Question

Rationalise denominator 3+√8/3-√8

Answer 1

Answer:

$\frac{3 + \sqrt{8} }{3 - \sqrt{8} } \times \frac{3 + \sqrt{8} }{3 + \sqrt{8} } \\ using \: \: 2 \: \: identities \: \\ 1.) \: (x + y {)}^{2} = {x}^{2} + {y}^{2} + 2xy \\ 2.) \: (x + y)(x - y) = {x}^{2} - {y}^{2} \\ = \frac{(3 + \sqrt{8} {)}^{2} }{ {3}^{2} - ( \sqrt{8} {)}^{2} } \\ = \frac{ {3}^{2} + ( \sqrt{8} {)}^{2} + 2 \times 3 \times \sqrt{8} }{9 - 8} \\ = \frac{9 + 8 + 6 \sqrt{8} }{1} \\ = 9 + 8 + 6 \sqrt{8} \\ = 17 + 6 \sqrt{8}$

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