To find the zeros of the function [tex]\( f(x) = x^2 + 2x - 24 \)[/tex], we need to determine the values of [tex]\( x \)[/tex] for which [tex]\( f(x) = 0 \)[/tex].
1. Start with the given equation:
[tex]\[
x^2 + 2x - 24 = 0
\][/tex]
2. This is a quadratic equation and we can solve it by using the quadratic formula:
[tex]\[
x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}
\][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -24 \)[/tex].
3. Calculate the discriminant ([tex]\( \Delta \)[/tex]):
[tex]\[
\Delta = b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot (-24) = 4 + 96 = 100
\][/tex]
4. Substitute the values into the quadratic formula:
[tex]\[
x = \frac{{-2 \pm \sqrt{100}}}{2 \cdot 1} = \frac{{-2 \pm 10}}{2}
\][/tex]
5. Solve for the two possible values of [tex]\( x \)[/tex]:
- When [tex]\( x = \frac{{-2 + 10}}{2} \)[/tex]:
[tex]\[
x = \frac{8}{2} = 4
\][/tex]
- When [tex]\( x = \frac{{-2 - 10}}{2} \)[/tex]:
[tex]\[
x = \frac{{-12}}{2} = -6
\][/tex]
6. Therefore, the zeros of the function [tex]\( f(x) = x^2 + 2x - 24 \)[/tex] are:
[tex]\[
x = -6 \quad \text{and} \quad x = 4
\][/tex]
So, the correct answer is:
-6 and 4
