Question

What is the solution set to the quadratic inequality x2+5x−14<0
?

Write the solution set in inequality notation.

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

Answer 1

To solve the quadratic inequality \( x^2 + 5x - 14 < 0 \), we first need to find the roots of the corresponding quadratic equation \( x^2 + 5x - 14 = 0 \).

We can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1 \), \( b = 5 \), and \( c = -14 \).

First, we calculate the discriminant:
\[
b^2 - 4ac = 5^2 - 4(1)(-14) = 25 + 56 = 81
\]
Now we apply the quadratic formula:
\[
x = \frac{-5 \pm \sqrt{81}}{2(1)} = \frac{-5 \pm 9}{2}
\]

Calculating the two roots:
1. \( x = \frac{-5 + 9}{2} = \frac{4}{2} = 2 \)
2. \( x = \frac{-5 - 9}{2} = \frac{-14}{2} = -7 \)

The roots are \( x = 2 \) and \( x = -7 \).

Now we can factor the quadratic:
\[
x^2 + 5x - 14 = (x - 2)(x + 7)
\]

Next, we need to determine where the product \((x - 2)(x + 7) < 0\). We will test the intervals determined by the roots \( x = -7 \) and \( x = 2 \):
1. Interval \( (-\infty, -7) \)
2. Interval \( (-7, 2) \)
3. Interval \( (2, \infty) \)

**Test each interval:**
- For \( x \in (-\infty, -7) \), pick \( x = -8 \):
\[
(-8 - 2)(-8 + 7) = (-10)(-1) = 10 > 0
\]

- For \( x \in (-7, 2) \), pick \( x = 0 \):
\[
(0 - 2)(0 + 7) = (-2)(7) = -14 < 0
\]

- For \( x \in (2, \infty) \), pick \( x = 3 \):
\[
(3 - 2)(3 + 7) = (1)(10) = 10 > 0
\]

The solution to the inequality \( (x - 2)(x + 7) < 0 \) is where the product is negative, which occurs in the interval \( (-7, 2) \).

In inequality notation, the solution set is:
\[
\boxed{(-7, 2)}
\]