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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