# How do you solve #sqrt(2x+20) + 2 = x# and find any extraneous solutions?

##### 1 Answer

Feb 17, 2017

#### Answer:

#### Explanation:

Always start by isolating the radical.

#sqrt(2x+ 20) = x - 2#

Square both sides.

#(sqrt(2x+ 20))^2 = (x - 2)^2#

#2x + 20 = x^2 - 4x + 4#

#0 = x^2 - 6x - 16#

#0 = (x - 8)(x + 2)#

#x = 8 and -2#

The last step to such problems is always to check your solutions within the original equation.

**Check: #x= 8#**

**Check: #x = -2#**

Therefore, the solution set is

Hopefully this helps!