Question

$\huge\mathfrak\pink{Solve}$
${the \: above \: equation \: }$
​

Answer 1

Given:

$x + \frac{1}{x} = 2 \times \frac{1}{20}$

Let's factorize:

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

Answer 2

• Answer

x + $\dfrac{1}{x}$= 2$\dfrac{1}{20}$

=> x + $\dfrac{1}{x}$= $\dfrac{41}{20}$

=> (x² + 1)20 = 41x

=> 20x² - 41x + 20 = 0

=> 20x² - 25x - 16x + 20

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

=> 5x -4 = 0 , 4x-5 = 0

=> x = $\dfrac{4}{5}$ , x = $\dfrac{5}{4}$

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Hence required solution is

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