What are the zeros of the quadratic function [tex]f(x)=8x^2-16x-15[/tex]?

A. [tex]x=-1-\sqrt{2}[/tex] and [tex]x=-1+\sqrt{2}[/tex]
B. [tex]x=-1-\sqrt{\frac{15}{8}}[/tex] and [tex]x=-1+\sqrt{\frac{15}{8}}[/tex]
C. [tex]x=1-\sqrt{\frac{23}{8}}[/tex] and [tex]x=1+\sqrt{\frac{23}{8}}[/tex]
D. [tex]x=1-\sqrt{7}[/tex] and [tex]x=1+\sqrt{7}[/tex]



Answer :

To find the zeros of the quadratic function [tex]\( f(x) = 8x^2 - 16x - 15 \)[/tex], we use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 8 \)[/tex], [tex]\( b = -16 \)[/tex], and [tex]\( c = -15 \)[/tex].

1. Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substitute [tex]\( a = 8 \)[/tex], [tex]\( b = -16 \)[/tex], and [tex]\( c = -15 \)[/tex]:
[tex]\[ \Delta = (-16)^2 - 4 \cdot 8 \cdot (-15) \][/tex]
[tex]\[ \Delta = 256 + 480 \][/tex]
[tex]\[ \Delta = 736 \][/tex]

2. Find the square root of the discriminant:
[tex]\[ \sqrt{736} \][/tex]
This is the value we need to evaluate the quadratic formula.

3. Apply the quadratic formula:
[tex]\[ x = \frac{-(-16) \pm \sqrt{736}}{2 \cdot 8} \][/tex]
Simplify:
[tex]\[ x = \frac{16 \pm \sqrt{736}}{16} \][/tex]

4. Separate into two solutions:
[tex]\[ x_1 = \frac{16 - \sqrt{736}}{16} \][/tex]
[tex]\[ x_2 = \frac{16 + \sqrt{736}}{16} \][/tex]

5. Simplify each solution:

[tex]\[ x_1 = 1 - \frac{\sqrt{736}}{16} \][/tex]
[tex]\[ x_2 = 1 + \frac{\sqrt{736}}{16} \][/tex]

Given the result:
- The discriminant [tex]\( \Delta = 736 \)[/tex]
- The approximate solutions are:
[tex]\[ x_1 \approx -0.695582495781317 \][/tex]
[tex]\[ x_2 \approx 2.6955824957813173 \][/tex]

Comparing these to the provided options:

- [tex]\( x=-1-\sqrt{2} \)[/tex] and [tex]\( x=-1+\sqrt{2} \)[/tex]
- [tex]\( x=-1-\sqrt{\frac{15}{8}} \)[/tex] and [tex]\( x=-1+\sqrt{\frac{15}{8}} \)[/tex]
- [tex]\( x=1-\sqrt{\frac{23}{8}} \)[/tex] and [tex]\( x=1+\sqrt{\frac{23}{8}} \)[/tex]
- [tex]\( x=1-\sqrt{7} \)[/tex] and [tex]\( x=1+\sqrt{7} \)[/tex]

The correct solution matches with:
[tex]\[ x = 1 - \sqrt{7} \quad \text{and} \quad x = 1 + \sqrt{7} \][/tex]

Hence, the zeros of the quadratic function [tex]\( f(x) = 8x^2 - 16x - 15 \)[/tex] are:

[tex]\[ x = 1 - \sqrt{7} \quad \text{and} \quad x = 1 + \sqrt{7} \][/tex]