Question

if x = √3+√2/√3−√2 , find x²+1/x
​

Answer 1

Question :-

$\sf if \: \: x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \:, \: find \: \: \: (1) \: {x}^{2} + \frac{1}{ {x}^{2} } \: \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (2) {x}^{4} + \frac{1}{ {x}^{4} }$

Answer :-

$(1) \: \to \boxed{ \sf{x}^{2} + \frac{1}{ {x}^{2} } = 98 \: \: \: } \: \\ (2)\to \boxed{\sf {x}^{4} + \frac{1}{ {x}^{4} } = 9602} \:$

Solution :-

We have ;

$\to \sf x \: = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ \sf rationalise \: by \: \sqrt{3} + \sqrt{2} \\ \\ \to \sf x = \bigg( \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \bigg) \times \bigg(\frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \bigg) \\ \\ \: \: \: \: \: \: \: \: \: = \frac{{( \sqrt{3} + \sqrt{2}) }^{2} }{ {( \sqrt{3}) }^{2} - {( \sqrt{2} )}^{2} } \\ \\ \: \: \: \: \: \: \: \: \: = \: \frac{3 + 2 + 2 \sqrt{6} }{1} \\ \\ \to \boxed{\sf x \: = 5 + 2 \sqrt{6} }$

Now ,

$\to \sf {x}^{2} = {(5 + 2 \sqrt{6}) }^{2} \\ \\ \: \: \: \: \: \: \: \: \: \: \: = 25 + {(2 \sqrt{6}) }^{2} + 20 \sqrt{6} \\ \\ \: \: \: \: \: \: \: \: \: \: \: = 25 \: + 24 + 20 \sqrt{6} \\ \\ \: \: \: \: \: \: \: \: \: \: \: = 49 + 20 \sqrt{6}$

Hence ,

$\implies \sf {x}^{2} + \frac{1}{ {x}^{2} } \\ \\ \implies \sf (49 + 20 \sqrt{6} ) + \frac{1}{(49 + 20 \sqrt{6} )} \\ \\ \implies \: 49 + 20 \sqrt{6} + \frac{(49 - 20 \sqrt{6}) }{ {(49)}^{2} - {(20 \sqrt{6} )}^{2} } \\ \\ \implies \: 49 + 20 \sqrt{6} + \frac{(49 - 20 \sqrt{6}) }{2401 - 2400} \\ \\ \implies \: 49 + 20 \sqrt{6} + 49 - 20 \sqrt{6} \\ \\ \implies \: 98 \\ \\ \to \boxed{ \sf{x}^{2} + \frac{1}{ {x}^{2} } = 98}$

Now ,

$\to \sf { \bigg( {x}^{2} + \frac{1}{ {x}^{2} } \bigg) }^{2} = {x}^{4} + \frac{1}{ {x}^{4} } + 2 \\ \\ \to \sf {(98)}^{2} = {x}^{4} + \frac{1}{ {x}^{4} } + 2 \\ \\ \to \sf {x}^{4} + \frac{1}{ {x}^{4} } = 9604 - 2 \\ \\ \to \boxed{\sf {x}^{4} + \frac{1}{ {x}^{4} } = 9602}$