Answer :
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]
where [tex]\(a = 2\)[/tex], [tex]\(b = 16\)[/tex], and [tex]\(c = -9\)[/tex].
1. Calculate the discriminant ([tex]\(b^2 - 4ac\)[/tex]):
[tex]\[ \text{Discriminant} = b^2 - 4ac = 16^2 - 4 \cdot 2 \cdot (-9) = 256 + 72 = 328 \][/tex]
2. Find the square root of the discriminant:
[tex]\[ \sqrt{328} = \sqrt{4 \cdot 82} = 2\sqrt{82} \][/tex]
3. Apply the quadratic formula to find the two solutions [tex]\(x_1\)[/tex] and [tex]\(x_2\)[/tex]:
[tex]\[ x_1 = \frac{-16 + 2\sqrt{82}}{4} = \frac{-16 + 2\sqrt{82}}{4} = -4 + \frac{\sqrt{82}}{2} \][/tex]
[tex]\[ x_2 = \frac{-16 - 2\sqrt{82}}{4} = -4 - \frac{\sqrt{82}}{2} \][/tex]
4. Convert the solutions into forms matching the provided options:
Since [tex]\(\frac{\sqrt{82}}{2}\)[/tex] is the same as [tex]\(\sqrt{\frac{82}{4}} = \sqrt{\frac{41}{2}}\)[/tex], we can rewrite the solutions as:
[tex]\[ x_1 = -4 + \sqrt{\frac{41}{2}} \][/tex]
[tex]\[ x_2 = -4 - \sqrt{\frac{41}{2}} \][/tex]
Thus, the zeros of the quadratic function [tex]\( f(x) = 2x^2 + 16x - 9 \)[/tex] are:
[tex]\[ x = -4 + \sqrt{\frac{41}{2}} \quad \text{and} \quad x = -4 - \sqrt{\frac{41}{2}} \][/tex]
Therefore, the correct answer is:
[tex]\[ x=-4-\sqrt{\frac{41}{2}} \quad \text{and} \quad x=-4+\sqrt{\frac{41}{2}} \][/tex]
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\(a = 2\)[/tex], [tex]\(b = 16\)[/tex], and [tex]\(c = -9\)[/tex].
1. Calculate the discriminant ([tex]\(b^2 - 4ac\)[/tex]):
[tex]\[ \text{Discriminant} = b^2 - 4ac = 16^2 - 4 \cdot 2 \cdot (-9) = 256 + 72 = 328 \][/tex]
2. Find the square root of the discriminant:
[tex]\[ \sqrt{328} = \sqrt{4 \cdot 82} = 2\sqrt{82} \][/tex]
3. Apply the quadratic formula to find the two solutions [tex]\(x_1\)[/tex] and [tex]\(x_2\)[/tex]:
[tex]\[ x_1 = \frac{-16 + 2\sqrt{82}}{4} = \frac{-16 + 2\sqrt{82}}{4} = -4 + \frac{\sqrt{82}}{2} \][/tex]
[tex]\[ x_2 = \frac{-16 - 2\sqrt{82}}{4} = -4 - \frac{\sqrt{82}}{2} \][/tex]
4. Convert the solutions into forms matching the provided options:
Since [tex]\(\frac{\sqrt{82}}{2}\)[/tex] is the same as [tex]\(\sqrt{\frac{82}{4}} = \sqrt{\frac{41}{2}}\)[/tex], we can rewrite the solutions as:
[tex]\[ x_1 = -4 + \sqrt{\frac{41}{2}} \][/tex]
[tex]\[ x_2 = -4 - \sqrt{\frac{41}{2}} \][/tex]
Thus, the zeros of the quadratic function [tex]\( f(x) = 2x^2 + 16x - 9 \)[/tex] are:
[tex]\[ x = -4 + \sqrt{\frac{41}{2}} \quad \text{and} \quad x = -4 - \sqrt{\frac{41}{2}} \][/tex]
Therefore, the correct answer is:
[tex]\[ x=-4-\sqrt{\frac{41}{2}} \quad \text{and} \quad x=-4+\sqrt{\frac{41}{2}} \][/tex]
