Question

Question 1 A) Determine whether the values -2 and 5 are solutions to the quadratic equation below. x2+3x+2=0x2+3x+2=0(2 points) Yes, this is a solution. No, this is not a solution. x=−2x=−2 x=5x=5 Question 2 A)Ben is trying to determine if x=23x=23 is a solution to the quadratic equation 9x2+3x−5=69x2+3x−5=6. Which explanation below demonstrates the correct reasoning?(1 point) Responses No, x=23x=23 is not a solution because substituting it back into the equation results in 6 = 6, which is a false statement. No, x is equal to 2 thirds is not a solution because substituting it back into the equation results in 6 = 6, which is a false statement. Yes, x=23x=23 is a solution because substituting it back into the equation results in 1 = 6, which is a true statement. Yes, x is equal to 2 thirds is a solution because substituting it back into the equation results in 1 = 6, which is a true statement. No, x=23x=23 is not a solution because substituting it back into the equation results in 1 = 6, which is a false statement. No, x is equal to 2 thirds is not a solution because substituting it back into the equation results in 1 = 6, which is a false statement. Yes, x=23x=23 is a solution because substituting it back into the equation results in 6 = 6, which is a true statement. Yes, x is equal to 2 thirds is a solution because substituting it back into the equation results in 6 = 6, which is a true statement. Question 3 A)How many solutions are in the solution set for each quadratic equation below?(3 points) 5x2=505x2=50 This quadratic equation would have solution(s). 3x2=−123x2=−12 This quadratic equation would have solution(s). 8x2=08x2=0 This quadratic equation would have solution(s). Question 4 A) Solve the quadratic equation below using square roots. (x−4)2=25(x−4)2=25(1 point) Responses {−1, 9}{−1, 9}{−1, 9}{−1, 9} {−2, 6}{−2, 6}{−2, 6}{−2, 6} {−29−−√, 29−−√}{−29, 29}{−29−−√, 29−−√}{−29, 29} {−3, 7}{−3, 7}{−3, 7}{−3, 7} Question 5 A) What should be the first step when solving the quadratic equation below? (x+10)2−25=36(x+10)2−25=36(1 point) Responses Add 25 to both sides of the equation. Add 25 to both sides of the equation. Divide both sides of the equation by 2. Divide both sides of the equation by 2. Subtract 10 from both sides of the equation. Subtract 10 from both sides of the equation. Take the square root of both sides of the equation. Take the square root of both sides of the equation. Question 6 A) Answer the questions about this quadratic equation shown below. (x−3)25=20(x−3)25=20(3 points) What should be the first step when solving the quadratic equation above? Which of the following is one of the solutions to the quadratic equation above? Which of the following is the other solution to the quadratic equation above? Question 7 A)Based on the Zero Product Property, which of the following statements must be true about the quadratic equation (5x+1)(x−8)=0(5x+1)(x−8)=0?(1 point) Responses Neither of the factors, (5x+1)(5x+1) or (x−8)(x−8), must equal 0 for the left side of the equation to equal 0. Neither of the factors, open paren 5 x plus 1 close paren or open paren x minus 8 close paren, must equal 0 for the left side of the equation to equal 0. No values of x will make the left side of the equation equal 0. No values of x will make the left side of the equation equal 0. At least one of the factors, (5x+1)(5x+1) or (x−8)(x−8), must equal 0 for the left side of the equation to equal 0. At least one of the factors, open paren 5 x plus 1 close paren or open paren x minus 8 close paren, must equal 0 for the left side of the equation to equal 0. Question 8 A) While using the Zero Product Property to find the solutions for the quadratic equation (x−2)(x−9)=0(x−2)(x−9)=0, a student completed the work below. Is the student's work correct? (x−2)(x−9)=0(x−2)(x−9)=0 original equation Solution #1: x−2=0x−2=0 x=2x=2 Solution #2: x+9=0x+9=0 x=−9x=−9 (1 point) Responses No, there should be a solution #3. No, there should be a solution #3. No, x−9=0x−9=0 should have been used to find solution #2. No, x minus 9 is equal to 0 should have been used to find solution #2. Yes, the student answered the problem correctly. Yes, the student answered the problem correctly. No, x+2=0x+2=0 should have been used to find solution #1. No, x plus 2 is equal to 0 should have been used to find solution #1. Question 9 A)Solve the quadratic equation (x+6)(x−4)=0(x+6)(x−4)=0 using the Zero Product Property.(1 point) Responses The solution set is {0, 2}{0, 2}. The solution set is {0, 2}{0, 2}. There are no solutions to this equation. There are no solutions to this equation. The solution set is {−4, 6}{−4, 6}. The solution set is {−4, 6}{−4, 6}. The solution set is {−6, 4}{−6, 4}. The solution set is {−6, 4}{−6, 4}. Question 10 A)Solve the quadratic equation 5x(x+12)=05x(x+12)=0 using the Zero Product Property.(1 point) Responses The solution set is x={5, 12}x={5, 12}. The solution set is x={5, 12}x={5, 12}. The solution set is x={0, 12}x={0, 12}. The solution set is x={0, 12}x={0, 12}. The solution set is x={−12, 0}x={−12, 0}. The solution set is x={−12, 0}x={−12, 0}.

