Question

rationalize the denominator of 5+root6/5-root6

Answer 1

Rationalizing factor of 5-root 6 is 5+root6.
So,
(5+root6)(5+root6)/ (5-root6)(5+root6)
= (5+root6)^2/ (5)^2-(root6)^2
=(5)^2 +2*5*root6+ (root6)^2/ 25- 6
=25+10* root6+ 6/ 19
=31+ 10* root6/19

Answer 2

Answer:

$\frac{5+\sqrt{6}}{5-\sqrt{6}}=\frac{31}{19}+\frac{+10\sqrt{6}}{19}$

Step-by-step explanation:

Given expression,

$\frac{5+\sqrt{6}}{5-\sqrt{6}}$

We have to rationalize the denominator of the given expression.

Consider,

$\frac{5+\sqrt{6}}{5-\sqrt{6}}$

Multiply and divide by 5 + √6

we get,

$\implies\frac{5+\sqrt{6}}{5-\sqrt{6}}\times\frac{5+\sqrt{6}}{5+\sqrt{6}}$

$\implies\frac{(5+\sqrt{6})^2}{(5-\sqrt{6})(5+\sqrt{6})}$

$\implies\frac{5^2+(\sqrt{6})^2+2\times5\times\sqrt{6}}{(5)^2-(\sqrt{6})^2}$

$\implies\frac{25+6+10\sqrt{6}}{25-6}$

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

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

Therefore, $\frac{5+\sqrt{6}}{5-\sqrt{6}}=\frac{31}{19}+\frac{+10\sqrt{6}}{19}$