Math, asked by theruler2126, 3 months ago

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

Answers

Answered by Flaunt
274

\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

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