Question

if x = √6+√5 then x^2+1/x^2-2 is equal to?​

Answer 1

the term written is( x-1/x )^2

retionalise denominator of second term then proceed you will get your answer

Answer 2

$i ) x = \sqrt{6} + \sqrt{5} \: --(1)$

$ii ) \frac{1}{x} \\= \frac{ 1}{ (\sqrt{6} + \sqrt{5})}\\= \frac{ (\sqrt{6} - \sqrt{5})}{ (\sqrt{6} + \sqrt{5})(\sqrt{6} - \sqrt{5})}\\= \frac{ (\sqrt{6} - \sqrt{5})}{ (\sqrt{6})^{2} - (\sqrt{5})^{2}}\\= \frac{ (\sqrt{6} - \sqrt{5})}{ 6 - 5}\\= \frac{ (\sqrt{6} - \sqrt{5})}{ 1}\\= \sqrt{6} - \sqrt{5} \: --(2)$

$iii ) x - \frac{1}{x} \\= \sqrt{6} + \sqrt{5} - (\sqrt{6} - \sqrt{5}) \\= \sqrt{6} + \sqrt{5} - \sqrt{6} + \sqrt{5} \\= 2\sqrt{5} \: --(3)$

$Now, \red{ x^{2} + \frac{1}{x^{2}} - 2 } \\= x^{2} + \frac{1}{x^{2}} - 2\times x \times \frac{1}{x} \\= \Big( x - \frac{1}{x} \Big)^{2} \\= (2\sqrt{5} )^{2} \:[ From \: (3) ]$

$= 4 \times 5 \\= 20$

Therefore.,

$Now, \red{Value \:of \: x^{2} + \frac{1}{x^{2}} - 2 }\\\green { = 20 }$

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