Respuesta :
8x^2 - 16x - 15 = 0
x = [ -(-16) +/- sqrt((-16)^2 - 4*8*-15) / ] 2*8
= 2.696, -0.696
Answer:
The zeroes of the given polynomial f(x) are
[tex]x=1-\sqrt{\dfrac{23}{8}},~~~x=1+\sqrt{\dfrac{23}{8}}[/tex]
Step-by-step explanation:
Given : Quadratic function [tex]f(x)=8x^2-16x-15[/tex]
To find : What are the zeros of the quadratic function?
Solution :
Using quadratic formula of the general equation [tex]ax^2+bx+c=0[/tex] to get the roots is [tex]x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
Comparing the given quadratic equation,
a = 8, b = -16 and c = - 15.
Substitute the value in the formula,
[tex]x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\\\Rightarrow x=\dfrac{-(-16)\pm\sqrt{(-16)^2-4\times 8\times (-15)}}{2\times 8}\\\\\\\Rightarrow x=\dfrac{16\pm\sqrt{256+480}}{16}\\\\\\\Rightarrow x=\dfrac{16\pm\sqrt{786}}{16}\\\\\\\Rightarrow x=\dfrac{16\pm4\sqrt{46}}{16}\\\\\\\Rightarrow x=\dfrac{4\pm\sqrt{46}}{4}\\\\\\\Rightarrow x=1\pm\sqrt{\dfrac{46}{16}}\\\\\\\Rightarrow x=1\pm\sqrt{\dfrac{23}{8}}\\\\\\\Rightarrow x=1-\sqrt{\dfrac{23}{8}},~~~x=1+\sqrt{\dfrac{23}{8}}.[/tex]
Therefore, The zeroes of the given polynomial f(x) are
[tex]x=1-\sqrt{\dfrac{23}{8}},~~~x=1+\sqrt{\dfrac{23}{8}}[/tex]
