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=
Remaining Attempts : 3

Answer 1

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

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

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

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 roots using the quadratic formula:

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

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: \( x = -3 \)

So, the final answers are:

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