Square root of (q^2+7q+6) minus q minus 3=0. Minus 3 and minus q are not under he radical sign.
I assume your problem is this:
√(q^2 + 7
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.
To solve the equation: √(q^2+7q+6) - q - 3 = 0, you need to follow these steps:
Step 1: Move the constants to the other side of the equation.
√(q^2+7q+6) = q + 3
Step 2: Square both sides of the equation to eliminate the square root.
(√(q^2+7q+6))^2 = (q + 3)^2
Simplifying the equation:
q^2 + 7q + 6 = (q + 3)^2
Step 3: Expand the right side of the equation.
q^2 + 7q + 6 = q^2 + 6q + 9
Step 4: Combine like terms.
7q = 6q + 9 - 6
Simplifying the equation:
7q - 6q = 3
Step 5: Solve for q.
q = 3
Therefore, the solution to the equation √(q^2+7q+6) - q - 3 = 0 is q = 3.