To solve the quadratic inequality \(-2x^2 + x + 15 \geq 0\), we first need to determine the roots of the corresponding quadratic equation \(-2x^2 + x + 15 = 0\).
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Find the roots using the quadratic formula:
The quadratic formula is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = -2\), \(b = 1\), and \(c = 15\).
First, compute the discriminant: \[ b^2 - 4ac = 1^2 - 4(-2)(15) = 1 + 120 = 121 \]
Now, substituting the values into the quadratic formula: \[ x = \frac{-1 \pm \sqrt{121}}{2(-2)} = \frac{-1 \pm 11}{-4} \]
This gives us the two roots: \[ x_1 = \frac{-1 + 11}{-4} = \frac{10}{-4} = -\frac{5}{2} \] \[ x_2 = \frac{-1 - 11}{-4} = \frac{-12}{-4} = 3 \]
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Determine the intervals generated by these roots: The critical points from the roots are \(x = -\frac{5}{2}\) and \(x = 3\). We will now test the intervals \((- \infty, -\frac{5}{2})\), \((- \frac{5}{2}, 3)\), and \((3, \infty)\).
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Test the intervals:
- For \(x < -\frac{5}{2}\), let's take \(x = -3\): \[ -2(-3)^2 + (-3) + 15 = -18 - 3 + 15 = -6 < 0 \]
- For \(-\frac{5}{2} < x < 3\), let's take \(x = 0\): \[ -2(0)^2 + 0 + 15 = 15 \geq 0 \]
- For \(x > 3\), let's take \(x = 4\): \[ -2(4)^2 + 4 + 15 = -32 + 4 + 15 = -13 < 0 \]
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Conclusion from testing intervals: The quadratic expression is \(\geq 0\) on the interval \([- \frac{5}{2}, 3]\) including the endpoints (where it equals zero).
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Write the solution set in inequality notation: The solution set in inequality notation is: \[ \boxed{[-\frac{5}{2}, 3]} \]
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