Question

Find a and b root 3+root 2/root 3-root 2=a+b root 6

Answer 1

$\frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3 } + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \frac{ {( \sqrt{3} + \sqrt{2)} }^{2} }{ { \sqrt{3} }^{2} - { \sqrt{2} }^{2} } \\ = \frac{3 + 2 + 2 \sqrt{6} }{3 - 2} \\ = 5 + 2 \sqrt{6}$
By comparing it with
$a + b \sqrt{6}$
we get
a=5
b=2

Answer 2

$\textbf{ Hello Mate! }$

LHS

$= \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$

Rationalize

$\frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ = \frac{ {( \sqrt{3} + \sqrt{2} )}^{2} }{ { \sqrt{3} }^{2} - { \sqrt{2} }^{2} } \\ = \frac{3 + 2 + 2 \sqrt{6} }{3 - 2} \\ = 5 + 2 \sqrt{6}$

LHS = RHS

$5 + 2 \sqrt{6} = a + b \sqrt{6}$
On comparing we get

a = 5 and b = 2

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