To determine which value is in the solution set of the quadratic equation \(-2x^2 - 3x + 2 = 0\), we will substitute each given \(x\) value into the equation and evaluate it. If the result equals zero, then that \(x\) value is a solution to the equation.
Let's evaluate each given value:
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For \(x = -34\): \[ -2(-34)^2 - 3(-34) + 2 = -2(1156) + 102 + 2 = -2312 + 102 + 2 = -2210 \quad (\text{Not a solution}) \]
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For \(x = -12\): \[ -2(-12)^2 - 3(-12) + 2 = -2(144) + 36 + 2 = -288 + 36 + 2 = -250 \quad (\text{Not a solution}) \]
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For \(x = 12\): \[ -2(12)^2 - 3(12) + 2 = -2(144) - 36 + 2 = -288 - 36 + 2 = -322 \quad (\text{Not a solution}) \]
After evaluating each value:
- None of the values \(-34\), \(-12\), or \(12\) satisfy the equation \(-2x^2 - 3x + 2 = 0\) as they do not yield a true statement (i.e., they do not equal zero).
Thus, none of the values is in the solution set for the equation provided.
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