c squared-4c-12=0
we have to solve each equation by completing the square.
when you added 4 on your second step, your 'c' dissapeared... where'd it go?
typo error, it should be there as 4c
If you expand (x-2)^2 you will see I did not lose it.
To solve the equation c^2 - 4c - 12 = 0 by completing the square, follow these steps:
Step 1: Move the constant term to the other side of the equation:
c^2 - 4c = 12
Step 2: Take half of the coefficient of the linear term (in this case, -4) and square it:
(-4/2)^2 = (-2)^2 = 4
Step 3: Add the squared value from Step 2 to both sides of the equation:
c^2 - 4c + 4 = 12 + 4
(c - 2)^2 = 16
Step 4: Take the square root of both sides of the equation:
√((c - 2)^2) = ±√16
(c - 2) = ±4
Step 5: Solve for c:
c - 2 = 4 or c - 2 = -4
For the first case:
c = 4 + 2 = 6
For the second case:
c = -4 + 2 = -2
So, the solutions to the equation c^2 - 4c - 12 = 0 by completing the square are c = 6 and c = -2.
c^2 - 4c = 12
c^2 - 4 + 4 = 12 + 4
(c-2)^2 = 16
c-2 = ±4
c = 6 or c = -2