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 $$.

1 answer

To solve the quadratic inequality \(-2x^2 + x + 15 \geq 0\), we first need to analyze the related quadratic equation:

\[ -2x^2 + x + 15 = 0 \]

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

  1. Calculate the discriminant:

\[ b^2 - 4ac = 1^2 - 4(-2)(15) = 1 + 120 = 121 \]

  1. Since the discriminant is positive, there are two distinct real roots. Now, calculate the roots:

\[ x = \frac{-1 \pm \sqrt{121}}{2(-2)} = \frac{-1 \pm 11}{-4} \]

Calculating the two roots:

  • For \(x = \frac{-1 + 11}{-4} = \frac{10}{-4} = -\frac{5}{2}\)
  • For \(x = \frac{-1 - 11}{-4} = \frac{-12}{-4} = 3\)

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

Next, we can determine the intervals where the quadratic \( -2x^2 + x + 15 \) is greater than or equal to zero. The roots divide the number line into three intervals:

  1. \( (-\infty, -\frac{5}{2}) \)
  2. \( \left(-\frac{5}{2}, 3\right) \)
  3. \( (3, \infty) \)

Now we will test a point from each interval to see where the quadratic is greater than or equal to zero.

  1. Interval \( (-\infty, -\frac{5}{2}) \): Test \(x = -3\) \[ -2(-3)^2 + (-3) + 15 = -18 - 3 + 15 = -6 < 0 \]

  2. Interval \( \left(-\frac{5}{2}, 3\right) \): Test \(x = 0\) \[ -2(0)^2 + (0) + 15 = 15 > 0 \]

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

From this analysis, we find that the quadratic is non-negative in the interval \( \left[-\frac{5}{2}, 3\right] \).

Including the roots where the quadratic equals zero, the final solution set in inequality notation is:

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