To determine which values satisfy the quadratic equation \(-2x^2 - 3x + 2 = 0\), we will substitute each value of \(x\) into the equation and evaluate.
Substituting Each Value:
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For \(x = -\frac{3}{4}\): \[ -2\left(-\frac{3}{4}\right)^2 - 3\left(-\frac{3}{4}\right) + 2 = -2\left(\frac{9}{16}\right) + \frac{9}{4} + 2 \] \[ = -\frac{18}{16} + \frac{9}{4} + 2 = -\frac{9}{8} + \frac{9}{4} + 2 \] \[ = -\frac{9}{8} + \frac{18}{8} + \frac{16}{8} = -\frac{9}{8} + \frac{34}{8} = \frac{25}{8} \neq 0 \]
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For \(x = -\frac{1}{2}\): \[ -2\left(-\frac{1}{2}\right)^2 - 3\left(-\frac{1}{2}\right) + 2 = -2\left(\frac{1}{4}\right) + \frac{3}{2} + 2 \] \[ = -\frac{1}{2} + \frac{3}{2} + 2 = \frac{2}{2} + 2 = 1 + 2 = 3 \neq 0 \]
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For \(x = \frac{1}{2}\): \[ -2\left(\frac{1}{2}\right)^2 - 3\left(\frac{1}{2}\right) + 2 = -2\left(\frac{1}{4}\right) - \frac{3}{2} + 2 \] \[ = -\frac{1}{2} - \frac{3}{2} + 2 = -2 + 2 = 0 \]
Table Completion:
| x | Substituted | Evaluate | True Statement? | |----------|-------------|---------------------|------------------------| | -3/4 | \(-2(-\frac{3}{4})^2 - 3(-\frac{3}{4}) + 2\) | \(\frac{25}{8}\) | False | | -1/2 | \(-2(-\frac{1}{2})^2 - 3(-\frac{1}{2}) + 2\) | \(3\) | False | | 1/2 | \(-2(\frac{1}{2})^2 - 3(\frac{1}{2}) + 2\) | \(0\) | True |
Conclusion:
Only \(x = \frac{1}{2}\) is in the solution set.