Question

someone please solve this​

Answer 1

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

$20 {x}^{2} + 20 = 41x$

20x^2-41x+20

Answer 2

Step-by-step explanation:

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

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

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

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

Applying Quadratic Formula,

$x = \sf \dfrac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac} }{2a}$

$x = \dfrac{41 \: \pm \: \sqrt{1681 - 1600} }{40} \\ x = \dfrac{50}{40} \: or \: x = \dfrac{32}{40} \: \: \: \\ x = \dfrac{5}{4} \: or \: x = \frac{4}{5}$