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.)
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:

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

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

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

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

Since the discriminant is positive, there are two distinct real roots. Now we can find the roots using the quadratic formula:

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

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

1. For the plus sign \( ( -15 + 9 ) \):

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

2. For the minus sign \( ( -15 - 9 ) \):

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

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

Now we can list them as follows:

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

Therefore, the final answer is:

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