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 find the roots of the corresponding quadratic equation \(-2x^2 + x + 15 = 0\).

We can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \(a = -2\), \(b = 1\), and \(c = 15\).

Calculating the discriminant:
\[
b^2 - 4ac = 1^2 - 4(-2)(15) = 1 + 120 = 121.
\]

Since the discriminant is positive, there are two distinct real roots:
\[
x = \frac{-1 \pm \sqrt{121}}{2(-2)} = \frac{-1 \pm 11}{-4}.
\]
Calculating the roots:
1. \(x_1 = \frac{-1 + 11}{-4} = \frac{10}{-4} = -\frac{5}{2}\),
2. \(x_2 = \frac{-1 - 11}{-4} = \frac{-12}{-4} = 3\).

So the roots are \(x = -\frac{5}{2}\) and \(x = 3\).

Next, we determine the intervals to test in the inequality \(-2x^2 + x + 15 \geq 0\):
1. \( (-\infty, -\frac{5}{2}) \)
2. \( (-\frac{5}{2}, 3) \)
3. \( (3, \infty) \)

We choose test points from each interval:
- For \(x = -3\) (in \( (-\infty, -\frac{5}{2}) \)):
\(-2(-3)^2 + (-3) + 15 = -18 - 3 + 15 = -6 < 0\) (not included).
- For \(x = 0\) (in \( (-\frac{5}{2}, 3) \)):
\(-2(0)^2 + 0 + 15 = 15 \geq 0\) (included).
- For \(x = 4\) (in \( (3, \infty) \)):
\(-2(4)^2 + 4 + 15 = -32 + 4 + 15 = -13 < 0\) (not included).

Thus, the inequality \(-2x^2 + x + 15 \geq 0\) holds true in the interval \((- \frac{5}{2}, 3)\).

Since the quadratic opens downwards (as the coefficient of \(x^2\) is negative), it also includes the roots, making the final solution set:

\[
[-\frac{5}{2}, 3].
\]

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