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Find the value of $a \times b$ if $ \frac{3+2 \sqrt{3}}{3-2 \sqrt{3}}=a+b \sqrt{3} $.
Given:
\( \frac{3+2 \sqrt{3}}{3-2 \sqrt{3}}=a+b \sqrt{3} \).
To do:
We have to find the value of $a \times b$.
Solution:
Rationalising factor of a fraction with denominator $a+\sqrt{b}$ is $a-\sqrt{b}$.
Therefore,
$\frac{3+2 \sqrt{3}}{3-2 \sqrt{3}}=\frac{3+2 \sqrt{3}}{3-2 \sqrt{3}}\times\frac{3+2 \sqrt{3}}{3+2 \sqrt{3}}$
$=\frac{(3+2 \sqrt{3})^2}{(3)^2-(2\sqrt3)^2}$
$=\frac{3^2+2(3)(2\sqrt3)+(2\sqrt3)^2}{9-4(3)}$
$=\frac{9+12\sqrt3+4(3)}{9-12}$
$=\frac{21+12\sqrt3}{-3}$
$=-(7+4\sqrt3)$
Comparing to with $a+b\sqrt3$, we get,
$a=-7$ and $b=-4$
Therefore,
$a \times b=(-7)\times(-4)$
$=28$
The value of $a \times b$ is $28$.
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