Question

There are two solutions to the following equation: 9x+9−−−−−√=x+1
(1 point)
Responses

True
True

False

Answer 1

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 \( x \):

1. \( x = \frac{7 + 9}{2} = \frac{16}{2} = 8 \)
2. \( x = \frac{7 - 9}{2} = \frac{-2}{2} = -1 \)

Now we have two potential solutions: \( x = 8 \) and \( x = -1 \).

However, we need to check if both solutions satisfy the original equation \( \sqrt{9x + 9} = x + 1 \):

1. For \( x = 8 \):
\[
\sqrt{9(8) + 9} = \sqrt{72 + 9} = \sqrt{81} = 9
\]
and
\[
8 + 1 = 9
\]
This works.

2. 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**