Question

The sum of a non zero number and its reciproal is $\frac{41}{20}$. Find the no.

Answer 1

Let the number be $x(x\ne 0)$

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

or, $20x^2+20=41x$

or, $20x^2-41x+20=0$

or, $20x^2-25x-16x+20=0$

or, $5x(4x-5)-4(4x-5)=0$

or, $(4x-5)(5x-4)=0$

either $4x-5=0$ OR $5x-4=0$

$\therefore x=\frac{5}{4}$ OR $\therefore x=\frac{4}{5}$