Question

if 3+root 7/3-4 root 7=a+b root 7 where a and b are rational numbers

Answer 1

$\frac{3+\sqrt7}{3 - 4 \sqrt7} = (a + b) \sqrt{7},\ \ \ \ cross-multiply, \\ \\ 3 + \sqrt7 = (3 - 4 \sqrt7)*[(a + b) \sqrt7]$

$3 + \sqrt7 = (3a+3b)\sqrt7-4(a+b)\sqrt7*\sqrt7 \\ \\ 3 + \sqrt7 = (3a+3b) \sqrt{7} - (4a+4b)*7 \\ \\ 3+\sqrt7 = -(28a+28b) + (3a+3b)\sqrt7 \\ \\ equate\ rational\ and\ irrational\ parts \\ \\ 28a+28b = -3,\ \ \ \ and\ \ \ 3a +3b = 1 \\ \\ There\ is\ inconsistency\ as\ a+b=-3/28\ according\ to\ 1st\ equation \\ \\ a+b=1/3\ according\ to\ 2nd\ equation \\ \\ there\ is\ no\ solution$

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$If\ RHS\ =\ a + b\ \sqrt7,\ \ then, \\ \\ 3a-28b=3\ \ \ and\ 3b-4a=1 \\ \\ Solving\ we\ get \\ \\ b = -15/103\ \ and\ \ \ a = -37/103$

Answer 2

Consider LHS

$\frac{3+ \sqrt{7} }{3-4 \sqrt{7} }$

$\frac{3+ \sqrt{7} }{3-4 \sqrt{7} } * \frac{3+4 \sqrt{7} }{3+4 \sqrt{7} }$

$\frac{9+12 \sqrt{7} + 3 \sqrt{7}+28 }{(3)^2 - (4 \sqrt{7})^2 }$

$\frac{37+15 \sqrt{7} }{9 - 112 }$

$\frac{37+15 \sqrt{7} }{-103 }$

$\frac{-37}{103} - \frac{15}{103} \sqrt{7}$

$\frac{-37}{103} - \frac{15}{103} \sqrt{7} = a + b \sqrt{7}$

On comparing both sides

$a = \frac{-37}{103}; b = - \frac{15}{103}$