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

2(4x - 5) = 494x - 5 = 49/24x = 24.5 + 5x = 29.5 / 4x = 7.375...

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

Ответ:

westball101

12.03.2022

Mathematics

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.

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

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