Question

The sum of a non-zero number and its reciprocal is $\frac{41}{20}$. Find that number.

Answer 1

let the number be $x$
then,

According to the qusetion

$x+\frac{1}{x}=\frac{41}{20}$

$\frac{x^2+1}{x}=\frac{41}{20}$
$20(x^2+1)=41x$
$\Rightarrow 20x^2+20-41x=0$
$\Rightarrow 20x^2-16x-25x+20=0$
$\Rightarrow 4x(5x-4)-5(5x-4)=0$
$\Rightarrow (4x-5)(5x-4)=0$
When, $4x-5=0\Rightarrow x=\frac{5}{4}$

and when $5x-4=0\Rightarrow x=\frac{4}{5}$

So, $x=\frac{5}{4},\frac{4}{5}$

Hence, the number is $\frac{5}{4}or\frac{4}{5}$