Question

What is the solution set to the quadratic inequality 2x2−x−3≤0
? Enter the answers as decimals.(1 point)
The solution set in inequality notation is $$
.

Answer 1

To solve the quadratic inequality \( 2x^2 - x - 3 \leq 0 \), we first need to find the roots of the corresponding quadratic equation \( 2x^2 - x - 3 = 0 \).

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = -1 \), and \( c = -3 \):

1. Calculate the discriminant:
\[
b^2 - 4ac = (-1)^2 - 4(2)(-3) = 1 + 24 = 25
\]

2. Now, find the roots:
\[
x = \frac{-(-1) \pm \sqrt{25}}{2(2)} = \frac{1 \pm 5}{4}
\]

This results in two roots:
\[
x_1 = \frac{6}{4} = 1.5 \quad \text{and} \quad x_2 = \frac{-4}{4} = -1
\]

So the roots are \( x = -1 \) and \( x = 1.5 \).

Next, we analyze the sign of the quadratic \( 2x^2 - x - 3 \) in the intervals defined by the roots:
- Interval \( (-\infty, -1) \)
- Interval \( (-1, 1.5) \)
- Interval \( (1.5, \infty) \)

### Testing the intervals:

1. **Interval \( (-\infty, -1) \)**:
Choose \( x = -2 \):
\[
2(-2)^2 - (-2) - 3 = 8 + 2 - 3 = 7 \quad (\text{positive})
\]

2. **Interval \( (-1, 1.5) \)**:
Choose \( x = 0 \):
\[
2(0)^2 - 0 - 3 = -3 \quad (\text{negative})
\]

3. **Interval \( (1.5, \infty) \)**:
Choose \( x = 2 \):
\[
2(2)^2 - 2 - 3 = 8 - 2 - 3 = 3 \quad (\text{positive})
\]

### Summary of sign analysis:
- \( 2x^2 - x - 3 > 0 \) in \( (-\infty, -1) \) and \( (1.5, \infty) \)
- \( 2x^2 - x - 3 < 0 \) in \( (-1, 1.5) \)
- It equals \( 0 \) at \( x = -1 \) and \( x = 1.5 \)

Thus, the solution set for the inequality \( 2x^2 - x - 3 \leq 0 \) is:
\[
[-1, 1.5]
\]

### Solution in interval notation:
The solution set in inequality notation is:
\[
[-1, 1.5]
\]