sing the Quadratic Formula, find the solution to 3x2+18x+24=0

Question

Responses

x=−2

x=4, x=2

x=−2, x=−4

x=−4

Answer 1

To solve the quadratic equation \(3x^2 + 18x + 24 = 0\) using the Quadratic Formula, we first identify the coefficients \(a\), \(b\), and \(c\) in the standard form \(ax^2 + bx + c = 0\):

• \(a = 3\)
• \(b = 18\)
• \(c = 24\)

The Quadratic Formula is given by:

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

Now, we can substitute the values of \(a\), \(b\), and \(c\) into the formula:

1. Calculate the discriminant (\(b^2 - 4ac\)): \[ b^2 - 4ac = 18^2 - 4 \cdot 3 \cdot 24 \] \[ = 324 - 288 \] \[ = 36 \]

2. Substitute the values into the quadratic formula: \[ x = \frac{-18 \pm \sqrt{36}}{2 \cdot 3} \] \[ = \frac{-18 \pm 6}{6} \]

3. Calculate the two possible solutions:

   • First solution: \[ x = \frac{-18 + 6}{6} = \frac{-12}{6} = -2 \]
   • Second solution: \[ x = \frac{-18 - 6}{6} = \frac{-24}{6} = -4 \]

Thus, the solutions to the equation \(3x^2 + 18x + 24 = 0\) are:

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

The correct response is:

x=−2, x=−4.

Answer 2

Which of the following is a solution to the quadratic equation −x2+x=−10
? Assume that the solution has been rounded to the nearest hundredth, if applicable.
Responses

x=0.27

x=1.7

x=3.7

x=0.77

Answer 3

To find the solution to the quadratic equation \(-x^2 + x = -10\), let's first rearrange it into standard form:

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

To simplify the calculations, we can multiply the entire equation by -1:

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

Now, we'll use the Quadratic Formula, which is given by:

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

In our equation, we have:

• \(a = 1\)
• \(b = -1\)
• \(c = -10\)

Now, we can calculate the discriminant (\(b^2 - 4ac\)):

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

Now substitute into the Quadratic Formula:

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

Calculating the two possible solutions:

1. First solution: \[ x = \frac{1 + \sqrt{41}}{2} \]

2. Second solution: \[ x = \frac{1 - \sqrt{41}}{2} \]

Now we can approximate \(\sqrt{41}\):

\[ \sqrt{41} \approx 6.403 \]

Now, calculating the first solution:

\[ x \approx \frac{1 + 6.403}{2} = \frac{7.403}{2} \approx 3.7015 \approx 3.70 \]

Now calculating the second solution:

\[ x \approx \frac{1 - 6.403}{2} = \frac{-5.403}{2} \approx -2.7015 \]

Since we are looking for the solution rounded to the nearest hundredth, the solution \(x \approx 3.70\) is relevant here.

Thus, the correct response is:

x=3.7.

Answer 4

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

Responses

(−4,−56)

(−1,−7)

(−1, 13)

(−4, 220)

Answer 5

To determine which coordinate pairs are points on the parabola defined by the equation \(y = -4x^2 - 53x - 56\), we need to substitute the x-coordinates of each of the given pairs into the equation and check if the resulting y-value matches the y-coordinate of the pair.

The given pairs are:

1. \((-4, -56)\)
2. \((-1, -7)\)
3. \((-1, 13)\)
4. \((-4, 220)\)

Let's calculate \(y\) for \(x = -4\) and \(x = -1\):

1. For \((−4,−56)\):

Substituting \(x = -4\) into the equation: \[ y = -4(-4)^2 - 53(-4) - 56 \] \[ = -4(16) + 212 - 56 \] \[ = -64 + 212 - 56 \] \[ = 148 \quad (\text{not } -56) \]

2. For \((−1,−7)\):

Substituting \(x = -1\): \[ y = -4(-1)^2 - 53(-1) - 56 \] \[ = -4(1) + 53 - 56 \] \[ = -4 + 53 - 56 \] \[ = -7 \quad (\text{matches } -7) \]

3. For \((−1, 13)\):

Substituting \(x = -1\): \[ y = -4(-1)^2 - 53(-1) - 56 \] Using the same calculation as above, we have: \[ y = -7 \quad (\text{not } 13) \]

4. For \((−4, 220)\):

We already calculated \(y\) for \(x = -4\) and found \(y = 148\) (not \(220\)).

