Question

solve
$\frac{1}{x} - \frac{1}{x - 2} equles3$
​

Answer 1

$\sf\huge\bold{\underline{\underline{{Solution}}}}$

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

$\sf \longmapsto \dfrac{1(x - 2) - 1(x)}{x(x - 2)}=3$

$\sf \longmapsto \dfrac{x - 2 - x}{ {x}^{2} - 2x} = 3$

$\sf \longmapsto \dfrac{ - 2}{ {x}^{2} - 2x } = 3$

$\sf \longmapsto3 {x}^{2} - 6x = - 2$

$\sf \longmapsto3 {x}^{2} - 6x + 2 = 0$

It is in the form of a quadratic equation so ,we simply factorise it.

We will use quadratic formula to factorise.

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

$\sf \longmapsto \: x = \dfrac{6 \pm \sqrt{ {(6)}^{2} - 4(3)(2)} }{6}$

$\sf \longmapsto \: x = \dfrac{6 \pm \sqrt{36 - 24} }{6}$

$\sf \longmapsto \: x = \dfrac{6 \pm \sqrt{12} }{6}$

$\sf \longmapsto x = \dfrac{6 \pm2 \sqrt{3} }{6} = \dfrac{2(3 \pm \sqrt{3} )}{6}$

$\sf \longmapsto \: x = \dfrac{3 \pm \sqrt{3} }{3}$

$\sf \longmapsto \: x = \dfrac{ \sqrt{3} ( \sqrt{3} \pm 1) }{ \sqrt{3} \times \sqrt{3} }$

$\sf \longmapsto \: x = \dfrac{ \sqrt{3} \pm 1}{ \sqrt{3} }$

$\sf \bold{\red{x = 1 + \dfrac{1}{ \sqrt{3} } \: or \: 1 - \dfrac{1}{ \sqrt{3} }}}$

Check

$\sf \longmapsto3 {x}^{2} - 6x + 2 = 0$

$\sf \longmapsto3 {(1 + \frac{1}{ \sqrt{3} }) }^{2} - 6(1 + \frac{1}{ \sqrt{3} } ) + 2$

$\sf \longmapsto3( {1}^{2} + \frac{1}{3} + 2 \times 1 \times \frac{1}{ \sqrt{3} } ) - 6( \frac{ \sqrt{3} + 1}{ \sqrt{3} } ) + 2$

$\sf \longmapsto3( \frac{3 + 1}{3} + \frac{2}{ \sqrt{3} } ) - 6( \frac{ \sqrt{3} + 1}{ \sqrt{3} } ) + 2$

$\sf \longmapsto3( \frac{4 + 2 \sqrt{3} }{3} ) - 6( \frac{ \sqrt{3} + 1}{ \sqrt{3} } ) + 2$

$\sf \longmapsto \frac{12 + 6 \sqrt{3} - 6(( \sqrt{3} + 1) \sqrt{3} )}{3} + 2$

$\sf \longmapsto \frac{12 + { \cancel{6 \sqrt{3} }} - 6 \times 3 - { \cancel{6 \sqrt{3}}} + 6}{3} = \frac{12 - 18 + 6}{3}$

$\sf \longmapsto \frac{ - 6 + 6}{3} = 0$