How do you solve #sqrt4=sqrt(x+2)# and check your solution?

1 Answer
David L.
Apr 20, 2017

#x=2#

Explanation:

Raise both sides to the power of #^2#, doing so reduces the square roots on both sides of the equation to #1# and makes this problem simplier without changing the question.

#(sqrt4)^2=(sqrt(x+2))^2#

#4=x+2#

#x=2#

Checking the solution by substituiting #x=2 # into the original problem,

#sqrt4=sqrt(2+2#

#sqrt4=sqrt4#

Therefore #x=2# is the solution

Related questions
See all questions in Radical Equations
Impact of this question
1338 views around the world
You can reuse this answer
Creative Commons License