Determine the zeros of the quadratic function y=x2+15x+36

Question

. (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, the coefficients are:

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

First, we calculate the discriminant:

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

Now, substitute the values into the quadratic formula:

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

Since \( \sqrt{81} = 9 \), we have:

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

This gives us two solutions:

1. \( x = \frac{-15 + 9}{2} = \frac{-6}{2} = -3 \)
2. \( x = \frac{-15 - 9}{2} = \frac{-24}{2} = -12 \)

Thus, the zeros of the quadratic function are:

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

To summarize:

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