Question

find the value of a and b if
$\frac{3 + \sqrt{2} }{3 - \sqrt{2} } = a + b \sqrt{2}$

Answer 1

Hi friend ✋✋✋✋
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Your answer
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$\frac{3 + \sqrt{2} }{3 - \sqrt{2} } = a \: + b \sqrt{2} \\ \\ now \: \\ - - - - \\ \\ we \: will \: have \: to \: rationalise \: the \: denominator \\ \\ \frac{(3 + \sqrt{2})}{(3 - \sqrt{2} )} \times \frac{(3 + \sqrt{2}) }{(3 + \sqrt{2}) } \\ \\ = > \frac{(3 + \sqrt{2} ) {}^{2} }{(3) {}^{2} - ( \sqrt{2} ) {}^{2} } \\ \\ = > \frac{(3) {}^{2} + 6 \sqrt{2} + ( \sqrt{2}) {}^{2} }{9 - 2} \\ \\ = > \frac{9 + 6 \sqrt{2} + 2 }{7} \\ \\ = > \frac{11 + 6 \sqrt{2} }{7} \\ \\ > \frac{11}{7} + \frac{6 \sqrt{2} }{7} = a \: + b \sqrt{2} \\ \\ = > a \: = \frac{11}{7} and \: b = \frac{6}{7}$
HOPE IT HELPS