Question

Quadratic Equations Unit Test:
For the quadratic equation - 2 x2 - 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 = 1/2 is in the solution set.

Answer 1

To determine which of the given values is a solution to the quadratic equation \(-2x^2 - 3x + 2 = 0\), we need to substitute each value into the equation and see if it satisfies the equation (i.e., whether it makes the left-hand side equal to zero).

Given the equation:

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

Let’s test the values one by one, including \(x = \frac{1}{2}\) to see if it is indeed a solution.

### Testing \(x = \frac{1}{2}\):

Substituting \(x = \frac{1}{2}\):

\[
-2\left(\frac{1}{2}\right)^2 - 3\left(\frac{1}{2}\right) + 2 = 0
\]

Calculating each term:

1. \(-2\left(\frac{1}{2}\right)^2 = -2 \cdot \frac{1}{4} = -\frac{1}{2}\)
2. \(-3\left(\frac{1}{2}\right) = -\frac{3}{2}\)
3. The constant: \(2\)

Now, put them together:

\[
-\frac{1}{2} - \frac{3}{2} + 2 = -\frac{1}{2} - \frac{3}{2} + \frac{4}{2} = -\frac{4}{2} + \frac{4}{2} = 0
\]

Since the left-hand side equals \(0\), \(x = \frac{1}{2}\) is indeed a solution.

### Now let's test some other values that might be provided:

For this example, let's test other common values such as \(x = 0\), \(x = 1\), and \(x = 2\).

### Testing \(x = 0\):

\[
-2(0)^2 - 3(0) + 2 = 0
\]
\[
0 + 0 + 2 = 2 \quad (\text{not a solution})
\]

### Testing \(x = 1\):

\[
-2(1)^2 - 3(1) + 2 = 0
\]
\[
-2 - 3 + 2 = -3 \quad (\text{not a solution})
\]

### Testing \(x = 2\):

\[
-2(2)^2 - 3(2) + 2 = 0
\]
\[
-2 \cdot 4 - 6 + 2 = -8 - 6 + 2 = -12 \quad (\text{not a solution})
\]

### Conclusion

Based on the calculations, we find:

- \(x = \frac{1}{2}\) is a solution (it makes the equation equal to zero).
- \(x = 0\), \(x = 1\), and \(x = 2\) are **not** solutions.

Thus, the only value in the solution set is \(x = \frac{1}{2}\).