To determine if there are two solutions to the equation \( \sqrt{9x + 9} = x + 1 \), we can first square both sides to eliminate the square root:
\[ 9x + 9 = (x + 1)^2 \]
Expanding the right side, we have:
\[ 9x + 9 = x^2 + 2x + 1 \]
Now, rearranging the equation gives:
\[ 0 = x^2 + 2x + 1 - 9x - 9 \]
Simplifying this, we obtain:
\[ 0 = x^2 - 7x - 8 \]
Next, we can rearrange this into standard form:
\[ x^2 - 7x - 8 = 0 \]
To solve for \( x \), we can apply the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -7 \), and \( c = -8 \):
\[ x = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} \]
Calculating the discriminant:
\[ (-7)^2 - 4 \cdot 1 \cdot (-8) = 49 + 32 = 81 \]
Since the discriminant \( 81 \) is positive, there are two distinct real solutions.
Next, we find the solutions:
\[ x = \frac{7 \pm \sqrt{81}}{2} = \frac{7 \pm 9}{2} \]
Calculating the two possible values for \( x \):
- \( x = \frac{7 + 9}{2} = \frac{16}{2} = 8 \)
- \( x = \frac{7 - 9}{2} = \frac{-2}{2} = -1 \)
Now we have two potential solutions: \( x = 8 \) and \( x = -1 \).
However, we need to check if both solutions satisfy the original equation \( \sqrt{9x + 9} = x + 1 \):
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For \( x = 8 \): \[ \sqrt{9(8) + 9} = \sqrt{72 + 9} = \sqrt{81} = 9 \] and \[ 8 + 1 = 9 \] This works.
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For \( x = -1 \): \[ \sqrt{9(-1) + 9} = \sqrt{-9 + 9} = \sqrt{0} = 0 \] and \[ -1 + 1 = 0 \] This also works.
Since both \( x = 8 \) and \( x = -1 \) are valid solutions, the statement "There are two solutions to the following equation" is True.
Final Answer: True