To determine which values solve the quadratic equation \(-2x^2 - 3x + 2 = 0\), we need to evaluate the equation for each given value of \(x\) and see if it results in zero.
Let's first rewrite the equation: \[ -2x^2 - 3x + 2 = 0 \]
We will test each proposed value in the equation. Here are the steps to check if each value is a solution:
- Substitute the value of \(x\) into the left-hand side of the quadratic equation.
- Evaluate the expression to see if it equals zero.
Let’s test the potential values (you'll need to provide the specific values you're checking if they are not provided):
Example Process:
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Let’s assume we want to test the values. For instance, let's say we want to test \(x = 1\): \[ -2(1)^2 - 3(1) + 2 = -2(1) - 3 + 2 = -2 - 3 + 2 = -3 \quad (\text{not a solution}) \]
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Now let’s test \(x = -1\): \[ -2(-1)^2 - 3(-1) + 2 = -2(1) + 3 + 2 = -2 + 3 + 2 = 3 \quad (\text{not a solution}) \]
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Now testing \(x = \frac{1}{2}\): \[ -2\left(\frac{1}{2}\right)^2 - 3\left(\frac{1}{2}\right) + 2 = -2\left(\frac{1}{4}\right) - \frac{3}{2} + 2 = -\frac{1}{2} - \frac{3}{2} + 2 = -2 + 2 = 0 \quad (\text{is a solution}) \]
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And testing \(x = 2\): \[ -2(2)^2 - 3(2) + 2 = -2(4) - 6 + 2 = -8 - 6 + 2 = -12 \quad (\text{not a solution}) \]
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Now testing \(x = -2\): \[ -2(-2)^2 - 3(-2) + 2 = -2(4) + 6 + 2 = -8 + 6 + 2 = 0 \quad (\text{is a solution}) \]
Solution Set Evaluation:
After testing the values, you would identify all the values that make the equation equal to zero.
For instance:
- If testing \(x = \frac{1}{2}\) results in zero, then this is in the solution set.
- If testing \(x = -2\) results in zero, then this is also in the solution set.
Based on the results:
- If only one value results in zero, identify it and complete the statement:
"Only \( x = \frac{1}{2} \) is in the solution set." (or however your findings conclude).
Please replace the example values provided here with the specific values you have to test against the original equation.
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