Question

complete the square to identify the two values that solve the following quadratic equation: x^2 + 8x = 9.

Answer 1

no it isn't it is just one and it is nor -9 or 1

Answer 2

To solve the quadratic equation \(x^2 + 8x = 9\) by completing the square, follow these steps:

1. **Move the constant to the other side**:
\[
x^2 + 8x - 9 = 0
\]
Now, we isolate the quadratic and linear terms:
\[
x^2 + 8x = 9
\]

2. **Complete the square**:
Take the coefficient of \(x\) (which is 8), divide it by 2, and square it:
\[
\left(\frac{8}{2}\right)^2 = 4^2 = 16
\]

3. **Add and subtract this square to the left side of the equation**:
\[
x^2 + 8x + 16 - 16 = 9
\]
This can be rewritten as:
\[
(x + 4)^2 - 16 = 9
\]

4. **Move the subtracted square back to the right**:
\[
(x + 4)^2 = 9 + 16
\]
\[
(x + 4)^2 = 25
\]

5. **Take the square root of both sides**:
\[
x + 4 = \pm 5
\]

6. **Solve for \(x\)**:
- For the positive square root:
\[
x + 4 = 5 \quad \Rightarrow \quad x = 5 - 4 \quad \Rightarrow \quad x = 1
\]
- For the negative square root:
\[
x + 4 = -5 \quad \Rightarrow \quad x = -5 - 4 \quad \Rightarrow \quad x = -9
\]

Thus, the two values that solve the equation \(x^2 + 8x = 9\) are:
\[
x = 1 \quad \text{and} \quad x = -9
\]

Answer 3

I apologize for the confusion. Let's carefully go through the steps to complete the square for the given equation \(x^2 + 8x = 9\) again to find the correct solutions.

1. **Start with the original equation**:
\[
x^2 + 8x = 9
\]

2. **Move the constant to the other side**:
\[
x^2 + 8x - 9 = 0
\]

3. **Complete the square on the left side**:
- Take the coefficient of \(x\) (which is 8), divide it by 2, and square it:
\[
\left(\frac{8}{2}\right)^2 = 4^2 = 16
\]
- Add and subtract this square inside the equation:
\[
x^2 + 8x + 16 - 16 = 9
\]
This gives us:
\[
(x + 4)^2 - 16 = 9
\]
Now add 16 to both sides:
\[
(x + 4)^2 = 25
\]

4. **Take the square root of both sides**:
\[
x + 4 = \pm 5
\]

5. **Solve for \(x\)**:
- For the positive root:
\[
x + 4 = 5 \quad \Rightarrow \quad x = 1
\]
- For the negative root:
\[
x + 4 = -5 \quad \Rightarrow \quad x = -9
\]

Therefore, the solutions are:
\[
x = 1 \quad \text{and} \quad x = -9
\]

If you meant that there is only one solution, we can check the discriminant of the equation to confirm whether it's a perfect square or if there's any other nuance to the problem.

The discriminant of the quadratic equation \(ax^2 + bx + c = 0\) is given by \(D = b^2 - 4ac\).

For our equation \(x^2 + 8x - 9 = 0\):
- \(a = 1\)
- \(b = 8\)
- \(c = -9\)

Calculating the discriminant:
\[
D = 8^2 - 4 \cdot 1 \cdot (-9) = 64 + 36 = 100
\]

Since the discriminant is positive, there are indeed two distinct real roots.

If there's something specific that might have led to the misunderstanding, please let me know!