Question

Find a and b in the following:-

root 11 - root7 / root 11 - root 7 = a - root 77 b​

Answer 1

Appropriate question :

Find a and b in the following :

$\sf \dfrac{\sqrt{11}-\sqrt{7}}{\sqrt{11}+\sqrt{7}}=a-\sqrt{77}b$

Solution :

Firstly, rationalizing the denominator of the LHS :

$\sf: \implies \dfrac{ \sqrt{11} - \sqrt{7} }{ \sqrt{11} + \sqrt{7} } \times \bigg( \cfrac{ \sqrt{11} - \sqrt{7} }{ \sqrt{11} - \sqrt{7} } \bigg)$

$\sf: \implies \dfrac{( \sqrt{11} -\sqrt{7} )^{2} }{( \sqrt{11} - \sqrt{7} )( \sqrt{11} + \sqrt{7} ) }$

Using the identities : (a-b) (a+b) = a²-b² and (a-b)² = a²+b²-2ab for the denominator and numerator :

$\sf: \implies \dfrac{ { (\sqrt{11} ) }^{2} + {( \sqrt{7} )}^{2} - 2( \sqrt{11} )( \sqrt{7} )}{ {( \sqrt{11} ) - }^{2} {( \sqrt{7} )}^{2} }$

$\sf: \implies \dfrac{11 + 7 -2( \sqrt{77}) }{11 - 7}$

$\sf: \implies \dfrac{18 - 2 \sqrt{77} }{4}$

Removing out 2 as the common term :

$\sf: \implies \dfrac{2(9 - \sqrt{77} )}{2(2)}$

$\sf: \implies \dfrac{ \cancel2(9 - \sqrt{77} )}{ \cancel2(2)}$

$\sf: \implies \dfrac{9 - \sqrt{77} }{2}$

On comparing this with RHS, we get :

$\red{ \bigstar} \: \: \: \bf{a = \cfrac{9}{2} \: \: \: \red{ \bigstar} } \\ \\ \red{ \bigstar} \: \: \: \bf{b = \dfrac{1}{2} \: \: \: \red{ \bigstar}}$

Answer 2

Answer:

sorry brother but urgent tha