Question

Complete the square to identify the two values that solve the following quadratic equation: x2+8x=9 .(1 point) x=blank

Answer 1

Using the quadratic formula, find the solution to x2+6x+9=0 .(1 point)

Answer 2

Solve −2x2+3x−1=−2 using the quadratic formula. Round the solution(s) to the nearest hundredths. Enter the smaller solution first.(1 point) x=blank or blank

Answer 3

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

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

2. **Isolate the quadratic and linear terms:**
\[
x^2 + 8x = 9
\]

3. **Complete the square:**
To complete the square, take half of the coefficient of \( x \) (which is 8), square it, and add it to both sides. Half of 8 is 4, and squaring it gives \( 4^2 = 16 \).

\[
x^2 + 8x + 16 = 9 + 16
\]

4. **Rewrite the left side as a squared term:**
\[
(x + 4)^2 = 25
\]

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

6. **Solve for \( x \):**
- For the positive case:
\[
x + 4 = 5 \implies x = 5 - 4 = 1
\]
- For the negative case:
\[
x + 4 = -5 \implies x = -5 - 4 = -9
\]

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

So, the final answer for \( x \) is:
\[
x = 1, -9
\]

Answer 4

To solve the quadratic equation \( x^2 + 6x + 9 = 0 \) using the quadratic formula, we first identify the coefficients \( a \), \( b \), and \( c \) from the standard form \( ax^2 + bx + c = 0 \):

- \( a = 1 \)
- \( b = 6 \)
- \( c = 9 \)

The quadratic formula is given by:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Now, we can substitute the values of \( a \), \( b \), and \( c \) into the formula:

1. **Calculate \( b^2 - 4ac \):**

\[
b^2 - 4ac = 6^2 - 4 \cdot 1 \cdot 9 = 36 - 36 = 0
\]

2. **Substitute into the quadratic formula:**

Since \( b^2 - 4ac = 0 \), there is exactly one solution (a double root):

\[
x = \frac{-6 \pm \sqrt{0}}{2 \cdot 1}
\]

3. **Simplify:**

\[
x = \frac{-6 \pm 0}{2} = \frac{-6}{2} = -3
\]

Thus, the solution to the equation \( x^2 + 6x + 9 = 0 \) is:

\[
x = -3
\]

Answer 5

To solve the equation \(-2x^2 + 3x - 1 = -2\) using the quadratic formula, we first rearrange it into standard form \(ax^2 + bx + c = 0\).

1. **Rearrange the equation:**

\[
-2x^2 + 3x - 1 + 2 = 0
\]
\[
-2x^2 + 3x + 1 = 0
\]

Now we can identify the coefficients:
- \(a = -2\)
- \(b = 3\)
- \(c = 1\)

2. **Apply the quadratic formula:**

The quadratic formula is:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

3. **Calculate the discriminant:**

First, compute \(b^2 - 4ac\):

\[
b^2 - 4ac = 3^2 - 4 \cdot (-2) \cdot 1 = 9 + 8 = 17
\]

4. **Use the quadratic formula:**

\[
x = \frac{-3 \pm \sqrt{17}}{2 \cdot (-2)} = \frac{-3 \pm \sqrt{17}}{-4}
\]

To separate the two solutions, we have:

\[
x = \frac{-3 + \sqrt{17}}{-4} \quad \text{and} \quad x = \frac{-3 - \sqrt{17}}{-4}
\]

5. **Calculate the two solutions:**

**First solution:**

\[
x_1 = \frac{-3 + \sqrt{17}}{-4}
\]

**Second solution:**

\[
x_2 = \frac{-3 - \sqrt{17}}{-4}
\]

Next, we will calculate the numerical values:

- **Calculate \( \sqrt{17} \) (approximately):**

\( \sqrt{17} \approx 4.1231 \)

- Now substituting back:

\[
x_1 = \frac{-3 + 4.1231}{-4} = \frac{1.1231}{-4} \approx -0.2808
\]

\[
x_2 = \frac{-3 - 4.1231}{-4} = \frac{-7.1231}{-4} \approx 1.7808
\]

6. **Round the solutions to the nearest hundredths:**

- \( x_1 \approx -0.28 \)
- \( x_2 \approx 1.78 \)

Thus, the solutions to the equation \(-2x^2 + 3x - 1 = -2\) are:

\[
x = -0.28 \quad \text{or} \quad x = 1.78
\]

### Final Answer
**x = -0.28 or 1.78**