To solve for the zeros of the quadratic function [tex]\( f(x) = 16x^2 + 32x - 9 \)[/tex], you need to find the values of [tex]\( x \)[/tex] that satisfy the equation [tex]\( f(x) = 0 \)[/tex]. Let's explore which, if any, of the given values of [tex]\( x \)[/tex] satisfy this equation.
1. Checking [tex]\( x = -5.25 \)[/tex]:
[tex]\[
f(-5.25) = 16(-5.25)^2 + 32(-5.25) - 9
\][/tex]
Simplifying inside the parentheses first:
[tex]\[
(-5.25)^2 = 27.5625
\][/tex]
Then,
[tex]\[
f(-5.25) = 16(27.5625) + 32(-5.25) - 9 = 441 - 168 - 9 = 264 \ne 0
\][/tex]
So, [tex]\( x = -5.25 \)[/tex] is not a zero of [tex]\( f(x) \)[/tex].
2. Checking [tex]\( x = -2.25 \)[/tex]:
[tex]\[
f(-2.25) = 16(-2.25)^2 + 32(-2.25) - 9
\][/tex]
Simplifying inside the parentheses first:
[tex]\[
(-2.25)^2 = 5.0625
\][/tex]
Then,
[tex]\[
f(-2.25) = 16(5.0625) + 32(-2.25) - 9 = 81 - 72 - 9 = 0
\][/tex]
So, [tex]\( x = -2.25 \)[/tex] is a zero of [tex]\( f(x) \)[/tex].
3. Checking [tex]\( x = -1.25 \)[/tex]:
[tex]\[
f(-1.25) = 16(-1.25)^2 + 32(-1.25) - 9
\][/tex]
Simplifying inside the parentheses first:
[tex]\[
(-1.25)^2 = 1.5625
\][/tex]
Then,
[tex]\[
f(-1.25) = 16(1.5625) + 32(-1.25) - 9 = 25 - 40 - 9 = -24 \ne 0
\][/tex]
So, [tex]\( x = -1.25 \)[/tex] is not a zero of [tex]\( f(x) \)[/tex].
4. Checking [tex]\( x = -0.25 \)[/tex]:
[tex]\[
f(-0.25) = 16(-0.25)^2 + 32(-0.25) - 9
\][/tex]
Simplifying inside the parentheses first:
[tex]\[
(-0.25)^2 = 0.0625
\][/tex]
Then,
[tex]\[
f(-0.25) = 16(0.0625) + 32(-0.25) - 9 = 1 - 8 - 9 = -16 \ne 0
\][/tex]
So, [tex]\( x = -0.25 \)[/tex] is not a zero of [tex]\( f(x) \)[/tex].
After evaluating each given value, we find that the only solution for [tex]\( f(x) = 16x^2 + 32x - 9 = 0 \)[/tex] from the provided options is:
[tex]\[
\boxed{x = -2.25}
\][/tex]
