Question

The sum of a number and its reciprocal is 41/20. Find the number. smaller value larger value

Answer 1

Step-by-step explanation:

Let the number is x. Its reciprocal will be 1/x. According to given condition, we get :

$x+\dfrac{1}{x}=\dfrac{41}{20}\\\\\dfrac{x^2+1}{x}=\dfrac{41}{20}\\\\20x^2+20=41x\\\\20x^2-41x+20=0$

It is a quadratic equation. The values of x are :

$20x^2-25x - 16x+ 20 =0\\\\5x(4x - 5) -4( 4x - 5) = 0\\\\(5x-4)(4x-5)=0\\\\5x-4=0\ \text{and}\ (4x-5)=0\\\\x=\dfrac{4}{5}\ \text{and}\ x=\dfrac{5}{4}$

The smaller number is 4/5 and the larger number is 5/4.