What are the zeros of the quadratic function [tex]f(x) = 2x^2 + 16x - 9[/tex]?

A. [tex]x = -4 - \sqrt{\frac{7}{2}}[/tex] and [tex]x = -4 + \sqrt{\frac{7}{2}}[/tex]

B. [tex]x = -4 - \sqrt{\frac{25}{2}}[/tex] and [tex]x = -4 + \sqrt{\frac{25}{2}}[/tex]

C. [tex]x = -4 - \sqrt{\frac{21}{2}}[/tex] and [tex]x = -4 + \sqrt{\frac{21}{2}}[/tex]

D. [tex]x = -4 - \sqrt{\frac{41}{2}}[/tex] and [tex]x = -4 + \sqrt{\frac{41}{2}}[/tex]



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]