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.)

Question

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:

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

For the quadratic equation \( ax^2 + bx + c \), we have:

• \( a = 1 \)
• \( b = 15 \)
• \( c = 36 \)

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

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

Now we can find the zeros:

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

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

This gives us two solutions:

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

2. For the negative case (\(-9\)): \[ x = \frac{-15 - 9}{2} = \frac{-24}{2} = -12 \]

Thus, the zeros of the quadratic function are:

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

In summary:

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