Question

find the value of a and b if 3+√2/3-√2=a+b√2

Answer 1

( 3 + √2 ) / ( 3 - √2 ) = a + b√2


LHS = ( 3 + √2 ) / ( 3 - √2 )



=> ( 3 + √2 ) / ( 3 - √2 ) × ( 3 + √2 ) / ( 3+√2 )



=> ( 3 + √2 ) ( 3 + √2 ) / ( 3 - √2 ) ( 3 + √2 )



=> ( 3 + √2 )² / (3)² - (√2 )²



=> (3)² + (√2)² + 2 × 3 × √2 / 9 - 2



=> 9 + 2 + 6√2 / 7




=> 11 + 6√2 / 7.





Now,

LHS = RHS


11 + 6√2/7 = a + b√2


Clearly,


a = 11/7 and b = 6/7

Answer 2

Identities used :

${(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \\ (x + y)(x - y) = {x}^{2} - {y}^{2}$

$\frac{3 + \sqrt{2} }{3 - \sqrt{2} } = a + b \sqrt{2} \\ \\ \bf \: L.H.S \\ \\ \bf \: On \: rationalizing \: the \: denominator \\ \bf \: we \: get \\ \\ = \frac{3 + \sqrt{2} }{3 - \sqrt{2} } \times \frac{ 3 + \sqrt{2} }{3 + \sqrt{2} } \\ \\ = \frac{ {(3)}^{2} + {( \sqrt{2} )}^{2} + 2(3)( \sqrt{2}) }{ {(3)}^{2} - {( \sqrt{2} )}^{2} } \\ \\ = \frac{9 + 2 + 6 \sqrt{2} }{9 - 2} \\ \\ = \frac{11 + 6 \sqrt{2} }{7} \\ \\ \bf \: On \: comparing \: both \: the \: sides \\ \bf \: we \: get \\ \\ \frac{11 + 6 \sqrt{2} }{7} = a + b \sqrt{2} \\ \\ \frac{11}{7} + \frac{6}{7} . \sqrt{2} = a + b \sqrt{2} \\ \\ \bf \: a = \frac{11}{7} \: \: : b = \frac{6}{7}$

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