Question

IF √2+√3/(3√2-2√3)=a+b√6,find the value of a and b​

Answer 1

$\sf {\underline {\underline {Answer :-}}}$

$<b>The values of a and b are 2 and 5/6 respectively </b>$

$\sf {\underline {\underline {Explanation :-}}}$

$\sf {\dfrac {\sqrt {2} + \sqrt {3}}{3\sqrt {2} - 2\sqrt {3}} = a + b\sqrt {6}}\\\\\\ \textsf {On rationalising the denominator, we get,}\\\\\\ </p><p>\sf {\implies \dfrac {(\sqrt {2} + \sqrt {3}) (3\sqrt {2} + 2\sqrt {3})}{(3\sqrt {2} - 2\sqrt {3}) (3\sqrt {2} + 2\sqrt {3})} \: = \: a + b\sqrt {6}}\\\\\\ \sf {\implies \dfrac {\sqrt {2}(3\sqrt {2} + 2\sqrt {3}) + \sqrt {3}(3\sqrt {2} + 2\sqrt {3})}{3\sqrt {2}(3\sqrt {2} + 2\sqrt {3}) - 2\sqrt {3}(3\sqrt {2} + 2\sqrt {3})} \: = \: a + b\sqrt {6}}\\\\\\ \sf {\implies \dfrac {12 + 5\sqrt {6}}{6} \: = \: a + b\sqrt {6}}\\\\\\ \sf {\implies \dfrac {12}{6} + \dfrac {5\sqrt {6}}{6} \: = \: a + b\sqrt {6}}\\\\\\ \sf {\implies 2 + \dfrac {5}{6}\times \sqrt {6}} \: = \: a + b\sqrt {6}\\\\\\ \textsf{On comparing LHS and RHS, we get,} \\\\\\ \sf {\implies a = 2, b = \dfrac {5}{6}}$

Answer 2

first we multiply the numerator & denominator by(3√2+2√3) then we use formula in the denominator = a+b)(a-b) = a^2-b&2

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The value of a= 2 & b= -5/6

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