Question

what are the zeros of the quadratic function f(x)=8x2-16x-15

Answer 1

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:  We are given to find the zeroes of the quadratic function below:

$f(x)=8x^2-16x-15.$

To find the zeroes, we must find the roots of the following equation

$f(x)=0\\\\\Rightarrow 8x^2-16x-15=0~~~~~~~~~~~~~~~~~~~~~~~(i)$.

We know that the roots of the quadratic equation of the form $ax^2+bx+c=0,~a\neq 0$ is given by

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.$

In the given quadratic equation , a = 8, b = -16  and  c = - 15.

Therefore, the roots of the equation are

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

Thus, the zeroes of the given polynomial f(x) are

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

Option (C) is correct.

Answer 2

The zeroes of the given polynomial f(x) = 8x^2-16x-15 are x = 1 + $\sqrt{23}$ / 8 and x = 1 -$\sqrt{23}$ / 8 Thus, Option C is correct.

What is a quadratic equation?

A quadratic equation is the second-order degree algebraic expression in a variable. the standard form of this expression is  ax² + bx + c = 0 where a. b are coefficients and x is the variable and c is a constant.

We are given to find the zeroes of the quadratic function f(x) = 8x^2 - 16x-15

f(x) = $8x^2-16x-15$ = 0

We know that the roots of the quadratic equation ax² + bx + c = 0 is

x = -b ± $\sqrt{b^2 - 4ac}$ / 2a

In the given quadratic equation , a = 8, b = -16  and  c = - 15.

So,

x = -b ± $\sqrt{b^2 - 4ac}$ / 2a

x = -(-16) ± $\sqrt{(-16)^2 - 4(8)(-15)}$ / 2(8)

x = 16  ± $\sqrt{256+ 480}$ / 16

x = 16  ± $\sqrt{786}$ / 16

x = 4  ± $\sqrt{46}$ / 4

x = 1 ± $\sqrt{23}$ / 8

Thus, the zeroes of the given polynomial f(x) are

x = 1 + $\sqrt{23}$ / 8 and x = 1 -$\sqrt{23}$ / 8. Option C is correct.

Learn more about quadratic equations;

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