Question

3 + under root 6 divided by root 3 + root 2 is equals to a + b root 3 find the value of integers a and b

Answer 1

Hey friend,
Here is the answer you were looking for:
$\frac{3 + \sqrt{6} }{ \sqrt{3} + \sqrt{2} } = a + b \sqrt{3} \\ \\ o n \: rationalizing \: the \: denominator \: we \: get \\ \\ = \frac{3 + \sqrt{6} }{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ using \: the \: identity \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{3 \times \sqrt{3} - 3 \times \sqrt{2} + \sqrt{6} \times \sqrt{3} - \sqrt{6} \times \sqrt{2} }{ {( \sqrt{3} )}^{2} - {( \sqrt{2} )}^{2} } \\ \\ = \frac{3 \sqrt{3} - 3 \sqrt{2} + \sqrt{18} - \sqrt{12} }{3 - 2} \\ \\ = 3 \sqrt{3} - 3 \sqrt{2} + \sqrt{2 \times 3 \times 3} - \sqrt{2 \times 2 \times 3} \\ \\ = 3 \sqrt{3} - 3 \sqrt{2} + 3 \sqrt{2} - 2 \sqrt{3} \\ \\ = 1 \sqrt{3} \\ \\ \sqrt{3} = a + b \sqrt{3} \\ \\ a = 0 \\ \\ b = 1$


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