Question

find b : √2+√3 / 3√2-2√3 = 2-b√6

Answer 1

Hello here is your sol.... hope this helps :)

Answer 2

Hey friend,
Here is the answer you were looking for:
$b : \frac{ \sqrt{2} + \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3} } = 2 - b \sqrt{6} \\ \\ on \: rationalizing \: the \: denominator \: we \: get \\ \\ = \frac{ \sqrt{2} + \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3} } \times \frac{3 \sqrt{2} + 2 \sqrt{3} }{3 \sqrt{2} + 2 \sqrt{3} } \\ \\ using \: the \: identity \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{ \sqrt{2} \times 3 \sqrt{2} + \sqrt{2} \times 2 \sqrt{3} + \sqrt{3} \times 3 \sqrt{2} + \sqrt{3} + 2 \sqrt{3} }{ {(3 \sqrt{2} )}^{2} - {(2 \sqrt{3}) }^{2} } \\ \\ = \frac{6 + 2 \sqrt{6} + 3 \sqrt{6} + 6 }{18 - 12} \\ \\ = \frac{12 + 5 \sqrt{6} }{6} \\ \\ 2 + \frac{5}{6} \sqrt{6} = 2 - b \sqrt{6} \\ \\ b = - \frac{5}{6}$

Hope this helps!!!

@Mahak24

Thanks...
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