I assume your problem is this:
√(q^2 + 7
Square root of (q^2+7q+6) minus q minus 3=0. Minus 3 and minus q are not under he radical sign.
2 answers
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.
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.
5 answers