# In each of the following determine rational numbers $a$ and $b$:$\frac{\sqrt{11}-\sqrt{7}}{\sqrt{11}+\sqrt{7}}=a-b \sqrt{77}$

Given:

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

To do:

We have to determine rational numbers $a$ and $b$.

Solution:

We know that,

Rationalising factor of a fraction with denominator ${\sqrt{a}}$ is ${\sqrt{a}}$.

Rationalising factor of a fraction with denominator ${\sqrt{a}-\sqrt{b}}$ is ${\sqrt{a}+\sqrt{b}}$.

Rationalising factor of a fraction with denominator ${\sqrt{a}+\sqrt{b}}$ is ${\sqrt{a}-\sqrt{b}}$.

LHS $=\frac{\sqrt{11}-\sqrt{7}}{\sqrt{11}+\sqrt{7}}=\frac{(\sqrt{11}-\sqrt{7})(\sqrt{11}-\sqrt{7})}{(\sqrt{11}+\sqrt{7})(\sqrt{11}-\sqrt{7})}$

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

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

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

$=\frac{9-\sqrt{77}}{2}$

$=\frac{9}{2}-\frac{1}{2}\sqrt{77}$

Therefore,

$a-b \sqrt{77}=\frac{9}{2}-\frac{1}{2}\sqrt{77}$

Comparing both sides, we get,

$a=\frac{9}{2}$ and $b=\frac{1}{2}$

Hence, $a=\frac{9}{2}$ and $b=\frac{1}{2}$.

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Updated on: 10-Oct-2022

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