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In each of the following determine rational numbers $a$ and $b$:$ \frac{4+\sqrt{2}}{2+\sqrt{2}}=a-\sqrt{b} $
Given:
\( \frac{4+\sqrt{2}}{2+\sqrt{2}}=a-\sqrt{b} \)
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 $=\frac{4+\sqrt{2}}{2+\sqrt{2}}=\frac{(4+\sqrt{2})(2-\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})}$
$=\frac{8-4 \sqrt{2}+2 \sqrt{2}-2}{(2)^{2}-(\sqrt{2})^{2}}$
$=\frac{6-2 \sqrt{2}}{4-2}$
$=\frac{6-2 \sqrt{2}}{2}$
$=3-\sqrt{2}$
Therefore,
$a-\sqrt{b}=3-\sqrt{2}$
Comparing both sides, we get,
$a=3$ and $b=2$
Hence, $a=3$ and $b=2$.
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