Question

Sum of no. and its reciprocal is 41/20 find the no.

Answer 1

x + 1/x = 41/20 

multiply both sides of equation by 20x 
20x²-41x+20 = 0 
(4x-5)(5x-4) = 0 
x = 5/4 or 4/5
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Answer 2

The number is $\frac{5}{4} , \frac{4}{5}$

Step-by-step explanation:

Let the number be x

Its reciprocal = $\frac{1}{x}$

We are given that the sum and its reciprocal is $\frac{41}{20}$

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

$\frac{x^2+1}{x}=\frac{41}{20}$

$20x^2+20=41x$

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

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

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

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

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Sum of a no. and its reciprocal is 3 41/80 find the no

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