Question

Use the table to answer the question.

x Substituted Evaluate True Statement?
−3/4
−1/2
1/2
​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.​

(1 point)

Answer 1

Only x= is in the solution set.

Answer 2

To determine which of the values is a solution to the quadratic equation \(-2x^2 - 3x + 2 = 0\), we will substitute each given value of \(x\) into the equation and evaluate.

### Step 1: Substitute each value into the equation

1. **For \(x = -\frac{3}{4}\)**:
\[
-2\left(-\frac{3}{4}\right)^2 - 3\left(-\frac{3}{4}\right) + 2
\]
\[
= -2\left(\frac{9}{16}\right) + \frac{9}{4} + 2
\]
\[
= -\frac{18}{16} + \frac{36}{16} + \frac{32}{16}
\]
\[
= \frac{-18 + 36 + 32}{16} = \frac{50}{16} = \frac{25}{8} \neq 0
\]

2. **For \(x = -\frac{1}{2}\)**:
\[
-2\left(-\frac{1}{2}\right)^2 - 3\left(-\frac{1}{2}\right) + 2
\]
\[
= -2\left(\frac{1}{4}\right) + \frac{3}{2} + 2
\]
\[
= -\frac{1}{2} + \frac{3}{2} + 2
\]
\[
= \frac{2}{2} + 2 = 1 + 2 = 3 \neq 0
\]

3. **For \(x = \frac{1}{2}\)**:
\[
-2\left(\frac{1}{2}\right)^2 - 3\left(\frac{1}{2}\right) + 2
\]
\[
= -2\left(\frac{1}{4}\right) - \frac{3}{2} + 2
\]
\[
= -\frac{1}{2} - \frac{3}{2} + 2
\]
\[
= -2 + 2 = 0
\]

### Summary

- For \(x = -\frac{3}{4}\): the evaluation resulted in \(\frac{25}{8} \neq 0\) (Not a solution).
- For \(x = -\frac{1}{2}\): the evaluation resulted in \(3 \neq 0\) (Not a solution).
- For \(x = \frac{1}{2}\): the evaluation resulted in \(0\) (Is a solution).

### Conclusion

The value in the solution set is \( \frac{1}{2} \).

Answer 3

Based on the evaluation of the quadratic equation \(-2x^2 - 3x + 2 = 0\) for the provided values of \(x\), the only value in the solution set is:

**\(x = \frac{1}{2}\)**

So, we can summarize:

Only \(x = \frac{1}{2}\) is in the solution set.