Question

$\frac{3 + \sqrt{2} }{3 - \sqrt{2} } \: = a +b \sqrt{2}$
rationalise the denominator and find the value of a and b. PLEASE FAST!!​

Answer 1

Answer:

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Answer 2

$\frac{3 + \sqrt{2} }{3 - \sqrt{2} } \: = a +b \sqrt{2} \\ \\ \frac{3 + \sqrt{2} }{3 - \sqrt{2} } \times \frac{3 + \sqrt{2} }{3 + \sqrt{2} } = a +b \sqrt{2} \\ \\ \frac{ ({3 + \sqrt{2} )}^{2} }{(3 - \sqrt{2} )(3 + \sqrt{2} )} = a +b \sqrt{2} \\ \\ \frac{ {3}^{2} + { \sqrt{2} }^{2} + 2 \times 3 \times \sqrt{2} }{ {3}^{2} - { \sqrt{2} }^{2} } = a +b \sqrt{2} \\ \\ \frac{9 + 2 + 6 \sqrt{2} }{9 - 2} = a +b \sqrt{2} \\ \\ \frac{11 + 6 \sqrt{2} }{7} = a +b \sqrt{2} \\ \\ \frac{11}{7} + \frac{6 \sqrt{2} }{7} = a +b \sqrt{2} \\ \\ on \: comparing \: both \: sides \\ \\ a = \frac{11}{7} \\ \\ b = \frac{6}{7}$

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