Question

find a and b
$\frac{2 + \sqrt{3} }{2 - \sqrt{3} } = a \sqrt{7} + b$
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Answer 1

taking l.h.s

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

rationalise

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

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

opening square

$4 + 3 + 4 \sqrt{3}$

$7 + 4\sqrt{3}$

some correct in question in right hand side a root 3 +b should be there

so a=4

b=7

by comparing LHS AND RHS