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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Updated on: 10-Oct-2022

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