To identify the zeros of the quadratic function [tex]\(f(x) = 2x^2 + 16x - 9\)[/tex], consider the quadratic formula, which states the solutions for a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] are given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For the quadratic function [tex]\(f(x) = 2x^2 + 16x - 9\)[/tex]:
1. Identify the coefficients:
- [tex]\(a = 2\)[/tex]
- [tex]\(b = 16\)[/tex]
- [tex]\(c = -9\)[/tex]
2. Compute the discriminant [tex]\(\Delta\)[/tex] using [tex]\(\Delta = b^2 - 4ac\)[/tex]:
[tex]\[
\Delta = 16^2 - 4 \cdot 2 \cdot (-9) = 256 + 72 = 328
\][/tex]
3. Employing the quadratic formula, the solutions [tex]\(x_1\)[/tex] and [tex]\(x_2\)[/tex] require calculating [tex]\(\sqrt{328}\)[/tex]:
[tex]\[
\sqrt{328} = \sqrt{4 \cdot 82} = 2\sqrt{82}
\][/tex]
4. Substitute these results into the quadratic formula:
[tex]\[
x_{1,2} = \frac{-16 \pm 2\sqrt{82}}{4} = \frac{-16 \pm 2\sqrt{82}}{4} = \frac{-16}{4} \pm \frac{2\sqrt{82}}{4} = -4 \pm \frac{\sqrt{82}}{2}
\][/tex]
5. Compare with the provided options to identify correct roots:
- [tex]\(x = -4 - \sqrt{\frac{41}{2}}\)[/tex] and [tex]\(x = -4 + \sqrt{\frac{41}{2}}\)[/tex], these correctly align with our solutions as [tex]\(\sqrt{\frac{41}{2}} = \frac{\sqrt{82}}{2}\)[/tex].
Thus, the correct option is:
[tex]\[ x = -4 - \sqrt{\frac{41}{2}} \quad \text{and} \quad x = -4 + \sqrt{\frac{41}{2}} \][/tex]
Hence, the zeros of the quadratic function [tex]\(f(x) = 2x^2 + 16x - 9\)[/tex] are:
[tex]\[
\boxed{4}
\][/tex]
