SOLUTION: I don't know if is the right topic. the problem just says to solve. ((( x=2 sqrt (x+3) ))) I tried to square it on both sides and then factor it but it doesn't workout

Algebra ->  Square-cubic-other-roots -> SOLUTION: I don't know if is the right topic. the problem just says to solve. ((( x=2 sqrt (x+3) ))) I tried to square it on both sides and then factor it but it doesn't workout       Log On


   

Question 338125: I don't know if is the right topic.
the problem just says to solve.
((( x=2 sqrt (x+3) )))
I tried to square it on both sides and then factor it but it doesn't workout
please help me.
thanks.
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
(Use triple braces "{ { {" not triple parentheses "( ( (" when posting algebra expressions and equations on here)
I don't know if is the right topic.
the problem just says to solve.
x=2sqrt%28x%2B3%29
I tried to square it on both sides and then factor it but it doesn't workout
please help me.
thanks. 
------------------

x=2sqrt%28x%2B3%29+

%28x%29%5E2=%282sqrt%28x%2B3%29%29%5E2+

x%5E2=2%5E2%28x%2B3%29

x%5E2=4%28x%2B3%29

x%5E2=4x%2B12

Get 0 on the right:





But radical equations must be checked because sometimes
they give extraneous "solutions", which are not solutions
at all.

Checking these two answers in the original equation:

Checking x=6

+x=2sqrt%28x%2B3%29+
+6=2sqrt%286%2B3%29+
+6=2sqrt%289%29+ 
+6=2%2A3+
6=6

That checks so 6 is a solution:

Checking x=-2

+x=2sqrt%28x%2B3%29+
+-2=2sqrt%28-2%2B3%29+
+-2=2sqrt%281%29+ 
+-2=2%281%29+
-2=2

That doesn't check, so -2 is not a solution.

So there is just one solution, 6 

Edwin