Question

if a and b are rational numbers and $\frac{(\sqrt{11}-\sqrt{7})}{(\sqrt{11}+\sqrt{7})}$ = a - b $\sqrt{77}$. find the value of a and b

Answer 1

ANSWER:--------

$\rm given :- \frac{ \sqrt{11} - \sqrt{7} }{ \sqrt{11} + \sqrt{7} } = a - b \sqrt{77} \\ \\ \rm LHS : - \\ \\ \rm = \frac{ \sqrt{11} - \sqrt{7} }{ \sqrt{11} + 7 } \times \frac{ \sqrt{11} - \sqrt{7} }{ \sqrt{11} - \sqrt{7} } \\ \\ \rm = \frac{ {( \sqrt{11} - \sqrt{7} ) }^{2} }{ ( \sqrt{11} + \sqrt{7})( \sqrt{11} - \sqrt{7} ) } \\ \\ \rm = \frac{ {( \sqrt{11} )}^{2} - 2( \sqrt{11})( \sqrt{7} ) + {( \sqrt{7}) }^{2} }{ {(\sqrt{11}) }^{2} - {( \sqrt{7} )}^{2} } \\ \\ \rm = \frac{11 - 2 \sqrt{77} + 7 }{11 - 7} \\ \\ \rm = \frac{18 - 2 \sqrt{77} }{4} = \frac{9 - 1 \sqrt{77} }{2} \\ \\ \rm on \: comparing \: with \: RHS : - \\ \\ \sf = > \frac{9 - 1 \sqrt{77} }{2} = a - b \sqrt{77} \\ \\ \sf \therefore \: a = \frac{9}{2} \: and \: b = \frac{1}{2}$

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