Question

find the value of a and b if
√3+√2/√3-√2 = a+b√5
$\sqrt{3} + \sqrt{2} \div \sqrt{3} - \sqrt{2} = a + b \sqrt{5}$
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Answer 1

Answer:

$= \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } . \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ = \frac{ \sqrt{3}( \sqrt{3} + \sqrt{2}) + \sqrt{2} ( \sqrt{3} + \sqrt{2} ) }{ \sqrt{3} ( \sqrt{3} + \sqrt{2} ) - \sqrt{2} ( \sqrt{3} + \sqrt{2} )} \\ = \frac{ \sqrt{9} + \sqrt{6} + \sqrt{6} + \sqrt{4} }{ \sqrt{9} + \sqrt{6} - \sqrt{6} - \sqrt{4} } \\ = \frac{3 + 2 \sqrt{6} + 2}{3 - 2} \\ = \frac{5 + 2 \sqrt{6} }{1} \\ = 5 + 2 \sqrt{6}$

so, for a is = 5 and b is = 2