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.)(1 point)
smaller zero: x=___; greater zero: x=___

Answer 1

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

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

Here, the coefficients are:
- \( a = 1 \)
- \( b = 15 \)
- \( c = 36 \)

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

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

Now we can substitute \( a \), \( b \), and the discriminant into the quadratic formula:

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

Calculating \( \sqrt{81} \):

\[
\sqrt{81} = 9
\]

Now substitute this back into the formula:

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

This gives us two solutions:

1. For the positive case:

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

2. For the negative case:

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

So the zeros of the quadratic function are:

- Smaller zero: \( x = -12 \)
- Greater zero: \( x = -3 \)

Thus, we can summarize the results:

- smaller zero: \( x = -12 \)
- greater zero: \( x = -3 \)