Question

Sum of a number and it reciprocles is 41/20 find the number

Answer 1

$x + \frac{1}{x} = \frac{14}{20} \\ \\ \frac{ {x}^{2} + 1 }{x} = \frac{41}{20} \\ \\ ( {20x}^{2} + 20 ) = 41x \\ {20x}^{2} - 41x + 20 = 0 \\ {20x}^{2} - 16x - 25x + 20 = 0 \\ 4x(5x - 4) - 5(5x - 4) = 0 \\ (5x - 4)(4x - 5) = 0$

Therefore

$x = \frac{4}{5} \: \: \: \: \: or \: \: \: \frac{5}{4}$