Join IDNLearn.com to access a wealth of knowledge and get your questions answered by experts. Join our interactive Q&A community and get reliable, detailed answers from experienced professionals across a variety of topics.
Sagot :
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]
[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]
Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. IDNLearn.com is your reliable source for answers. We appreciate your visit and look forward to assisting you again soon.