Question

What is the solution set to the quadratic inequality −2x2+x+15≥0
?

Write the solution set in inequality notation.

(1 point)
The solution set in inequality notation is $$
.

Answer 1

To solve the quadratic inequality \(-2x^2 + x + 15 \geq 0\), we first need to determine the roots of the corresponding quadratic equation \(-2x^2 + x + 15 = 0\).

1. **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:
\[
x_1 = \frac{-1 + 11}{-4} = \frac{10}{-4} = -\frac{5}{2}
\]
\[
x_2 = \frac{-1 - 11}{-4} = \frac{-12}{-4} = 3
\]

2. **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 \((3, \infty)\).

3. **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 \(x = 4\):
\[
-2(4)^2 + 4 + 15 = -32 + 4 + 15 = -13 < 0
\]

4. **Conclusion from testing intervals:**
The quadratic expression is \(\geq 0\) on the interval \([- \frac{5}{2}, 3]\) including the endpoints (where it equals zero).

5. **Write the solution set in inequality notation:**
The solution set in inequality notation is:
\[
\boxed{[-\frac{5}{2}, 3]}
\]