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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 $b=\frac{1}{2}$.