Question

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

x=2
, x=−4
x equals 2 , x equals negative 4

x=10
, x=−8
x equals 10 , x equals negative 8

x=−2
, x=4
x equals negative 2 , x equals 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+315−−√, 11−315−−√
x equals 11 plus 3 Start Root 15 End Root comma 11 minus 3 Start Root 15 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+14−−√, −11−14−−√

Answer 2

Using the quadratic formula, find the solution to x2−6x−16=0
.(1 point)
Responses

x=8
, x=−2
x equals 8 , x equals negative 2

x=8
x equals 8

x=8
, x=−8
x equals 8 , x equals negative 8

x=−2

Answer 3

Which of the following is a solution to the quadratic equation 2x2−5x=6
? Assume that the solution has been rounded to the nearest hundredth, if applicable.(1 point)
Responses

x=−1.03
x equals negative 1.03

x=−0.89
x equals negative 0.89

x=1.03
x equals 1.03

x=0.89

Answer 4

How many real solutions does the following quadratic equation have?

4x2+x+3=0


(1 point)
Responses

one real solution
one real solution

three real solutions
three real solutions

no real solutions
no real solutions

two real solutions

Answer 5

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

1. Start with the equation:
\[
x^2 + 2x - 8 = 0
\]

2. Move the constant term to the other side:
\[
x^2 + 2x = 8
\]

3. 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 \( 1 \).
- Squaring \( 1 \) gives \( 1 \).

4. Add \( 1 \) to both sides:
\[
x^2 + 2x + 1 = 8 + 1
\]
\[
x^2 + 2x + 1 = 9
\]

5. The left-hand side can be factored as a perfect square:
\[
(x + 1)^2 = 9
\]

6. Take the square root of both sides:
\[
x + 1 = 3 \quad \text{or} \quad x + 1 = -3
\]

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

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

The correct response is:
**x=2, x=−4**.

Answer 6

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

1. Start with the equation:
\[
x^2 - 22x + 1 = 15
\]

2. Move the constant term \( 1 \) to the right side:
\[
x^2 - 22x = 15 - 1
\]
\[
x^2 - 22x = 14
\]

3. To 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 \( -11 \).
- Squaring \( -11 \) gives \( 121 \).

4. Add \( 121 \) to both sides:
\[
x^2 - 22x + 121 = 14 + 121
\]
\[
x^2 - 22x + 121 = 135
\]

5. The left-hand side can be factored as a perfect square:
\[
(x - 11)^2 = 135
\]

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

7. Solve for \( x \):
- From \( x - 11 = \sqrt{135} \):
\[
x = 11 + \sqrt{135}
\]
- From \( x - 11 = -\sqrt{135} \):
\[
x = 11 - \sqrt{135}
\]

Thus, the solutions are:
\[
x = 11 + \sqrt{135}, \quad x = 11 - \sqrt{135}
\]

Now, checking the options provided, the correct response is:
**x=11+√135, 11−√135**.

Answer 7

To find the solutions to the quadratic equation \( x^2 - 6x - 16 = 0 \) using the quadratic formula, we use the formula:

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

where \( a = 1 \), \( b = -6 \), and \( c = -16 \).

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

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

3. **Calculate the two possible solutions**:
- First solution:
\[
x = \frac{6 + 10}{2} = \frac{16}{2} = 8
\]
- Second solution:
\[
x = \frac{6 - 10}{2} = \frac{-4}{2} = -2
\]

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

The correct response is:
**x=8, x=−2**.

Answer 8

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

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

Next, we can apply the quadratic formula:

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

where \( a = 2 \), \( b = -5 \), and \( c = -6 \).

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

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

3. **Calculate the two possible solutions**:
- For \( x = \frac{5 + \sqrt{73}}{4} \):
- First, find \( \sqrt{73} \).
\[
\sqrt{73} \approx 8.544
\]
- Now substitute:
\[
x_1 = \frac{5 + 8.544}{4} \approx \frac{13.544}{4} \approx 3.386
\]

- For \( x = \frac{5 - \sqrt{73}}{4} \):
- Using the same approximation for \( \sqrt{73} \):
\[
x_2 = \frac{5 - 8.544}{4} \approx \frac{-3.544}{4} \approx -0.886
\]

- Rounding \( -0.886 \) to the nearest hundredth:
\[
x_2 \approx -0.89
\]

Therefore, the two solutions rounded to two decimal places are:
1. \( x \approx 3.39 \)
2. \( x \approx -0.89 \)

Among the provided options, the correct response is:
**x=−0.89**.

Answer 9

To determine the number of real solutions for the quadratic equation \( 4x^2 + x + 3 = 0 \), we can use the discriminant of a quadratic equation, which is given by:

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

In this case, for the equation \( ax^2 + bx + c = 0 \):
- \( a = 4 \)
- \( b = 1 \)
- \( c = 3 \)

Now, plug in these values into the discriminant formula:

\[
D = (1)^2 - 4(4)(3)
\]
\[
D = 1 - 48
\]
\[
D = 1 - 48 = -47
\]

The discriminant \( D \) is negative (\( -47 \)).

### Conclusion:
When the discriminant is negative, it indicates that the quadratic equation has **no real solutions**.

Thus, the correct response is:
**no real solutions**.