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)
smaller zero: x=
; greater zero: x=

Answer 1

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

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

where \( a = 1 \), \( b = 15 \), and \( c = 36 \).

First, we calculate the discriminant:

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

Next, we apply the quadratic formula:

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

Since \( \sqrt{81} = 9 \), we have:

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

Now, we calculate the two possible values for \( x \):

1. For the positive case:

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

2. For the negative case:

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

Thus, the zeros of the function are \( x = -12 \) and \( x = -3 \).

Finally, we list the smaller and greater zeros:

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

So, the final answer is:

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