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:

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

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

1. **Calculate the discriminant**:
\[
b^2 - 4ac = 15^2 - 4 \cdot 1 \cdot 36 = 225 - 144 = 81
\]

2. **Substitute values into the quadratic formula**:
\[
x = \frac{-15 \pm \sqrt{81}}{2 \cdot 1} = \frac{-15 \pm 9}{2}
\]

3. **Calculate the two possible solutions**:
\[
x_1 = \frac{-15 + 9}{2} = \frac{-6}{2} = -3
\]
\[
x_2 = \frac{-15 - 9}{2} = \frac{-24}{2} = -12
\]

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

When listing the smaller zero and greater zero:
- Smaller zero: \( x = -12 \)
- Greater zero: \( x = -3 \)

So the final answer is:
- smaller zero: \( x = -12 \)
- greater zero: \( x = -3 \)