Question

Question

Each of the following quadratic equations can be solved by factoring. Which equation has only one distinct solution?(1 point)
Responses

x2−4=0
x squared minus 4 equals 0

x2−5x+4=0
x squared minus 5 x plus 4 equals 0

x2+5x+4=0
x squared plus 5 x plus 4 equals 0

x2+4x+4=0

Answer 1

To determine which quadratic equation has only one distinct solution, we look for equations that can be factored into a perfect square form, which would indicate that the solutions are repeated (i.e., the same root).

1. Equation: \( x^2 - 4 = 0 \)
   This factors to \( (x - 2)(x + 2) = 0 \)
   Solutions: \( x = 2 \) and \( x = -2 \) (two distinct solutions)

2. Equation: \( x^2 - 5x + 4 = 0 \)
   This factors to \( (x - 1)(x - 4) = 0 \)
   Solutions: \( x = 1 \) and \( x = 4 \) (two distinct solutions)

3. Equation: \( x^2 + 5x + 4 = 0 \)
   This factors to \( (x + 1)(x + 4) = 0 \)
   Solutions: \( x = -1 \) and \( x = -4 \) (two distinct solutions)

4. Equation: \( x^2 + 4x + 4 = 0 \)
   This factors to \( (x + 2)(x + 2) = 0 \)
   Solutions: \( x = -2 \) (one distinct solution, repeated root)

Since only the last equation, \( x^2 + 4x + 4 = 0 \), yields only one distinct solution, the answer is:

\( x^2 + 4x + 4 = 0 \)

Answer 2

Question
Solve the following quadratic equation using the perfect square trinomial pattern: 14x2+2x+4=0
.(1 point)
x=

Answer 3

To solve the quadratic equation \(14x^2 + 2x + 4 = 0\) using the perfect square trinomial pattern, we can start by rearranging and simplifying the equation.

1. First, divide the entire equation by 2 to make the numbers simpler: \[ 7x^2 + x + 2 = 0 \]

2. Rearrange the equation: \[ 7x^2 + x = -2 \]

3. To complete the square, we first divide all terms involving \(x\) by \(7\): \[ x^2 + \frac{1}{7}x = -\frac{2}{7} \]

4. Now, we'll complete the square. The coefficient of \(x\) is \(\frac{1}{7}\). To complete the square, take half of that coefficient, square it, and add it to both sides:

   • Half of \(\frac{1}{7}\) is \(\frac{1}{14}\).
   • Squaring \(\frac{1}{14}\) gives \(\frac{1}{196}\).

So, we add \(\frac{1}{196}\) to both sides: \[ x^2 + \frac{1}{7}x + \frac{1}{196} = -\frac{2}{7} + \frac{1}{196} \]

5. We need to find a common denominator to add the fractions on the right side. The common denominator will be \(196\):

   • Convert \(-\frac{2}{7}\) to \(-\frac{56}{196}\).
   • So, we have: \[ -\frac{2}{7} + \frac{1}{196} = -\frac{56}{196} + \frac{1}{196} = -\frac{55}{196} \]

6. Now our equation is: \[ x^2 + \frac{1}{7}x + \frac{1}{196} = -\frac{55}{196} \]

7. The left-hand side can be factored: \[ \left(x + \frac{1}{14}\right)^2 = -\frac{55}{196} \]

Since the right side is negative, there are no real solutions to this equation.

The solutions are complex. We can express the square root of the negative number: \[ \left(x + \frac{1}{14}\right)^2 = \frac{55}{196} i \]

Now take the square root of both sides: \[ x + \frac{1}{14} = \pm \sqrt{\frac{55}{196}} i \] \[ x + \frac{1}{14} = \pm \frac{\sqrt{55}}{14} i \]

8. Finally, solving for \(x\): \[ x = -\frac{1}{14} \pm \frac{\sqrt{55}}{14} i \]

Thus, the solutions to the quadratic equation \(14x^2 + 2x + 4 = 0\) are: \[ x = -\frac{1}{14} + \frac{\sqrt{55}}{14} i \quad \text{and} \quad x = -\frac{1}{14} - \frac{\sqrt{55}}{14} i \]

