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Complete the square to solve the following quadratic equation: x2−6x−20=20.(1 point)

Responses

x=52, x=−46
x equals 52 , x equals negative 46

x=10, x=−4
x equals 10 , x equals negative 4

x=3+210−−√, 3−210−−√
x equals 3 plus 2 Start Root 10 End Root comma 3 minus 2 Start Root 10 End Root

x=−10, x=4

11 answers

To solve the equation \( x^2 - 6x - 20 = 20 \) by completing the square, first, let's rewrite the equation in standard form:

  1. Move 20 to the left side: \[ x^2 - 6x - 20 - 20 = 0 \quad \Rightarrow \quad x^2 - 6x - 40 = 0 \]

  2. Now, we will complete the square for \( x^2 - 6x \):

    • Take half of the coefficient of \( x \) (which is -6), square it, and add it.
    • Half of -6 is -3, and squaring it gives us 9.

  3. Rewrite the quadratic: \[ x^2 - 6x + 9 - 9 - 40 = 0 \quad \Rightarrow \quad (x - 3)^2 - 49 = 0 \]

  4. Set the completed square equal to 49: \[ (x - 3)^2 = 49 \]

  5. Take the square root of both sides: \[ x - 3 = 7 \quad \text{or} \quad x - 3 = -7 \]

  6. Solve for \( x \):

    • \( x - 3 = 7 \) gives \( x = 10 \)
    • \( x - 3 = -7 \) gives \( x = -4 \)

So the solutions to the quadratic equation are: \[ x = 10 \quad \text{and} \quad x = -4 \]

Therefore, the correct response is: x=10, x=−4 x equals 10 , x equals negative 4

Using the quadratic formula, find the solution to 3x^2+18x+24=0.(1 point)
Responses

x=4, x=2
x equals 4 , x equals 2

x=−4
x equals negative 4

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

x=−2

To solve the quadratic equation \(3x^2 + 18x + 24 = 0\) using the quadratic formula, we start by identifying the coefficients \(a\), \(b\), and \(c\) from the standard form \(ax^2 + bx + c = 0\).

Here:

  • \(a = 3\)
  • \(b = 18\)
  • \(c = 24\)

The quadratic formula is given by:

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

Now, let's calculate \(b^2 - 4ac\):

\[ b^2 = 18^2 = 324 \] \[ 4ac = 4 \cdot 3 \cdot 24 = 288 \] \[ b^2 - 4ac = 324 - 288 = 36 \]

Since the discriminant (\(b^2 - 4ac\)) is positive, we will have two real solutions. Now, we can substitute back into the quadratic formula:

\[ x = \frac{-18 \pm \sqrt{36}}{2 \cdot 3} \] \[ x = \frac{-18 \pm 6}{6} \]

Now we find the two possible solutions:

  1. \(x = \frac{-18 + 6}{6} = \frac{-12}{6} = -2\)
  2. \(x = \frac{-18 - 6}{6} = \frac{-24}{6} = -4\)

Therefore, the solutions to the equation are:

\[ x = -2 \quad \text{and} \quad x = -4 \]

The correct response is: x=−2, x=−4 x equals negative 2 , x equals negative 4

Which of the following is a solution to the quadratic equation −x^2+x=−10? Assume that the solution has been rounded to the nearest hundredth, if applicable.(1 point)
Responses

x=0.77
x equals 0.77

x=0.27
x equals 0.27

x=3.7
x equals 3.7

x=1.7

To solve the quadratic equation \(-x^2 + x = -10\), we first need to rearrange it into standard form.

  1. We can add \(10\) to both sides of the equation:

    \[ -x^2 + x + 10 = 0 \]

  2. To make calculations easier, let's multiply the entire equation by \(-1\) to switch the signs:

    \[ x^2 - x - 10 = 0 \]

Now, we can use the quadratic formula, which is:

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

Here, \(a = 1\), \(b = -1\), and \(c = -10\).

