Question

Determine the zeros of the quadratic function y = x^2 + 15x + 36. (When listing the smaller zero, remember that negative values are always smaller than positive values.)

Answer 1

To find the zeros of the quadratic function \( y = x^2 + 15x + 36 \), we can set \( y \) equal to zero and solve for \( x \):

\[
x^2 + 15x + 36 = 0
\]

We can use the quadratic formula, which is given by:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For our quadratic equation, the coefficients are:
- \( a = 1 \)
- \( b = 15 \)
- \( c = 36 \)

Now, we will calculate the discriminant (\( b^2 - 4ac \)):

\[
b^2 - 4ac = 15^2 - 4 \cdot 1 \cdot 36 = 225 - 144 = 81
\]

Since the discriminant is positive, we will have two distinct real zeros.

Now we can substitute the values into the quadratic formula:

\[
x = \frac{-15 \pm \sqrt{81}}{2 \cdot 1} = \frac{-15 \pm 9}{2}
\]

This gives us two solutions:

1. For the positive square root:

\[
x = \frac{-15 + 9}{2} = \frac{-6}{2} = -3
\]

2. For the negative square root:

\[
x = \frac{-15 - 9}{2} = \frac{-24}{2} = -12
\]

Thus, the zeros of the function \( y = x^2 + 15x + 36 \) are:

\[
x = -12 \quad \text{and} \quad x = -3
\]

When listing the smaller zero first, we have:

\[
\boxed{-12} \quad \text{and} \quad \boxed{-3}
\]