1. Use the table to complete the statement.

Question

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x Substituted Evaluate True Statement?
−34

−12

12



(1 point)
For the quadratic equation −2x2−3x+2=0
, complete the table by testing each of the given values to determine whether it is a solution. Identify which one of the values is in the solution set. Only x =

is in the solution set.
Question 2
2. Solve x2=289
by inspection. There are two real solutions. Enter the lesser number first.

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(2 points)
{
,
}
Question 3
3. Which values are in the solution set of the quadratic equation, −9+(x−3)2−4=−4
? Solve using square roots and select all that apply.

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(2 points)
Responses

−2
negative 2

16
16

28
28

−10
negative 10

8
8

2
2

−8
negative 8

−22
negative 22
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Answer 1

To analyze and complete the tasks, we will proceed step by step for each question.

Question 1: Analyze the quadratic equation

We need to check if any of the provided values satisfy the quadratic equation:

\[ -2x^2 - 3x + 2 = 0 \]

Let's evaluate each value:

1. For \( x = -34 \): \[ -2(-34)^2 - 3(-34) + 2 = -2(1156) + 102 + 2 = -2312 + 102 + 2 = -2210 \quad \text{(not a solution)} \]

2. For \( x = -12 \): \[ -2(-12)^2 - 3(-12) + 2 = -2(144) + 36 + 2 = -288 + 36 + 2 = -250 \quad \text{(not a solution)} \]

3. For \( x = 12 \): \[ -2(12)^2 - 3(12) + 2 = -2(144) - 36 + 2 = -288 - 36 + 2 = -322 \quad \text{(not a solution)} \]

None of these substitutions yield a true statement, meaning none satisfy the equation. Therefore, we conclude:

Only \( x = \) (none of the given values) is in the solution set.

Question 2: Solve \( x^2 = 289 \)

To solve this equation by inspection, we recognize that:

\[ x^2 = 289 \implies x = \pm \sqrt{289} \]

Calculating the square root gives:

\[ \sqrt{289} = 17 \]

Thus, the solutions are:

\[ x = 17 \quad \text{and} \quad x = -17 \]

Enter them in order: \[{ -17, 17 }\]

Question 3: Solve the quadratic equation \( -9 + (x-3)^2 - 4 = -4 \)

First, simplify the equation:

\[ -9 + (x-3)^2 - 4 = -4 \]

This simplifies to:

\[ (x-3)^2 - 13 = 0 \] \[ (x-3)^2 = 13 \]

Now take the square root:

\[ x - 3 = \pm \sqrt{13} \]

Thus:

\[ x = 3 + \sqrt{13} \quad \text{or} \quad x = 3 - \sqrt{13} \]

Now, let’s approximate \( \sqrt{13} \approx 3.605 \):

1. \( x \approx 3 + 3.605 \approx 6.605 \)
2. \( x \approx 3 - 3.605 \approx -0.605 \)

Now we check the values provided in the options:

• \( -2 \) -> Not a solution
• \( 16 \) -> Not a solution
• \( 28 \) -> Not a solution
• \( -10 \) -> Not a solution
• \( 8 \) -> Not a solution
• \( 2 \) -> Not a solution
• \( -8 \) -> Not a solution
• \( -22 \) -> Not a solution

None of the values listed in the responses are in the solution set since the approximate solutions are not among those values.

Final Answers:

1. Only \( x = \) (none).
2. { -17, 17 }
3. None of the provided options are in the solution set.