Question

3+root2/3-root2=a+b root2..

Answer 1

Hey friend,
Here is the answer you were looking for:
$\frac{3 + \sqrt{2} }{3 - \sqrt{2} } = a + b \sqrt{2} \\ \\ on \: rationalizing \: the \: denominator \: we \: get \\ \\ = \frac{3 + \sqrt{2} }{3 - \sqrt{2} } \times \frac{3 + \sqrt{2} }{3 + \sqrt{2} } \\ \\ using \: the \: identity \\ {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{ {(3)}^{2} + {( \sqrt{2} )}^{2} + 2 \times 3 \times \sqrt{2} }{ {(3)}^{2} - {( \sqrt{2}) }^{2} } \\ \\ = \frac{9 + 2 + 6 \sqrt{2} }{9 - 2} \\ \\ \frac{11 + 6 \sqrt{2} }{7} = a + b \sqrt{2} \\ \\ a = \frac{11}{7} \\ \\ b = \frac{6}{7}$


Hope this helps!!!!

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Answer 2

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


                    By Rationalization,


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

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

$\frac{ 9 + 2 + 6 \sqrt{2} }{9 - 2 } = a +b \sqrt{2}$

$\frac{11 + 6 \sqrt{2} }{7} = a + b \sqrt{2}$


a = $\frac{11}{7}$

b = $\frac{6}{7}$




i hope this will help you



-by ABHAY