Question

Complete the square to solve the following quadratic equation: x2+2x−8=0 .(1 point) Responses x=−2, x=4 x equals negative 2 , x equals 4 x=−10, x=8 x equals negative 10 , x equals 8 x=2, x=−4 x equals 2 , x equals negative 4 x=10, x=−8

Answer 1

Complete the square to solve the following quadratic equation: x2−22x+1=15 .(1 point) Responses x=−11+14−−√, −11−14−−√ x equals negative 11 plus Start Root 14 End Root comma negative 11 minus Start Root 14 End Root x=−11+315−−√, −11−315−−√ x equals negative 11 plus 3 Start Root 15 End Root comma negative 11 minus 3 Start Root 15 End Root x=11+14−−√, 11−14−−√ x equals 11 plus Start Root 14 End Root comma 11 minus Start Root 14 End Root x=11+315−−√, 11−315−−√

Answer 2

To solve the quadratic equation \( x^2 + 2x - 8 = 0 \) by completing the square, follow these steps:

1. Move the constant to the right side of the equation:
\[
x^2 + 2x = 8
\]

2. To complete the square, take half of the coefficient of \( x \) (which is \( 2 \)), square it, and add it to both sides. Half of \( 2 \) is \( 1 \), and \( 1^2 = 1 \):
\[
x^2 + 2x + 1 = 8 + 1
\]
\[
x^2 + 2x + 1 = 9
\]

3. The left side can now be factored:
\[
(x + 1)^2 = 9
\]

4. Now, take the square root of both sides:
\[
x + 1 = 3 \quad \text{or} \quad x + 1 = -3
\]

5. Solve for \( x \):
\[
x + 1 = 3 \quad \Rightarrow \quad x = 2
\]
\[
x + 1 = -3 \quad \Rightarrow \quad x = -4
\]

Thus, the solutions are:
\[
x = 2 \quad \text{and} \quad x = -4
\]

So, the correct answer is:
**x equals 2, x equals negative 4**.

Answer 3

To solve the given quadratic equation \( x^2 - 22x + 1 = 15 \) by completing the square, we will follow these steps:

1. **Rearrange the equation**:
Start by moving the constant term from the right side:
\[
x^2 - 22x + 1 - 15 = 0
\]
Simplifying gives:
\[
x^2 - 22x - 14 = 0
\]

2. **Move the constant to the right side**:
\[
x^2 - 22x = 14
\]

3. **Complete the square**:
Take half of the coefficient of \( x \) (which is \( -22 \)), square it, and add it to both sides. Half of \( -22 \) is \( -11 \), and \( -11^2 = 121 \):
\[
x^2 - 22x + 121 = 14 + 121
\]
\[
x^2 - 22x + 121 = 135
\]

4. **Factor the left side**:
\[
(x - 11)^2 = 135
\]

5. **Take the square root of both sides**:
\[
x - 11 = \sqrt{135} \quad \text{or} \quad x - 11 = -\sqrt{135}
\]

6. **Solve for \( x \)**:
\[
x = 11 + \sqrt{135} \quad \text{or} \quad x = 11 - \sqrt{135}
\]

Now, since \( \sqrt{135} = \sqrt{9 \cdot 15} = 3\sqrt{15} \), the final solutions are:
\[
x = 11 + 3\sqrt{15} \quad \text{and} \quad x = 11 - 3\sqrt{15}
\]

Therefore, the correct response is:
**x equals 11 plus 3 Start Root 15 End Root, 11 minus 3 Start Root 15 End Root**.