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