Answer 4

Gwen was asked to solve 3x2−20=7
by factoring using the difference of squares pattern. What is/are the correct solution(s)?(1 point)
Responses

x=−3
, x=3
x equals negative 3 , x equals 3

x=3
x equals 3

x=−3
x equals negative 3

x=9
, x=−9

Answer 5

To solve the equation \(3x^2 - 20 = 7\) by factoring using the difference of squares pattern, we can follow these steps:

1. Rearrange the equation:
   Start by moving 7 to the left side: \[ 3x^2 - 20 - 7 = 0 \] Simplifying this, we get: \[ 3x^2 - 27 = 0 \]

2. Factor out the common term:
   Factor out \(3\): \[ 3(x^2 - 9) = 0 \]

3. Recognize the difference of squares:
   Notice that \(x^2 - 9\) can be factored further using the difference of squares: \[ 3(x - 3)(x + 3) = 0 \]

4. Set each factor to zero:
   Set the factors equal to zero: \[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \] \[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \]

Thus, the solutions to the equation \(3x^2 - 20 = 7\) are:

• \(x = 3\)
• \(x = -3\)

Therefore, the correct response is: \(x = -3, x = 3\)

Answer 6

Complete the square to identify the two values that solve the following quadratic equation: x2−16x−13=4
.(1 point)
x=

Answer 7

To solve the quadratic equation \(x^2 - 16x - 13 = 4\) by completing the square, follow these steps:

Step 1: Rearrange the equation

Move \(4\) to the left side of the equation: \[ x^2 - 16x - 13 - 4 = 0 \] This simplifies to: \[ x^2 - 16x - 17 = 0 \]

Step 2: Isolate the quadratic and linear terms

We'll work with the equation in this form: \[ x^2 - 16x = 17 \]

Step 3: Complete the square

To complete the square, take half of the coefficient of \(x\), square it, and add it to both sides:

• The coefficient of \(x\) is \(-16\).
• Half of \(-16\) is \(-8\).
• Squaring \(-8\) gives \(64\).

Add \(64\) to both sides: \[ x^2 - 16x + 64 = 17 + 64 \] This simplifies to: \[ x^2 - 16x + 64 = 81 \]

Step 4: Factor the left side

Now the left side can be factored as a perfect square: \[ (x - 8)^2 = 81 \]

Step 5: Take the square root of both sides

Taking the square root of both sides gives us: \[ x - 8 = \pm 9 \]

Step 6: Solve for \(x\)

Now we solve for \(x\) by setting up two equations:

1. \(x - 8 = 9\) \[ x = 9 + 8 = 17 \]

2. \(x - 8 = -9\) \[ x = -9 + 8 = -1 \]

Final Solutions

Thus, the two solutions to the equation are: \[ x = 17 \quad \text{and} \quad x = -1 \]

The solutions are \(x = 17\) and \(x = -1\).

Answer 8

Which of the following is a solution to the quadratic equation 4x2+2x=10
? Assume that the solution has been rounded to the nearest hundredth, if applicable.(1 point)
Responses

x=−1.85
x equals negative 1.85

x=1.04
x equals 1.04

x=−0.77
x equals negative 0.77

x=1.15

Answer 9

To find the solution to the quadratic equation \(4x^2 + 2x = 10\), we start by rearranging the equation into the standard form \(ax^2 + bx + c = 0\):

1. Subtract 10 from both sides: \[ 4x^2 + 2x - 10 = 0 \]

Next, we can either factor or use the quadratic formula to solve for \(x\). Since this equation does not easily factor, we'll use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

where \(a = 4\), \(b = 2\), and \(c = -10\).

