Question

if 2+2✓3/2-✓3=a+b✓3 find the value of a and b​

Answer 1

$\sf given : - \frac{2 + 2 \sqrt{3} }{2 - \sqrt{3} } = a + b \sqrt{3} \\ \\ \sf first \: we \: have \: to \: rationalize \: the \\ \sf denominator \: of \: \frac{2 + 2 \sqrt{3} }{2 - \sqrt{3} } \\ \\ \sf = \frac{2 + 2 \sqrt{3} }{2 - \sqrt{3} } \times \frac{2 + \sqrt{3} }{2 + \sqrt{3} } \\ \\ = \frac{(2 + 2 \sqrt{3})(2 + \sqrt{3} ) }{(2 - \sqrt{3})(2 + \sqrt{3}) } \\ \\ \sf = \frac{2(2 + \sqrt{3} ) + 2 \sqrt{3}(2 + \sqrt{3}) }{( {2)}^{2} - ( { \sqrt{3}) }^{2} } \\ \\ \sf = \frac{4 + 2 \sqrt{3} + 4 \sqrt{3} + 6 }{4 - 3} \\ \\ \sf = \boxed{10 + 6 \sqrt{3} } - - - - (i)\\ \\ \sf comparing \: (i) \: with \: a + b \sqrt{3} \\ \sf \\ = > 10 + 6 \cancel{\sqrt{3}} = a + b \cancel{\sqrt{3} }$

hence, a = 10 and b = 6

Answer 2

$\huge\underline\textbf{Answer:-}$

Refer the given attachment:-