Question

A number is greater than twice its reciprocal by 31/4. Find the number.

Answer 1

Answer:

$x=\frac{31+\sqrt{833}}{16}, \frac{31\-\sqrt{833}}{16}$

Step-by-step explanation:

Given : A number is greater than twice its reciprocal by 31/4.

To find : The number

Solution :

Let the number be x,

So, according to question

$x+\frac{2}{x}=\frac{31}{4}$

$\frac{x^2+2}{x}=\frac{31}{4}$

$4x^2+8=31x$

$4x^2-31x+8=0$

Now, solving quadratic equation,

Solution is $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

a=4 , b=-31 , c=8

$x=\frac{-(-31)\pm\sqrt{(-31)^2-4(4)(8)}}{2(8)}$

$x=\frac{31\pm\sqrt{961-128}}{16}$

$x=\frac{31\pm\sqrt{833}}{16}$

Therefore, The value of x is

$x=\frac{31+\sqrt{833}}{16}, \frac{31\-\sqrt{833}}{16}$

Answer 2

Answer:

write the reciprocal of the following numbers 31/4 is