Question

x-2/x+2+x+2/x-2=4 find the quadratic equations ​

Answer 1

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

$\sf \longmapsto \dfrac{x - 2}{x + 2} = \dfrac{x + 2}{x - 2} = 4$

Identity used here :

• (a+b)(a-b)=a²-b²
• (a+b)²=a²+b²+2ab
• (a-b)²=a²+b²-2ab

$\sf \longmapsto \dfrac{(x - 2)(x - 2) + (x + 2)(x + 2)}{(x + 2)(x - 2)}$

$\sf \longmapsto \dfrac{ {(x - 2)}^{2} + {(x + 2)}^{2} }{ {x}^{2} - {(2)}^{2} } = 4$

$\sf \longmapsto \dfrac{ {x}^{2} + {2}^{2} - 2(x)(2) + {x}^{2} + {2}^{2} + 2(x)(2)}{ {x}^{2} - {2}^{2} } = 4$

$\sf \longmapsto \dfrac{ {x}^{2} + 4{ \cancel{ - 4x }}+ {x}^{2} + 4 + { \cancel{4x}}}{ {x}^{2} - {2}^{2} } = 4$

$\sf \longmapsto \dfrac{2 {x}^{2} + 8 }{ {x}^{2} - 4 } = 4$

$\sf \longmapsto2 {x}^{2} + 8 = 4 {x}^{2} - 16$

$\sf \longmapsto2( {x}^{2} + 4) = 4( {x}^{2} - 4)$

$\sf \longmapsto {x}^{2} + 4 = 2( {x}^{2} - 4)$

$\sf \longmapsto {x}^{2} + 4 = 2 {x}^{2} - 8$

$\sf \longmapsto4 + 8 = 2 {x}^{2} - {x}^{2}$

$\sf \longmapsto {x}^{2} = 12$

$\sf \longmapsto { \bold{ x = { \bold{ \sqrt{12} = \pm2 \sqrt{3} }}}}$

Check:-

$\sf \longmapsto \dfrac{x - 2}{x + 2} + \dfrac{x + 2}{x - 2} = 4$

$\sf \longmapsto \dfrac{2 \sqrt{3} - 2 }{2 \sqrt{3} + 2} + \dfrac{2 \sqrt{3} + 2 }{2 \sqrt{3} - 2}$

$\sf \longmapsto \dfrac{(2 \sqrt{3} - 2)(2 \sqrt{3} - 2) + (2 \sqrt{3} + 2)(2 \sqrt{3} + 2) }{ {(2 \sqrt{3} )}^{2} - {2}^{2} }$

$\sf \longmapsto \dfrac{ {(2 \sqrt{3} - 2)}^{2} + {(2 \sqrt{3} + 2)}^{2} }{12 - 4}$

$\sf \longmapsto \dfrac{ {(2 \sqrt{3}) }^{2} + {2}^{2} - 2(2 \sqrt{3} )(2) + {(2 \sqrt{3} )}^{2} + {2}^{2} + 2(2)(2 \sqrt{3} )}{8}$

$\sf \longmapsto \dfrac{12 + 4 { \cancel{- 8 \sqrt{3}}} + 12 + 4 + { \cancel{8 \sqrt{3} }} }{8}$

$\sf \longmapsto \dfrac{12 + 4 + 12 + 4}{8} = \dfrac{32}{8} = 4$