Question

Please explain the answer

Answer 1

Step-by-step explanation:

Given that:

$1)\frac{3+\sqrt{2}}{3-\sqrt{2}}=a+b\sqrt{2}\\\\2)\frac{5-\sqrt{6}}{5+\sqrt{6}}=a-b\sqrt{6}$

To find: Value of a and b in both cases

Solution: In such type of questions,first rationalize the denominator,then compare both LHS and RHS.

$1)\frac{3+\sqrt{2}}{3-\sqrt{2}}\times\frac{3+\sqrt{2}}{3+\sqrt{2}}=a+b\sqrt{2}$$\frac{{(3+\sqrt{2})}^{2}}{{(3)}^2-({\sqrt{2})}^{2}}=a+b\sqrt{2}$$\\\frac{9+2+6\sqrt{2}}{9-2}=a+b\sqrt{2}$$\frac{11+6\sqrt{2}}{7}=a+b\sqrt{2}$$\frac{11}{7}+\frac{6\sqrt{2}}{7}=a+b\sqrt{2}\\\\compare\:LHS\:and\:RHS\\\bold{a=\frac{11}{7}}\\\\\bold{b=\frac{6}{7}}$

$2)\frac{5-\sqrt{6}}{5+\sqrt{6}}\times\frac{5-\sqrt{6}}{5-\sqrt{6}}=a-b\sqrt{6}$$\frac{{(5-\sqrt{6})}^{2}}{{(5)}^2-({\sqrt{6})}^{2}}=a-b\sqrt{6}$$\\\frac{25+6-10\sqrt{6}}{25-6}=a-b\sqrt{6}$$\frac{31-10\sqrt{6}}{19}=a-b\sqrt{6}$$\frac{31}{19}-\frac{10\sqrt{6}}{19}=a-b\sqrt{6}\\\\compare\:LHS\:and\:RHS\\\bold{a=\frac{31}{19}}\\\\\bold{b=\frac{10}{19}}$

Hope it helps you.