Step 1: Calculate the discriminant

\[ b^2 - 4ac = (2)^2 - 4(4)(-10) = 4 + 160 = 164 \]

Step 2: Apply the quadratic formula

Now plug in the values: \[ x = \frac{-2 \pm \sqrt{164}}{2 \times 4} = \frac{-2 \pm \sqrt{164}}{8} \]

Step 3: Simplify \(\sqrt{164}\)

We can simplify \(\sqrt{164}\): \[ \sqrt{164} = \sqrt{4 \cdot 41} = 2\sqrt{41} \] Thus, we have: \[ x = \frac{-2 \pm 2\sqrt{41}}{8} \] This simplifies to: \[ x = \frac{-1 \pm \sqrt{41}}{4} \]

Step 4: Calculate the two possible values for \(x\)

Now we will find both possible solutions:

1. First solution: \[ x = \frac{-1 + \sqrt{41}}{4} \]
2. Second solution: \[ x = \frac{-1 - \sqrt{41}}{4} \]

Now calculating these values:

1. Calculate \(\sqrt{41} \approx 6.4\):
   • \(x_1 = \frac{-1 + 6.4}{4} = \frac{5.4}{4} \approx 1.35\)
   • \(x_2 = \frac{-1 - 6.4}{4} = \frac{-7.4}{4} \approx -1.85\)

Step 5: Round the values as necessary

• The approximate values are \(x_1 \approx 1.35\) and \(x_2 \approx -1.85\).

Final Step: Compare with the given options

Now comparing with the available options:

• \(x = -1.85\)
• \(x = 1.04\)
• \(x = -0.77\)
• \(x = 1.15\)

The option that is closest to a calculated value is: \[ \boxed{x = -1.85} \]

Thus, the correct answer is \(x = -1.85\).

Answer 10

What are the key points on the graph of y=x2−16x+28
? Name the vertex, x-intercepts, and y-intercept.(2 points)
vertex: (
)

(smaller) x-intercept: (
)

(larger) x-intercept: (
)

y-intercept: (
)

Answer 11

To find the key points on the graph of the quadratic function \(y = x^2 - 16x + 28\), we will determine the vertex, x-intercepts, and y-intercept.

Step 1: Find the Vertex

The vertex of a quadratic function in the form \(y = ax^2 + bx + c\) can be found using the formula: \[ x = -\frac{b}{2a} \] Here, \(a = 1\) and \(b = -16\).

Calculating the x-coordinate of the vertex: \[ x = -\frac{-16}{2 \cdot 1} = \frac{16}{2} = 8 \] Next, we substitute \(x = 8\) back into the function to find the y-coordinate: \[ y = (8)^2 - 16(8) + 28 = 64 - 128 + 28 = -36 \] Thus, the vertex is: \[ \text{Vertex: } (8, -36) \]

Step 2: Find the x-intercepts

To find the x-intercepts, set \(y = 0\): \[ 0 = x^2 - 16x + 28 \] We can solve this quadratic using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In our case, \(a = 1\), \(b = -16\), and \(c = 28\).

Calculating the discriminant: \[ b^2 - 4ac = (-16)^2 - 4(1)(28) = 256 - 112 = 144 \] Now applying the quadratic formula: \[ x = \frac{-(-16) \pm \sqrt{144}}{2(1)} = \frac{16 \pm 12}{2} \] Calculating the two solutions:

1. \(x = \frac{16 + 12}{2} = \frac{28}{2} = 14\)
2. \(x = \frac{16 - 12}{2} = \frac{4}{2} = 2\)

Thus, the x-intercepts are: \[ \text{Smaller x-intercept: } (2, 0) \] \[ \text{Larger x-intercept: } (14, 0) \]

Step 3: Find the y-intercept

To find the y-intercept, set \(x = 0\): \[ y = (0)^2 - 16(0) + 28 = 28 \] Thus, the y-intercept is: \[ \text{Y-intercept: } (0, 28) \]

Summary of Key Points

• Vertex: \( (8, -36) \)
• Smaller x-intercept: \( (2, 0) \)
• Larger x-intercept: \( (14, 0) \)
• Y-intercept: \( (0, 28) \)

So the final answer is:

• Vertex: \( (8, -36) \)
• Smaller x-intercept: \( (2, 0) \)
• Larger x-intercept: \( (14, 0) \)
• Y-intercept: \( (0, 28) \)