# How do you solve sqrt(2x+20)+2=x?

Aug 8, 2015

$x = 8$

#### Explanation:

The trick to solve these kind of problems is to take all the non-radical expressions on one side, and then square both sides to remove the radical. Let us try out our strategy for this particular problem.

$\sqrt{2 x + 20} + 2 = x$
$\implies \sqrt{2 x + 20} = x - 2$
$\implies 2 x + 20 = {\left(x - 2\right)}^{2}$ (Squaring both sides)
$\implies 2 x + 20 = {x}^{2} - 4 x + 4$
$\implies {x}^{2} - 6 x - 16 = 0$
$\implies {x}^{2} - 8 x + 2 x - 16 = 0$
$\implies x \left(x - 8\right) + 2 \left(x - 8\right) = 0$
$\implies \left(x + 2\right) \left(x - 8\right) = 0$
$\implies x = - 2$ or $x = 8$

Now, we need to check the validity of our solution.
First, we check $x = 8$. Putting the value in the LHS, we get $8$, so it is a solution.
Next we check $x = - 2$. Putting the value in the LHS, we get $6$. But the RHS$= x = - 2$. So $- 2$ is not a valid solution of the equation. These kind of roots are called extraneous roots in literature.