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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

9 answers

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.

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−−√

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 \( 11 + 3\sqrt{15} \), \( 11 - 3\sqrt{15} \).

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

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.

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

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.

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

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.