Question

The sum of a number and its reciprocal 41/20 find the number​

Answer 1

Answer:

Step-by-step explanation:

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

Step-by-step explanation:

Let the number be x.

Therefore, reciprocal is

$\frac{1}{x}$

According to the problem,

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

Therefore,

either

$4x - 5 = 0 \\ = > 4x = 5 \\ = > x = \frac{5}{4}$

or,

$5x - 4 = 0 \\ = > 5x = 4 \\ = > x = \frac{4}{5}$

Therefore, number is either 4/5 or 5/4.

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