Question

$\sf \: Solve : \: \dfrac{1}{x} - \dfrac{1}{x - 2} = 3$​

Answer 1

Answer:

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Answer 2

Required Answer:-

According to the question

$\red \implies \boxed {\frac{1}{x} - \frac{1}{x} - 2 = 3}$

Calculation:-

$\pink \rightarrow \frac{1}{x } - \frac{1}{x} - 2 = 3$

$\pink \rightarrow(x - 2) - x$

$\pink \rightarrow3x(x - 2)$

$\pink \rightarrow x - 2 - x$

$\pink \rightarrow {3x}^{2} - 6x$

$\pink \rightarrow {3x}^{2} - 6x + 2 = 0$

$\pink \rightarrow x = \frac{ - b \: + \: \sqrt{ {b}^{2} \: - \: 4ac } }{29}$

${ \blue{ \fbox{ \pink{a = 3, \: b = - 6, \: c = 2}}}}$

$\boxed{x = - \frac{ - ( - 6) + ( - 6) ^{2} - 4 \times 3 \times 2 }{2 \times 3} }$

$\pink \rightarrow x = \frac{6 + \sqrt{36 - 24} }{6}$

$\pink \rightarrow x = \frac{6 + \sqrt{12} }{6}$

$\pink \rightarrow x = \frac{6 + 2 \sqrt{3} }{6}$

$\pink \rightarrow x = \frac{3 + 1 \sqrt{3} }{3}$

$\pink \rightarrow \sf Thus, \mathbb x = \frac{3 + 1 \sqrt{3} }{3}$

Therefore, roots of quadratic equation

$\boxed{\frac{1}{x} - \frac{1}{x} - 2 = 3 \sf \: are \: \mathbb x = \frac{3 + 1 \sqrt{3} }{3}, x= \frac{3 - 1 \sqrt{3} }{3} }$

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