To solve the equation \(-2x^2 - 3x + 2 = 0\) and evaluate the given \(x\) values, I'll first find the solutions to the quadratic equation using the quadratic formula.
The quadratic formula is given by:
\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \]
For the equation \(-2x^2 - 3x + 2 = 0\), the coefficients are:
- \(a = -2\)
- \(b = -3\)
- \(c = 2\)
Calculating the discriminant \((b^2 - 4ac)\):
\[ b^2 - 4ac = (-3)^2 - 4(-2)(2) = 9 + 16 = 25 \]
Now we calculate \(x\):
\[ x = \frac{{-(-3) \pm \sqrt{25}}}{2 \cdot (-2)} = \frac{{3 \pm 5}}{-4} \]
Calculating the two possible values for \(x\):
-
When using the positive square root: \[ x = \frac{{3 + 5}}{-4} = \frac{8}{-4} = -2 \]
-
When using the negative square root: \[ x = \frac{{3 - 5}}{-4} = \frac{-2}{-4} = \frac{1}{2} \]
Thus, the two solutions for the equation are \(x = -2\) and \(x = \frac{1}{2}\).
Now, I will evaluate the three specified \(x\) values in the equation:
For \(x = -\frac{3}{4}\):
Substituting into the equation:
\[ -2\left(-\frac{3}{4}\right)^2 - 3\left(-\frac{3}{4}\right) + 2 \] \[ = -2\left(\frac{9}{16}\right) + \frac{9}{4} + 2 \] \[ = -\frac{18}{16} + \frac{36}{16} + \frac{32}{16} \] \[ = \frac{-18 + 36 + 32}{16} = \frac{50}{16} \neq 0 \text{ (False)} \]
For \(x = -\frac{1}{2}\):
Substituting into the equation:
\[ -2\left(-\frac{1}{2}\right)^2 - 3\left(-\frac{1}{2}\right) + 2 \] \[ = -2\left(\frac{1}{4}\right) + \frac{3}{2} + 2 \] \[ = -\frac{1}{2} + \frac{3}{2} + 2 \] \[ = \frac{2}{2} + 2 = 1 + 2 = 3 \neq 0 \text{ (False)} \]
For \(x = \frac{1}{2}\):
Substituting into the equation:
\[ -2\left(\frac{1}{2}\right)^2 - 3\left(\frac{1}{2}\right) + 2 \] \[ = -2\left(\frac{1}{4}\right) - \frac{3}{2} + 2 \] \[ = -\frac{1}{2} - \frac{3}{2} + 2 \] \[ = -\frac{1}{2} - \frac{3}{2} + \frac{4}{2} = 0 \text{ (True)} \]
Conclusion:
- The values you are checking yielded:
- For \(x = -\frac{3}{4}\): False
- For \(x = -\frac{1}{2}\): False
- For \(x = \frac{1}{2}\): True
The only solutions to the equation are \(x = -2\) and \(x = \frac{1}{2}\).
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