Question

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)
Only x=
is in the solution set.

Answer 1

To determine whether the given values are solutions to the quadratic equation \(-2x^2 - 3x + 2 = 0\), we need to substitute each value of \(x\) into the equation and evaluate whether the result equals zero.

Let's evaluate each value step by step:

1. **For \(x = -\frac{3}{4}\)**:
\[
-2\left(-\frac{3}{4}\right)^2 - 3\left(-\frac{3}{4}\right) + 2 = -2\left(\frac{9}{16}\right) + \frac{9}{4} + 2
\]
\[
= -\frac{18}{16} + \frac{36}{16} + \frac{32}{16} = -\frac{18}{16} + \frac{68}{16} = \frac{50}{16} \neq 0
\]
Therefore, **not a solution**.

2. **For \(x = -\frac{1}{2}\)**:
\[
-2\left(-\frac{1}{2}\right)^2 - 3\left(-\frac{1}{2}\right) + 2 = -2\left(\frac{1}{4}\right) + \frac{3}{2} + 2
\]
\[
= -\frac{1}{2} + \frac{3}{2} + \frac{4}{2} = -\frac{1}{2} + \frac{7}{2} = \frac{6}{2} = 3 \neq 0
\]
Therefore, **not a solution**.

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

After testing all values, we conclude:

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