Question

if X and y are rational numbers , √3+√2/√3-√2 =x+y√6 find the values of X and y​

Answer 1

Answer:

$\sf{The \ values \ of \ x \ and \ y \ are \ 5 \ and}$

$\sf{2 \ respectively.}$

Given:

$\sf{\leadsto{\dfrac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}=x+y\sqrt6}}$

To find:

$\sf{The \ values \ of \ x \ and \ y \ if \ they \ are}$

$\sf{rational \ numbers.}$

Solution:

$\sf{\leadsto{x+y\sqrt6=\dfrac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}}}$

$\sf{Rationalising \ the \ denominator}$

$\sf{\leadsto{x+y\sqrt6=\dfrac{(\sqrt3+\sqrt2)^{2}}{(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}}}$

$\sf{\leadsto{x+y\sqrt6=\dfrac{3+2+2\sqrt6}{\sqrt3^{2}-\sqrt2^{2}}}}$

$\sf{\leadsto{x+y\sqrt6=\dfrac{5+2\sqrt6}{3-2}}}$

$\sf{\leadsto{x+y\sqrt6=5+2\sqrt6}}$

$\sf{On \ comparing \ we \ get}$

$\boxed{\sf{x=5}}$

$\sf{y\sqrt6=2\sqrt6}$

$\boxed{\sf{\therefore{y=2}}}$

$\sf\purple{\tt{\therefore{The \ values \ of \ x \ and \ y \ are \ 5 \ and}}}$

$\sf\purple{\tt{2 \ respectively.}}$