Question

find a and b if 3+√6/√3+√2= a+b√3
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Answer 1

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Answer 2

Given :-

$\sf \: if \: \dfrac{3 + \sqrt{6} }{ \sqrt{3} + \sqrt{2} } = a + b \sqrt{3}$

To find :-

the value of a and b

Solution :-

$\bf \: \dfrac{3 + \sqrt{6} }{ \sqrt{3} + \sqrt{2} }$

rationalize the denominator we get ---

$\implies \sf \: \dfrac{3 + \sqrt{6} }{ \sqrt{3} - \sqrt{2} } \times \dfrac{ \sqrt{3} - \sqrt{2} }{\sqrt{3} - \sqrt{2} } \\ \\ \implies \sf \: \dfrac{3 \sqrt{3} - 3 \sqrt{2} + \sqrt{18} - \sqrt{12} }{ { \sqrt{(3)} }^{2} - { \sqrt{(2)} }^{2} } \\ \\ \implies \sf \: \dfrac{3 \sqrt{3} - 3 \sqrt{2} + 3 \sqrt{2} - 2 \sqrt{3} }{3 - 2} \\ \\ \implies \sf \: \dfrac{3 \sqrt{3} \: \: \: \cancel{ - 3 \sqrt{2} } \: \: \: \cancel{ + 3 \sqrt{2}} - 2 \sqrt{3}}{1} \\ \\ \implies \sf \: 3 \sqrt{3} + 0- 2 \sqrt{3} \\ \\\implies \sf \:0 + 3 \sqrt{3} - 2 \sqrt{3} \\ \\ \implies \sf \:0 + 1\sqrt{3}$

and this is equal to a + b√3

0 + 1√3 = a + b√3

therefore,

the value of a = 0

the value of b = 1

basic instructions :-

$\blue \bigstar \tt \: the \: value \: of \: \sqrt{18} = 3 \sqrt{2} \\ \blue \bigstar \tt \: the \: value \: of \: \sqrt{12} = 2 \sqrt{3}$

here, in the Solution one identity is used .(a + b) (a - b) = (a)² - (b)²

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