Posted by Jay on Wednesday, April 25, 2012 at 7:01pm.
Square root of (q^2+7q+6) minus q minus 3=0. Minus 3 and minus q are not under he radical sign.

Algebra  MathGuru, Thursday, April 26, 2012 at 7:16pm
I assume your problem is this:
√(q^2 + 7

Algebra  MathGuru, Thursday, April 26, 2012 at 7:25pm
Oops! Let's try this again.
I assume your problem is this:
√(q^2 + 7q + 6)  q  3 = 0
Add q and 3 to both sides of the equation to get the radical by itself (whatever operation you do to one side of an equation you must do to the other side as well):
√(q^2 + 7q + 6) = q + 3
Now square both sides to get rid of the radical:
q^2 + 7q + 6 = (q + 3)^2
q^2 + 7q + 6 = q^2 + 6q + 9
q = 3
Test the answer with the original equation to see if it checks out. It always helps to check your work!
I hope this helps and is what you were asking.
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