraneous as putting $x = - 2$ in given equation gives us $2 = - 2$.

Hence solution is $x = 5$.