To find the zeros of the quadratic function [tex]\( f(x) = 2x^2 + 16x - 9 \)[/tex], we need to solve the quadratic equation [tex]\( 2x^2 + 16x - 9 = 0 \)[/tex].
The general form of a quadratic equation is [tex]\( ax^2 + bx + c = 0 \)[/tex], where [tex]\( a = 2 \)[/tex], [tex]\( b = 16 \)[/tex], and [tex]\( c = -9 \)[/tex]. The solutions for this equation can be found using the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
First, let's compute the discriminant [tex]\(\Delta\)[/tex]:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
Substituting the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[
\Delta = 16^2 - 4 \cdot 2 \cdot (-9)
\][/tex]
[tex]\[
\Delta = 256 + 72
\][/tex]
[tex]\[
\Delta = 328
\][/tex]
Next, we calculate the square root of the discriminant [tex]\(\sqrt{\Delta}\)[/tex]:
[tex]\[
\sqrt{328} = 2 \sqrt{82}
\][/tex]
Now, we substitute [tex]\(\Delta\)[/tex] back into the quadratic formula:
[tex]\[
x = \frac{-16 \pm 2\sqrt{82}}{4}
\][/tex]
Simplifying:
[tex]\[
x = \frac{-16 \pm 2\sqrt{82}}{4} = \frac{-16 \pm \sqrt{82}}{2}
\][/tex]
[tex]\[
x = -4 \pm \sqrt{\frac{41}{2}}
\][/tex]
This means the solutions are:
[tex]\[
x_1 = -4 - \sqrt{\frac{41}{2}}
\][/tex]
[tex]\[
x_2 = -4 + \sqrt{\frac{41}{2}}
\][/tex]
Therefore, the correct pair of zeros for the quadratic function [tex]\( f(x) = 2x^2 + 16x - 9 \)[/tex] is:
[tex]\[
x = -4 - \sqrt{\frac{41}{2}} \quad \text{and} \quad x = -4 + \sqrt{\frac{41}{2}}
\][/tex]
Hence, the correct choice is:
[tex]\[
x = -4 - \sqrt{\frac{41}{2}}, \quad x = -4 + \sqrt{\frac{41}{2}}
\][/tex]
