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=
Remaining Attempts : 3

Answer 1

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

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

For the given quadratic \( ax^2 + bx + c \):
- \( a = 1 \)
- \( b = 15 \)
- \( c = 36 \)

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

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

Next, we apply the quadratic formula:

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

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

1. **For the positive root:**
\[
x = \frac{-15 + 9}{2} = \frac{-6}{2} = -3
\]

2. **For the negative root:**
\[
x = \frac{-15 - 9}{2} = \frac{-24}{2} = -12
\]

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

Now, we identify the smaller and greater zeros:

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

Therefore, the final answers are:

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