Question

What are the zeros of the quadratic function f(x) = 8x2 – 16x – 15?

Answer 1

8x^2 - 16x - 15 = 0

x = [ -(-16) +/- sqrt((-16)^2 - 4*8*-15) / ] 2*8

= 2.696, -0.696

Answer 2

Answer:

The zeroes of the given polynomial f(x) are

$x=1-\sqrt{\dfrac{23}{8}},~~~x=1+\sqrt{\dfrac{23}{8}}$

Step-by-step explanation:

Given : Quadratic function $f(x)=8x^2-16x-15$

To find : What are the zeros of the quadratic function?

Solution :

Using quadratic formula of the general equation $ax^2+bx+c=0$ to get the roots is $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

Comparing the given quadratic equation,

a = 8, b = -16  and  c = - 15.

Substitute the value in the formula,

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\\\Rightarrow x=\dfrac{-(-16)\pm\sqrt{(-16)^2-4\times 8\times (-15)}}{2\times 8}\\\\\\\Rightarrow x=\dfrac{16\pm\sqrt{256+480}}{16}\\\\\\\Rightarrow x=\dfrac{16\pm\sqrt{786}}{16}\\\\\\\Rightarrow x=\dfrac{16\pm4\sqrt{46}}{16}\\\\\\\Rightarrow x=\dfrac{4\pm\sqrt{46}}{4}\\\\\\\Rightarrow x=1\pm\sqrt{\dfrac{46}{16}}\\\\\\\Rightarrow x=1\pm\sqrt{\dfrac{23}{8}}\\\\\\\Rightarrow x=1-\sqrt{\dfrac{23}{8}},~~~x=1+\sqrt{\dfrac{23}{8}}.$

Therefore, The zeroes of the given polynomial f(x) are

$x=1-\sqrt{\dfrac{23}{8}},~~~x=1+\sqrt{\dfrac{23}{8}}$