Question

please answer fast
Q15

Answer 1

$\bf \: Identities \: used : \\ \\ \bf \: {(a + b ) }^2 = {a}^2 + {b}^2 + 2ab \\ \bf \: {(a-b)}^2 = {a}^2 + {b}^2 - 2ab \\ \bf \: ( a + b ) ( a - b ) = {a}^2 - {b}^2$
$\frac{3 + \sqrt{8} }{3 - \sqrt{8} } + \frac{3 - \sqrt{8} }{3 + \sqrt{8} } = a + b \sqrt{2}$

L.H.S.

On rationalizing the denominator we get,

$= \frac{3 + \sqrt{8} }{3 - \sqrt{8} } \times \frac{3 + \sqrt{8} }{3 + \sqrt{8} } + \frac{ 3 - \sqrt{8} }{3 + \sqrt{8} } \times \frac{3 - \sqrt{8} }{3 - \sqrt{8} } \\ \\ = \frac{ {(3)}^{2} + {( \sqrt{8}) }^{2} + 2(3)( \sqrt{8} )}{{(3)}^{2} - {( \sqrt{8}) }^{2} } + \frac{ {(3)}^{2} + {( \sqrt{8}) }^{2} - 2(3)( \sqrt{8} ) }{ {(3)}^{2} - {( \sqrt{8}) }^{2} } \\ \\ = \frac{9 + 8 + 6 \sqrt{8} }{9 - 8} + \frac{9 + 8 - 6 \sqrt{8} }{9 - 8} \\ \\ = (17 + 6 \sqrt{ {2}^{2} \times 2 }) + (17 - 6 \sqrt{ {2}^{2} \times 2 } ) \\ \\ = 17 + 6 \times 2 \sqrt{2} + 17 - 6 \times 2 \sqrt{2} \\ \\ = 17 + 12 \sqrt{2} + 17 - 12 \sqrt{2} \\ \\ = 34$

Comparing both the sides we get,

$34 = a + b \sqrt{2} \\ \\ \bf \: a = 34 \: and \: b = 0$

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