Question

Use the table to answer the question.

x Substituted Evaluate True Statement?
−34
−12
12
​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 each value is a solution to the quadratic equation \(-2x^2 - 3x + 2 = 0\), we will substitute each of the values of \(x\) into the equation and evaluate.

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 the provided values satisfy the equation \(-2x^2 - 3x + 2 = 0\). Thus, **none of the provided values are in the solution set.**

If you would like to solve for the exact solutions of the quadratic equation, we can use the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \(a = -2\), \(b = -3\), and \(c = 2\). Plugging in these values:

\[
x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(-2)(2)}}{2(-2)} = \frac{3 \pm \sqrt{9 + 16}}{-4} = \frac{3 \pm \sqrt{25}}{-4} = \frac{3 \pm 5}{-4}
\]

Calculating the two possible \(x\) values:

1. \(x = \frac{8}{-4} = -2\)
2. \(x = \frac{-2}{-4} = \frac{1}{2}\)

**So the solution set is \(x = -2\) and \(x = \frac{1}{2}\).**

Please confirm if I incorrectly interpreted the question. If strictly looking for the values in the table, none are solutions.