Solve $x + \frac{1}{x} = 25 \frac{1}{25}$

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Solution

$x + \frac{1}{x} = 25 \frac{1}{25}$
$\frac{x^{2} + 1}{x} = \frac{626}{25}$
$25 x^{2} + 25 = 626 x$
$25 x^{2} - 626 x + 25 = 0$
using Shridharacharya rule
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
$x = \frac{626 \pm \sqrt{626^{2} - 4 \times 25 \times 25}}{2 \times 25}$
$x = \frac{626 \pm 624}{50}$
$x = \frac{626 + 624}{50} = 25$
$x = \frac{626 - 624}{50} = \frac{1}{25}$
Therefore $x = 25 o r \frac{1}{25}$

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Similar questions

x + \frac{1}{x} = 25 \frac{1}{25}

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