  1. Calculate the discriminant:

    \[ b^2 - 4ac = (-1)^2 - 4(1)(-10) = 1 + 40 = 41 \]

  2. Substitute into the quadratic formula:

    \[ x = \frac{-(-1) \pm \sqrt{41}}{2 \cdot 1} \]

    \[ x = \frac{1 \pm \sqrt{41}}{2} \]

  3. Now, calculate the two potential solutions:

    • For \(x = \frac{1 + \sqrt{41}}{2}\):

      \( \sqrt{41} \approx 6.403 \)

      \[ x \approx \frac{1 + 6.403}{2} \approx \frac{7.403}{2} \approx 3.70 \]

    • For \(x = \frac{1 - \sqrt{41}}{2}\):

      \[ x \approx \frac{1 - 6.403}{2} \approx \frac{-5.403}{2} \approx -2.70 \]

Thus, rounding to the nearest hundredth, the solutions are approximately:

  1. \(x \approx 3.70\)
  2. \(x \approx -2.70\) (not amongst the choices)

The only solution from the provided options is:

x=3.7 x equals 3.7

Which of the following coordinate pairs is a point on the parabola y=−4x^2−53x−56?(1 point)
Responses

(−4,−56)
left parenthesis negative 4 comma negative 56 right parenthesis

(−4,220)
left parenthesis negative 4 comma 220 right parenthesis

(−1,13)
left parenthesis neagtive 1 comma 13 right parenthesis

(−1,−7)
left parenthesis negative1 comma negative 7 right parenthesis

To determine which of the given coordinate pairs is a point on the parabola represented by the equation \(y = -4x^2 - 53x - 56\), we will substitute the \(x\) values from each coordinate pair into the equation and see if the corresponding \(y\) value matches.

Let's evaluate each coordinate pair.

  1. For \((-4, -56)\):

    \[ y = -4(-4)^2 - 53(-4) - 56 \] \[ = -4(16) + 212 - 56 \] \[ = -64 + 212 - 56 \] \[ = -64 + 156 = 92 \quad (\text{not } -56) \]

  2. For \((-4, 220)\):

    \[ y = -4(-4)^2 - 53(-4) - 56 \] As calculated above, \(y = 92\) (not \(220\)).

  3. For \((-1, 13)\):

    \[ y = -4(-1)^2 - 53(-1) - 56 \] \[ = -4(1) + 53 - 56 \] \[ = -4 + 53 - 56 \] \[ = -4 - 3 = -7 \quad (\text{not } 13) \]

  4. For \((-1, -7)\):

    Using the same calculation: \[ y = -4(-1)^2 - 53(-1) - 56 \] \[ = -4(1) + 53 - 56 \] \[ = -4 + 53 - 56 \] \[ = -4 - 3 = -7 \quad (\text{this matches } -7) \]

Thus, the coordinate pair \((-1, -7)\) is a point on the parabola \(y = -4x^2 - 53x - 56\).

The correct response is: (−1,−7) left parenthesis negative 1 comma negative 7 right parenthesis

Determine the zero(s) of the quadratic function y=−5x^2+16x.(1 point)
Responses

(0,5)(0,16)
left parenthesis 0 comma 5 right parenthesis left parenthesis 0 comma 16 right parenthesis

(0,0)(3.2,0)
left parenthesis 0 comma 0 right parenthesis left parenthesis 3.2 comma 0 right parenthesis

(−3.2,0)
left parenthesis negative 3.2 comma 0 right parenthesis

There are no zeros.

To determine the zero(s) of the quadratic function given by \( y = -5x^2 + 16x \), we need to set \( y \) to zero and solve for \( x \):

\[ -5x^2 + 16x = 0 \]

We can factor this equation:

\[ x(-5x + 16) = 0 \]

Setting each factor equal to zero gives:

  1. \( x = 0 \)

  2. \(-5x + 16 = 0\)

    Solving for \( x \):

    \[ -5x = -16 \quad \Rightarrow \quad x = \frac{16}{5} = 3.2 \]

Thus, the zeros of the function are:

\[ x = 0 \quad \text{and} \quad x = 3.2 \]

In coordinate form, these zeros can be represented as points:

\((0, 0)\) and \((3.2, 0)\).

Therefore, the correct response is: (0,0)(3.2,0)
left parenthesis 0 comma 0 right parenthesis left parenthesis 3.2 comma 0 right parenthesis

Identify the vertex of y=4(x−25)^2−61 .(1 point)
Responses

(−100,−61)
left parenthesis negative 100 comma negative 61 right parenthesis

(4,−25)
left parenthesis 4 comma negative 25 right parenthesis

(25,−61)
left parenthesis 25 comma negative 61 right parenthesis

(−25,−61)

The given quadratic function is:

\[ y = 4(x - 25)^2 - 61 \]

This is in the vertex form of a parabola, which is represented as:

\[ y = a(x - h)^2 + k \]

where \((h, k)\) is the vertex of the parabola.

From your equation:

  • \(h = 25\) (the value that \(x\) is shifted by)
  • \(k = -61\) (the vertical shift)

Thus, the vertex of the parabola is:

\[ (25, -61) \]

Therefore, the correct response is: (25,−61)
left parenthesis 25 comma negative 61 right parenthesis