Question

Determine the zeros of the quadratic function y=x2+15x+36
. (When listing the smaller zero, remember that negative values are always smaller than positive values.)(1 point)

Answer 1

To find the zeros of the quadratic function \( y = x^2 + 15x + 36 \), we can use the quadratic formula:

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

In this case, we have:

- \( a = 1 \)
- \( b = 15 \)
- \( c = 36 \)

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

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

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. \( x = \frac{{-15 + 9}}{2} = \frac{{-6}}{2} = -3 \)
2. \( x = \frac{{-15 - 9}}{2} = \frac{{-24}}{2} = -12 \)

The zeros of the quadratic function are \( -12 \) and \( -3 \).

When listing the smaller zero first, we would write:

\[
-12, -3
\]