Answer 1

Here are the answers for each of your questions based on the quadratic equations or contexts you've provided:

Question 1

A) To determine if -2 and 5 are solutions to the equation \( x^2 + 3x + 2 = 0 \):

• Substitute \( x = -2 \): \[ (-2)^2 + 3(-2) + 2 = 4 - 6 + 2 = 0 \quad \text{(This is a solution)} \]
• Substitute \( x = 5 \): \[ (5)^2 + 3(5) + 2 = 25 + 15 + 2 = 42 \quad \text{(This is not a solution)} \]

Answer: Yes, -2 is a solution. No, 5 is not a solution.

Question 2

A) Checking if \( x = \frac{2}{3} \) is a solution to \( 9x^2 + 3x - 5 = 6 \):

• Substitute \( x = \frac{2}{3} \): \[ 9\left(\frac{2}{3}\right)^2 + 3\left(\frac{2}{3}\right) - 5 = 9\left(\frac{4}{9}\right) + 2 - 5 = 4 + 2 - 5 = 1 \] This results in \( 1 = 6 \), which is false.

Answer: No, \( x = \frac{2}{3} \) is not a solution because substituting it back into the equation results in \( 1 = 6 \), which is a false statement.

Question 3

How many solutions are in the solution set for each quadratic equation?

• \( 5x^2 = 50 \): Divide by 5 → \( x^2 = 10 \) → 2 solutions (\( x = \pm\sqrt{10} \)).
• \( 3x^2 = -12 \): Divide by 3 → \( x^2 = -4 \) → 0 solutions (no real solutions).
• \( 8x^2 = 0 \): Divide by 8 → \( x^2 = 0 \) → 1 solution (\( x = 0 \)).

Question 4

A) Solve \( (x - 4)^2 = 25 \): Taking the square root of both sides: \[ x - 4 = 5 \quad \text{or} \quad x - 4 = -5 \] Thus, \( x = 9 \) or \( x = -1 \).

Answer: {−1, 9}

Question 5

A) For \( (x + 10)^2 - 25 = 36 \): First step: Add 25 to both sides of the equation.

Answer: Add 25 to both sides of the equation.

Question 6

A) For \( (x - 3)^2 = 20 \):

1. First step: Take the square root of both sides.
2. One solution: \( x - 3 = \sqrt{20} \) → \( x = 3 + \sqrt{20} \).
3. Other solution: \( x - 3 = -\sqrt{20} \) → \( x = 3 - \sqrt{20} \).

Question 7

Based on the Zero Product Property, at least one of the factors must equal zero. Answer: At least one of the factors, (5x + 1) or (x - 8), must equal 0 for the left side of the equation to equal 0.

Question 8

A) The student's work for \( (x-2)(x-9) = 0 \): Analysis: "Solution #2: \( x + 9 = 0 \)" is incorrect; it should have been \( x - 9 = 0 \).

Answer: No, \( x - 9 = 0 \) should have been used to find solution #2.

Question 9

A) Solve \( (x + 6)(x - 4) = 0 \):

1. Set each factor to zero:
   • \( x + 6 = 0 \) → \( x = -6 \)
   • \( x - 4 = 0 \) → \( x = 4 \)

Answer: The solution set is {−6, 4}.

Question 10

A) Solve \( 5x(x + 12) = 0 \) using the Zero Product Property:

1. Set each factor to zero:
   • \( 5x = 0 \) → \( x = 0 \)
   • \( x + 12 = 0 \) → \( x = -12 \)

Answer: The solution set is \( x = {0, -12} \).