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 \( c = -14 \).

Calculating the discriminant:

\[
b^2 - 4ac = 5^2 - 4(1)(-14) = 25 + 56 = 81.
\]

Now we can find the roots:

\[
x = \frac{-5 \pm \sqrt{81}}{2(1)} = \frac{-5 \pm 9}{2}.
\]

Calculating the two roots:

1. \( x = \frac{4}{2} = 2 \)
2. \( x = \frac{-14}{2} = -7 \)

So the roots of the quadratic equation are \( x = 2 \) and \( x = -7 \).

Next, we can test the intervals created by these roots to find where the inequality \( x^2 + 5x - 14 < 0 \) is satisfied. The intervals are:

1. \( (-\infty, -7) \)
2. \( (-7, 2) \)
3. \( (2, \infty) \)

We can select test points in each interval:

- For \( x = -8 \) in \( (-\infty, -7) \):
\[
(-8)^2 + 5(-8) - 14 = 64 - 40 - 14 = 10 \quad (\text{not } < 0)
\]

- For \( x = 0 \) in \( (-7, 2) \):
\[
0^2 + 5(0) - 14 = -14 \quad (< 0)
\]

- For \( x = 3 \) in \( (2, \infty) \):
\[
3^2 + 5(3) - 14 = 9 + 15 - 14 = 10 \quad (\text{not } < 0)
\]

The quadratic \( x^2 + 5x - 14 \) is negative in the interval \( (-7, 2) \).

Thus, the solution set in inequality notation is:

\[
\boxed{(-7, 2)}
\]