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

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

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

Next, we plug the values into the quadratic formula:

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

Now we can calculate the two zeros:

1. First zero (using the \( + \)):
\[
x_1 = \frac{{-15 + 9}}{2} = \frac{{-6}}{2} = -3
\]

2. Second zero (using the \( - \)):
\[
x_2 = \frac{{-15 - 9}}{2} = \frac{{-24}}{2} = -12
\]

Thus, the smaller and greater zeros of the quadratic function are:

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

So the final answer is:

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