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In each of the following determine rational numbers $a$ and $b$:$ \frac{5+3 \sqrt{3}}{7+4 \sqrt{3}}=a+b \sqrt{3} $
Given:
\( \frac{5+3 \sqrt{3}}{7+4 \sqrt{3}}=a+b \sqrt{3} \)
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 $=\frac{5+3 \sqrt{3}}{7+4 \sqrt{3}}=\frac{(5+3 \sqrt{3})(7-4 \sqrt{3})}{(7+4 \sqrt{3})(7-4 \sqrt{3})}$
$=\frac{35-20 \sqrt{3}+21 \sqrt{3}-12 \times 3}{(7)^{2}-(4 \sqrt{3})^{2}}$
$=\frac{35+\sqrt{3}-36}{49-48}$
$=\frac{\sqrt{3}-1}{1}$
$=-1+\sqrt{3}$
Therefore,
$a+b \sqrt{3}=-1+\sqrt{3}$
Comparing both sides, we get,
$a=-1$ and $b=1$
Hence, $a=-1$ and $b=1$.