Question

If a & b are rational numbers and
$\frac{ \sqrt{11} - \sqrt{7} }{ \sqrt{11} + \sqrt{7} } = a \: - b \sqrt{77}$
Then find a & b​

Answer 1

√11-√7/√11+√7 = a-b√77

Rationalisation of denominator (L.H.S.):

We get, √11-√7/√11+√7×√11-√7/√11-√7

=> (√11-√7)²/(√11-√7)(√11+√7)

Using identity: (a+b)(a-b) = a²-b² and (a-b)² = a²+b²-2ab

=> (11+7-2√77)/(√11)²-(√7)²

=> 18-2√77/11-7

=> 18-2√77/4

=> 9-√77/2 = L.H.S.

Now, 9-√77/2 = a-b√77

=> a = 9/2 and b = 1/2

Hence, The value of a = 9/2 and b = 1/2.

Hope it helps you.

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Answer 2

$\frac{ \sqrt{11} - \sqrt{7} }{ \sqrt{11} + \sqrt{7} } = a - b \sqrt{77} \\ =>\frac{ \sqrt{11} - \sqrt{7} }{ \sqrt{11} + \sqrt{7} } \times \frac{ \sqrt{11} - \sqrt{7} }{ \sqrt{11} - \sqrt{7} } = a - b \sqrt{77} \\ => \frac{ ({ \sqrt{11}) }^{2} - 2 \times \sqrt{11} \times \sqrt{7} + ({ \sqrt{7} )}^{2} }{ ({ \sqrt{11} )}^{2} - ( { \sqrt{7} )}^{2} } = a - b \sqrt{77} \\ => \frac{11 - 2 \sqrt{77} + 7 }{11 - 7} = a - b \sqrt{77} \\ => \frac{18 - 2 \sqrt{77} }{4} = a - b \sqrt{77} \\ => \frac{2(9 - \sqrt{77} )}{2(2)} = a - b \sqrt{77} \\ => \frac{9 - \sqrt{77} }{2} = a - b \sqrt{77}$

On comparing both sides, we get,

$a = \frac{9}{2} = 4 \frac{1}{2}$

and,

$b \sqrt{77} = \frac{ \sqrt{77} }{2} \\ => b = \frac{1}{2}$