Complete the square to solve the following quadratic equation: x2+2x−8=0

Question

.(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

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 2

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 3

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 4

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 5

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 6

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 7

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 8

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 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.