Question

or 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.(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 choose all that apply.(2 points)
Responses

-8
-8

-22
-22

8
8

2
2

16
16

-2
-2

28
28

-10
-10
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Answer 1

Let's tackle each question step by step.

### Question 1
**Given Quadratic Equation:**
\[ -2x^2 - 3x + 2 = 0\]

We will test the given values (you didn't specify which values to test, so you will need to input the values you're given). Let's assume we test with \( x = -2, -1, 0, 1, 2 \).

1. **If \( x = -2 \):**
\[
-2(-2)^2 - 3(-2) + 2 = -2(4) + 6 + 2 = -8 + 6 + 2 = 0
\text{ (solution)}
\]

2. **If \( x = -1 \):**
\[
-2(-1)^2 - 3(-1) + 2 = -2(1) + 3 + 2 = -2 + 3 + 2 = 3
\text{ (not a solution)}
\]

3. **If \( x = 0 \):**
\[
-2(0)^2 - 3(0) + 2 = 0 + 0 + 2 = 2
\text{ (not a solution)}
\]

4. **If \( x = 1 \):**
\[
-2(1)^2 - 3(1) + 2 = -2 - 3 + 2 = -3
\text{ (not a solution)}
\]

5. **If \( x = 2 \):**
\[
-2(2)^2 - 3(2) + 2 = -2(4) - 6 + 2 = -8 - 6 + 2 = -12
\text{ (not a solution)}
\]

The solution set from this example:
- Only \( x = -2 \) is in the solution set.

### Question 2
**Solve:**
\[ x^2 = 289\]

Taking the square root on both sides:
\[
x = \sqrt{289} \quad \text{or} \quad x = -\sqrt{289}
\]
Thus, the solutions are:
\[
x = 17 \quad \text{and} \quad x = -17
\]

The lesser number first:
\[
\{-17, 17\}
\]

### Question 3
**Given Quadratic Equation:**
\[ -9 + (x - 3)^2 - 4 = -4\]

First, simplify the equation:
\[
-9 + (x - 3)^2 - 4 + 4 = 0
\]
So we have:
\[
-9 + (x - 3)^2 = 0
\]
This simplifies to:
\[
(x - 3)^2 = 9
\]

Now, take the square root of both sides:
\[
x - 3 = 3 \quad \text{or} \quad x - 3 = -3
\]
Solving these gives:
1. \(x - 3 = 3 \) ⇒ \( x = 6\)
2. \( x - 3 = -3 \) ⇒ \( x = 0\)

The values that are in the solution set from given options are:
- **8** (not in the set)
- **2** (not in the set)
- **16** (not in the set)
- **-2** (not in the set)
- **28** (not in the set)
- **-10** (not in the set)

So the values **0** and **6** would be in the solution set if included.

**Final Responses:**
- For Question 1: Only \( x = -2 \) is in the solution set.
- For Question 2: \( \{-17, 17\} \).
- For Question 3: \( x = 0 \) and \( x = 6\).