Question

Complete the square to solve the following quadratic equation: x^2+2x−8=0 .(1 point)

x=−10 , x=8
x=2, x=-4
x=-2, x=4
x=10, x=-8

Answer 1

Complete the square to solve the following quadratic equation: x^2−22x+1=15 .(1 point) Responses x=11+315−−√, 11−315−−√ x equals 11 plus 3 Start Root 15 End Root comma 11 minus 3 Start Root 15 End Root 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+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

Using the quadratic formula, find the solution to x^2−6x−16=0 .(1 point) Responses x=8 , x=−2 x equals 8 , x equals negative 2 x=8 x equals 8 x=−2 x equals negative 2 x=8 , x=−8 x equals 8 , x equals negative 8

Answer 3

Which of the following is a solution to the quadratic equation 2x^2−5x=6 ? Assume that the solution has been rounded to the nearest hundredth, if applicable.(1 point) Responses x=−0.89 x equals negative 0.89 x=0.89 x equals 0.89 x=−1.03 x equals negative 1.03 x=1.03

Answer 4

How many real solutions does the following quadratic equation have? 4x^2+x+3=0 (1 point) Responses two real solutions two real solutions no real solutions no real solutions one real solution one real solution three real solutions

Answer 5

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

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

2. **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 \( 1^2 = 1 \).

So we add 1 to both sides:
\[
x^2 + 2x + 1 = 8 + 1
\]
This simplifies to:
\[
(x + 1)^2 = 9
\]

3. **Take the square root of both sides**:
\[
x + 1 = \pm 3
\]

4. **Solve for \( x \)**:
- For \( x + 1 = 3 \):
\[
x = 3 - 1 = 2
\]
- For \( x + 1 = -3 \):
\[
x = -3 - 1 = -4
\]

Thus, the solutions to the equation are \( x = 2 \) and \( x = -4 \).

From the given options, the correct pair is:
**x = 2, x = -4**.

Answer 6

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

1. **Move the constant term to the right side of the equation**:
\[
x^2 - 22x + 1 - 15 = 0 \implies x^2 - 22x - 14 = 0
\]

2. **Rewrite the equation**:
\[
x^2 - 22x = 14
\]

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 \( (-11)^2 = 121 \).

So we add 121 to both sides:
\[
x^2 - 22x + 121 = 14 + 121
\]
This simplifies to:
\[
(x - 11)^2 = 135
\]

4. **Take the square root of both sides**:
\[
x - 11 = \pm \sqrt{135}
\]

5. **Solve for \( x \)**:
\[
x = 11 \pm \sqrt{135}
\]
Since \( \sqrt{135} = \sqrt{9 \cdot 15} = 3\sqrt{15} \), we have:
\[
x = 11 + 3\sqrt{15} \quad \text{and} \quad x = 11 - 3\sqrt{15}
\]

Thus, the solutions to the equation are:
\[
x = 11 + 3\sqrt{15}, \quad x = 11 - 3\sqrt{15}
\]

From the given responses, the correct answer is:
**x equals \( 11 + 3\sqrt{15} \), \( 11 - 3\sqrt{15} \)**.

Answer 7

To solve the quadratic equation \( x^2 - 6x - 16 = 0 \) using the quadratic formula, we start with the formula:

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

In this case, the coefficients are:
- \( a = 1 \)
- \( b = -6 \)
- \( c = -16 \)

1. **Calculate the discriminant** (\( b^2 - 4ac \)):
\[
b^2 = (-6)^2 = 36
\]
\[
4ac = 4 \cdot 1 \cdot (-16) = -64
\]
\[
b^2 - 4ac = 36 - (-64) = 36 + 64 = 100
\]

2. **Substitute into the quadratic formula**:
\[
x = \frac{-(-6) \pm \sqrt{100}}{2 \cdot 1} = \frac{6 \pm 10}{2}
\]

3. **Calculate the two possible values for \( x \)**:
- For the positive root:
\[
x = \frac{6 + 10}{2} = \frac{16}{2} = 8
\]
- For the negative root:
\[
x = \frac{6 - 10}{2} = \frac{-4}{2} = -2
\]

Thus, the solutions to the equation are \( x = 8 \) and \( x = -2 \).

From the provided responses, the correct options are:
**x = 8, x = -2** and **x equals 8, x equals negative 2**.

Answer 8

To solve the quadratic equation \( 2x^2 - 5x = 6 \), we first rearrange it into standard form:

\[
2x^2 - 5x - 6 = 0
\]

Next, we can use the quadratic formula:

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

In this case:
- \( a = 2 \)
- \( b = -5 \)
- \( c = -6 \)

1. **Calculate the discriminant** (\( b^2 - 4ac \)):
\[
b^2 = (-5)^2 = 25
\]
\[
4ac = 4 \cdot 2 \cdot (-6) = -48
\]
\[
b^2 - 4ac = 25 - (-48) = 25 + 48 = 73
\]

2. **Substitute into the quadratic formula**:
\[
x = \frac{-(-5) \pm \sqrt{73}}{2 \cdot 2} = \frac{5 \pm \sqrt{73}}{4}
\]

3. **Calculate the two possible values for \( x \)**:
- Compute \( \sqrt{73} \) (approximately):
\[
\sqrt{73} \approx 8.54
\]
- Thus, the two potential solutions are:
\[
x_1 = \frac{5 + \sqrt{73}}{4} \approx \frac{5 + 8.54}{4} = \frac{13.54}{4} \approx 3.39
\]
\[
x_2 = \frac{5 - \sqrt{73}}{4} \approx \frac{5 - 8.54}{4} = \frac{-3.54}{4} \approx -0.89
\]

So the solutions rounded to the nearest hundredth are approximately \( x \approx 3.39 \) and \( x \approx -0.89 \).

From the given options, the correct answer is:
**x = −0.89** and **x equals negative 0.89**.

Answer 9

To determine how many real solutions the quadratic equation \( 4x^2 + x + 3 = 0 \) has, we can use the discriminant, which is given by the formula:

\[
D = b^2 - 4ac
\]

In this equation, the coefficients are:
- \( a = 4 \)
- \( b = 1 \)
- \( c = 3 \)

1. **Calculate the discriminant**:
\[
D = (1)^2 - 4(4)(3) = 1 - 48 = 1 - 48 = -47
\]

2. **Interpret the discriminant**:
- If \( D > 0 \), there are two distinct real solutions.
- If \( D = 0 \), there is one real solution (a repeated solution).
- If \( D < 0 \), there are no real solutions (the solutions are complex).

Since \( D = -47 \), which is less than zero, the quadratic equation \( 4x^2 + x + 3 = 0 \) has **no real solutions**.

From the provided responses, the correct answer is:
**no real solutions**.