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

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

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

2. Now we can apply the values into the quadratic formula:

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

Calculating the two possible values for \( x \):

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

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

Thus, the zeros of the quadratic function are:

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

So we write:

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

Answer 2

Determine the zeros of the quadratic function y=−x2+16x−39
. (When listing the smaller zero, remember that negative values are always smaller than positive values.)(1 point)
smaller zero: x=
; greater zero: x=