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If \(\frac{3 - \sqrt3}{2 + \sqrt 3} = a + b\sqrt3\), what are the values a and b?

If \(\frac {3 - \sqrt 3}{2 + \sqrt 3} = a + b\sqrt 3\), what are the values a and b?
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  • A A = 9, b = -5
  • B A = 5, b = 9
  • C A = 9, b = 5
  • D A = -5, b = 9
Correct Answer: Option A
Explanation:
\(\frac {3 - \sqrt 3}{2 + \sqrt 3} = a + b\sqrt 3\)

Rationalize

= \(\frac {3 - \sqrt 3}{2 + \sqrt 3} \times \frac {2 - \sqrt 3}{2 - \sqrt 3}\)

= \(\frac {(3 - \sqrt 3)}{(2 + \sqrt 3)} \frac {(2 - \sqrt 3)}{(2 - \sqrt 3)}\)

= \(\frac {6 - 3 \sqrt 3 - 2 \sqrt 3 + (\sqrt 3)^2}{4 - 2 \sqrt 3 + 2 \sqrt 3 - (\sqrt 3)^2}\)

= \(\frac {6 - 5 \sqrt 3 + 3}{4 - 3}\)

= \(\frac {9 - 5 \sqrt 3}{1} = 9 - 5 \sqrt 3\)

= 9 + (-5) \(\sqrt 3\)

\(\therefore a = 9, b = - 5\)

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