Question

(√2+√3) / (3√2-2√3) = a -b√6 find a,b​

Answer 1

Question :

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

then find a and b

Answer :

Given :

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

To find :

The value of a and b

According to the question,

$\sf{ \dfrac{ \sqrt{2} + \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3} } = a - b \sqrt{6} } \\ \\ \sf{ \dfrac{ \sqrt{2} + \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3} } \times \dfrac{3 \sqrt{2} + 2 \sqrt{3} }{3 \sqrt{2 } + 2 \sqrt{3} } = a - b \sqrt{6} } \\ \\ \sf{ \dfrac{ \sqrt{2} (3 \sqrt{2} + 2 \sqrt{3} ) + \sqrt{3} (3 \sqrt{2} + 2 \sqrt{3}) }{18 - 12} } \\ \\ \sf{ \dfrac{6 + 2 \sqrt{6} + 3 \sqrt{6} + 6 }{6} = a - b \sqrt{6} } \\ \\ \sf{ \dfrac{12 + 5 \sqrt{6} }{6} = a - b \sqrt{6} } \\ \\ \sf{ \cancel\dfrac{12}{6} \: {}^{2} + \bigg( - \dfrac{ 5 \sqrt{6} }{6} \bigg) = a - b \sqrt{6} } \\ \\ \sf{ 2 + \bigg( \frac{ - 5 \sqrt{6} }{6} \bigg) = a - b \sqrt{6} } \\ \\ \therefore {\boxed {\sf \red{a = 2 , \: \: \: b = - \dfrac{ 5}{6} }}}$

Answer 2

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