Question

Find a and b= √11-√7by√11+√7=a-b√77

Answer 1

Solution:-

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

Now,

By Rationalization,

$\frac{ \sqrt{11} - \sqrt{7} }{ \sqrt{11} + \sqrt{7} }$×$\frac{ \sqrt{11}- \sqrt{7} }{ \sqrt{11}- \sqrt{7} }$

Now,
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On Denominator

By Formula :-$(a+b)(a-b) = a^2 - b^2$

On  Nominator,

By Formula :- $a^2 + b^2+2ab = (a+b)^2$
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Then,

$\frac{( \sqrt{11}- \sqrt{7})^2 }{( \sqrt{11})^2 -( \sqrt{ 7 })^2 }$


$\frac{11+7-2 \sqrt{77} }{11-7}$


$\frac{18- 2\sqrt{77} }{4}$ = $a-b \sqrt{77}$

$\frac{18}{4} - \frac{2 \sqrt{77} }{4}$

Then,

a = $\frac{18}{4}$

a=$\frac{9}{2}$ 

b$\sqrt{77}$= $\frac{2 \sqrt{77} }{4} = \frac{ \sqrt{77}}{2}$

b = $\frac{1}{2}$



i hope this will help you


-by ABHAY

Answer 2

Hi friend, Harish here.

Here is your answer:

Given that,

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

To find,  

The value of a & b.

Solution:

First we must rationalize the denominator of the given number by multiplying and dividing with it's conjugate.

Conjugate of denominator is √11 - √7.

Then,

 $\frac{\sqrt{11}- \sqrt{7} }{\sqrt{11}+ \sqrt{7} } \times \frac{\sqrt{11}- \sqrt{7} }{\sqrt{11}- \sqrt{7} } = \frac{(\sqrt{11}- \sqrt{7} )^{2} }{(\sqrt{11}- \sqrt{7})(\sqrt{11}+ \sqrt{7}) }$

Here in the denominator to multiply we can use the identity:

(x+y)(x-y) = x² - y².

Then,

(√11 + √7) (√11 - √7) = (√11)² - (√7)² = 11 - 7 = 4

So,

$\frac{(\sqrt{11}- \sqrt{7} )^{2} }{(\sqrt{11}- \sqrt{7})(\sqrt{11}+ \sqrt{7}) } = \frac{11+7-2( \sqrt{11})( \sqrt{7}) }{4} = \frac{18 - 2 \sqrt{77} }{4}$

⇒  $\frac{18 - 2 \sqrt{77} }{4} = a-b \sqrt{77}$

⇒  $\frac{18}{4} - \frac{2 \sqrt{77} }{4} = a-b \sqrt{77}$

Now by comparing the above equation , We get:

$a= \frac{18}{4} = \frac{9}{2} : b= \frac{2}{4} = \frac{1}{2}$

Hence these are the values of a & b.
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Hope my answer is helpful to you