Question

$solve \: the \: euation \: \frac{1}{x} - \frac{1}{x - 2} = 3 \:$
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Answer 1

Solution:

Given –

$\tt \longrightarrow \dfrac{1}{x} - \dfrac{1}{x - 2} = 3$

LCM of x and (x - 2) is x(x - 2),

$\tt \longrightarrow \dfrac{x - 2 - x}{x(x - 2)} = 3$

$\tt \longrightarrow \dfrac{- 2}{ {x}^{2} - 2x} = 3$

Moving x² - 2x to the right side, we get,

$\tt \longrightarrow - 2 = 3( {x}^{2} - 2x)$

$\tt \longrightarrow 3{x}^{2} - 6x + 2 = 0$

Now,

→ a = 3

→ b = -6

→ c = 2

Therefore,

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

$\tt \longrightarrow x = \dfrac{ -( -6 ) \pm \sqrt{ { (- 6)}^{2} - 4 \times 3 \times 2} }{2 \times 3}$

$\tt \longrightarrow x = \dfrac{6\pm \sqrt{36- 24} }{6}$

$\tt \longrightarrow x = \dfrac{6\pm \sqrt{12} }{6}$

$\tt \longrightarrow x = \dfrac{6\pm2 \sqrt{3} }{6}$

$\tt \longrightarrow x = \dfrac{2(3\pm\sqrt{3}) }{6}$

$\tt \longrightarrow x = \dfrac{(3\pm\sqrt{3}) }{3}$

Therefore,

$\tt \longrightarrow x_{1} = \dfrac{3 + \sqrt{3}}{3}$

$\tt \longrightarrow x_{2} = \dfrac{3 - \sqrt{3}}{3}$

• So, the roots of the given equation are - (3 + √3)/3 and (3 - √3)/3

Answer:

• x = (3 + √3)/3 and (3 - √3)/3