To find the zeros of the quadratic function [tex]\( f(x) = 2x^2 + 16x - 9 \)[/tex], we can use the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
First, identify the coefficients [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] from the quadratic equation [tex]\( 2x^2 + 16x - 9 = 0 \)[/tex]:
[tex]\[
a = 2, \quad b = 16, \quad c = -9
\][/tex]
Next, calculate the discriminant, which is given by [tex]\( \Delta = b^2 - 4ac \)[/tex]:
[tex]\[
\Delta = 16^2 - 4 \cdot 2 \cdot (-9) = 256 + 72 = 328
\][/tex]
Now, use the quadratic formula to find the roots:
[tex]\[
x_{1,2} = \frac{-16 \pm \sqrt{328}}{2 \cdot 2} = \frac{-16 \pm \sqrt{328}}{4}
\][/tex]
Simplify the expression under the square root:
[tex]\[
\sqrt{328} = \sqrt{4 \cdot 82} = 2\sqrt{82}
\][/tex]
Substitute [tex]\( \sqrt{328} \)[/tex] into the quadratic formula:
[tex]\[
x_{1,2} = \frac{-16 \pm 2\sqrt{82}}{4} = \frac{-16}{4} \pm \frac{2\sqrt{82}}{4} = -4 \pm \frac{\sqrt{82}}{2}
\][/tex]
So, the roots or zeros of the quadratic function are:
[tex]\[
x_1 = -4 - \frac{\sqrt{82}}{2}, \quad x_2 = -4 + \frac{\sqrt{82}}{2}
\][/tex]
Now let's see which of the given options match these zeros:
[tex]\[
x = -4 - \sqrt{\frac{41}{2}} \quad \text{and} \quad x = -4 + \sqrt{\frac{41}{2}}
\][/tex]
Given that [tex]\(\frac{41}{2} = 20.5\)[/tex] and [tex]\(\sqrt{41/2} = \sqrt{20.5}\)[/tex]:
[tex]\[
-4 - \sqrt{20.5} \quad \text{and} \quad -4 + \sqrt{20.5}
\][/tex]
Since [tex]\(\sqrt{82}/2 \approx \sqrt{20.5}\)[/tex], the zeros are:
[tex]\[
x = -4 - \sqrt{\frac{41}{2}} \quad \text{and} \quad x = -4 + \sqrt{\frac{41}{2}}
\][/tex]
Thus, the correct matching option is:
[tex]\[
x = -4 - \sqrt{\frac{41}{2}} \quad \text{and} \quad x = -4 + \sqrt{\frac{41}{2}}
\][/tex]
