Question

5+(underoot6) over 5-(underoot6 ) =a+bunderoot6​

Answer 1

$\frac{5 + \sqrt{6}}{5 - \sqrt{6}} = a + b \sqrt{6}$

On Rationalizing :-

$\frac{5 + \sqrt{6}}{5 - \sqrt{6}} = \frac{ {(5 + \sqrt{6})}^{2}}{ {5}^{2} - {( \sqrt{6})}^{2}} \\ \\ \\ \\ = \frac{25 + 6 + 10 \sqrt{6}}{25 - 6} \\ \\ \\ \\= \frac{31 + 10 \sqrt{6}}{19} \\ \\ \\ = \frac{31}{19} + \frac{10 \sqrt{6}}{19}$

$\rule{200}{2}$

On comparing the LHS and the RHS we have the following :-

$\frac{31}{19} + \frac{10 \sqrt{6}}{19} = a + b \sqrt{6} \\ \\ \\ \implies a = \frac{31}{19} \\ \\ \\b = \frac{10}{19}$

$\rule{200}{2}$