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In each of the following determine rational numbers $a$ and $b$:$ \frac{3+\sqrt{2}}{3-\sqrt{2}}=a+b \sqrt{2} $
Given:
\( \frac{3+\sqrt{2}}{3-\sqrt{2}}=a+b \sqrt{2} \)
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 $=\frac{3+\sqrt{2}}{3-\sqrt{2}}=\frac{(3+\sqrt{2})(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}$
$=\frac{(3+\sqrt{2})^{2}}{(3)^{2}-(\sqrt{2})^{2}}$
$=\frac{9+2+2 \times 3 \sqrt{2}}{9-2}$
$=\frac{11+6 \sqrt{2}}{7}$
$=\frac{11}{7}+\frac{6}{7} \sqrt{2}$
Therefore,
$a+b \sqrt{2}=\frac{11}{7}+\frac{6}{7} \sqrt{2}$
Comparing both sides, we get,
$a=\frac{11}{7}$ and $b=\frac{6}{7}$
Hence, $a=\frac{11}{7}$ and $b=\frac{6}{7}$.
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