Summary of Results

1. \((-4, -56)\) does not work.
2. \((-1, -7)\) is a point on the parabola.
3. \((-1, 13)\) does not work.
4. \((-4, 220)\) does not work.

Thus, the correct coordinate pair that is a point on the parabola is:

(−1,−7).

Answer 6

Determine the zero(s) of the quadratic function y=−5x2+16x

Responses

(0, 0) (3.2, 0)

There are no zeros.

(−3.2, 0)

(0, 5) (0, 16)

Answer 7

To find the zeros of the quadratic function \(y = -5x^2 + 16x\), we need to set the equation to zero and solve for \(x\):

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

We can factor out \(x\):

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

Setting each factor equal to zero gives us:

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

Solving the second equation for \(x\):

\[ -5x + 16 = 0 \quad \Rightarrow \quad -5x = -16 \quad \Rightarrow \quad x = \frac{16}{5} = 3.2 \]

Thus, the zeros of the function are:

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

The coordinates of the zeros in terms of (x, y) are:

\[ (0, 0) \quad \text{and} \quad (3.2, 0) \]

So the correct response from the options provided is:

(0, 0) (3.2, 0).

Answer 8

identify the vertex of y=4(x−25)^2−61

Responses

(−100, −61)

(−25, −61)

(25, -61)

(4, −25)

Answer 9

The vertex form of a quadratic function is given by the equation:

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

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

In your function \(y = 4(x - 25)^2 - 61\), we can identify the values:

• \(h = 25\)
• \(k = -61\)

Thus, the vertex of the function is:

\[ (25, -61) \]

Therefore, the correct response is:

(25, -61).

Answer 10

What are the key points on the graph of y=x^2−2x−120

? Name the vertex, x-intercepts, and y-intercepts.

x-intercepts: (, ) and (1

Answer 11

To find the key points on the graph of the quadratic function \(y = x^2 - 2x - 120\), we will determine the vertex, x-intercepts, and y-intercept.

1. Finding the Vertex

The vertex of a parabola given in standard form \(y = ax^2 + bx + c\) can be found using the vertex formula:

\[ h = -\frac{b}{2a} \quad \text{and} \quad k = f(h) \]

In our equation, the coefficients are:

• \(a = 1\)
• \(b = -2\)
• \(c = -120\)

Calculating \(h\):

\[ h = -\frac{-2}{2 \cdot 1} = \frac{2}{2} = 1 \]

Now, substitute \(h\) back into the equation to find \(k\):

\[ k = f(1) = 1^2 - 2(1) - 120 = 1 - 2 - 120 = -121 \]

Thus, the vertex is:

\[ (1, -121) \]

2. Finding the X-Intercepts

To find the x-intercepts, we set \(y = 0\):

\[ 0 = x^2 - 2x - 120 \]

Now we can factor the quadratic equation:

\[ 0 = (x - 12)(x + 10) \]

Setting each factor to zero gives us the x-intercepts:

1. \(x - 12 = 0 \Rightarrow x = 12\)
2. \(x + 10 = 0 \Rightarrow x = -10\)

Thus, the x-intercepts are:

\[ (12, 0) \quad \text{and} \quad (-10, 0) \]

3. Finding the Y-Intercept

To find the y-intercept, we set \(x = 0\):

\[ y = 0^2 - 2(0) - 120 = -120 \]

This gives the y-intercept:

\[ (0, -120) \]

Summary of Key Points

• Vertex: \((1, -121)\)
• X-Intercepts: \((12, 0)\) and \((-10, 0)\)
• Y-Intercept: \((0, -120)\)

So putting it all together, the key points on the graph of \(y = x^2 - 2x - 120\) are:

• Vertex: (1, -121)
• X-Intercepts: (12, 0) and (-10, 0)
• Y-Intercept: (0, -120)