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:
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For \(x = -\frac{3}{4}\): \[ -2\left(-\frac{3}{4}\right)^2 - 3\left(-\frac{3}{4}\right) + 2 = -2\left(\frac{9}{16}\right) + \frac{9}{4} + 2 \] \[ = -\frac{18}{16} + \frac{36}{16} + \frac{32}{16} = -\frac{18}{16} + \frac{68}{16} = \frac{50}{16} \neq 0 \] Therefore, not a solution.
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For \(x = -\frac{1}{2}\): \[ -2\left(-\frac{1}{2}\right)^2 - 3\left(-\frac{1}{2}\right) + 2 = -2\left(\frac{1}{4}\right) + \frac{3}{2} + 2 \] \[ = -\frac{1}{2} + \frac{3}{2} + \frac{4}{2} = -\frac{1}{2} + \frac{7}{2} = \frac{6}{2} = 3 \neq 0 \] Therefore, not a solution.
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For \(x = \frac{1}{2}\): \[ -2\left(\frac{1}{2}\right)^2 - 3\left(\frac{1}{2}\right) + 2 = -2\left(\frac{1}{4}\right) - \frac{3}{2} + 2 \] \[ = -\frac{1}{2} - \frac{3}{2} + \frac{4}{2} = -\frac{1}{2} - \frac{3}{2} + 2 = -\frac{4}{2} + 2 = 0 \] Therefore, is a solution.
After testing all values, we conclude:
Only \(x = \frac{1}{2}\) is in the solution set.
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