Answers

2014-10-14T15:51:43+05:30
Given that
 \frac{ \sqrt{3}+\sqrt{2} }{\sqrt{3}-\sqrt{2}} = a + b\sqrt{6}

Rationalising the denominator by its conjugate

 \frac{ \sqrt{3}+\sqrt{2} }{\sqrt{3}-\sqrt{2}} * \frac{ \sqrt{3}+\sqrt{2} }{\sqrt{3}+\sqrt{2}} = a + b\sqrt{6}

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

\frac{3+2+2\sqrt{3}.\sqrt{2}}{3-2}<span>=a+b\sqrt{6}</span>

5 + 2\sqrt{6} = a + b\sqrt{6}

Therefore, a = 5 and b = 2
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