To solve the equation \(\sqrt{6x + 6} = x + 1\), let's follow these steps:
-
Square both sides to remove the square root: \[ (\sqrt{6x + 6})^2 = (x + 1)^2 \] This simplifies to: \[ 6x + 6 = (x + 1)(x + 1) \] \[ 6x + 6 = x^2 + 2x + 1 \]
-
Rearrange the equation to set it to zero: \[ 0 = x^2 + 2x + 1 - 6x - 6 \] \[ 0 = x^2 - 4x - 5 \]
-
Factor the quadratic: \[ 0 = (x - 5)(x + 1) \]
-
Set each factor to zero: \[ x - 5 = 0 \implies x = 5 \] \[ x + 1 = 0 \implies x = -1 \]
-
Check both solutions in the original equation \(\sqrt{6x + 6} = x + 1\):
-
For \(x = 5\): \[ \sqrt{6(5) + 6} = \sqrt{30 + 6} = \sqrt{36} = 6 \] \[ 5 + 1 = 6 \] This is valid.
-
For \(x = -1\): \[ \sqrt{6(-1) + 6} = \sqrt{-6 + 6} = \sqrt{0} = 0 \] \[ -1 + 1 = 0 \] This is also valid.
-
Thus, the solutions to the equation \(\sqrt{6x + 6} = x + 1\) are: \[ \boxed{5 \text{ and } -1} \]