SOLUTION: \/x+4=3 This is what I did: (\/x+4)^2=3^2 x+8=9 x=1 Is this correct or did I miss a step? Thank you! ac

Algebra ->  Radicals -> SOLUTION: \/x+4=3 This is what I did: (\/x+4)^2=3^2 x+8=9 x=1 Is this correct or did I miss a step? Thank you! ac      Log On


   

Question 57858: \/x+4=3
This is what I did:
(\/x+4)^2=3^2
x+8=9
x=1
Is this correct or did I miss a step?
Thank you!
ac
Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
I'm having trouble interpretting your problems because I know where your radicals begin, but I don't know where they end. If I misinterpret, let me know and I'll change it.
sqrt%28x%2B4%29=3
%28sqrt%28x%2B4%29%29%5E2=%283%29%5E2
x%2B4=9
x%2B4-4=9-4
x=5
Check your answers everytime that you do square root problems for extraneous soltuions:
sqrt%285%2B4%29=3
sqrt%289%29=3
3=3 Our solution is fine, x=5
Happy Calculating!!!