Question

what is the solution of this question​

Answer 1

Question :–

▪︎ If $\: \: { \bold{x = \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \: \: and \: \: y \: = 1 \: \: }}$ , then find the value of $\: \: { \bold{ \dfrac{x - y}{x - 3y} = ? }} \: \:$

ANSWER :–

$\\ \longrightarrow \: \: { \red{ \boxed{ \bold{ \dfrac{x - y}{x - 3y} = \dfrac{ 4 + \sqrt{6} }{ 5 } }}}} \: \:$

EXPLANATION :–

GIVEN :–

$\\ \: \:{ \huge{.}} \: \: \: { \bold{x = \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \: \: and \: \: y \: = 1 \: \: }}$

TO FIND :–

$\\ \: \: { \huge{. \: \: \: }}{ \bold{ \dfrac{x - y}{x - 3y} = ? }} \: \:$

SOLUTION :–

$\\ \: \implies { \bold{x = \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \: \: }}$

• Now Rationalization –

$\\ \implies { \bold{x = \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } }}$

$\\ \: \implies { \bold{x = \dfrac{( \sqrt{3} + \sqrt{2} )^{2} }{ (\sqrt{3})^{2} - ( \sqrt{2} )^{2} } }}$

• We know that –

$\\ \: { \huge { \star }}\: \: \large { \orange{ \boxed{ \bold{ {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab}}}}$

$\\ \: { \huge { \star }}\: \: \large { \orange{ \boxed{ \bold{ (a + b) (a - b) = {a}^{2} - {b}^{2} }}}}$

• So that –

$\\ \: \implies { \bold{x = \dfrac{( \sqrt{3})^{2} + (\sqrt{2} )^{2} + 2( \sqrt{3} )( \sqrt{2} )}{ 3 - 2 } }}$

$\\ \: \implies { \bold{x = 3+ 2 + 2( \sqrt{6} )}}$

$\\ \: \implies { \bold{x = 5 + 2( \sqrt{6} )}}$

• Now put the values –

$\\ \: \implies { \bold{ \dfrac{x - y}{x - 3y} = \frac{5 + 2 \sqrt{6} - 1}{5 + 2 \sqrt{6} - 3 } }} \: \:$

$\\ \: { \bold{ \: \: \: \: \: \: \: \: = \frac{4 + 2 \sqrt{6} }{2 + 2 \sqrt{6} } }} \: \:$

$\\ \: { \bold{ \: \: \: \: \: \: \: \: = \frac{ \cancel2(2+ \sqrt{6} )}{ \cancel2(1 + \sqrt{6} ) } }} \: \:$

• Now rationalization –

$\\ \: { \bold{ \: \: \: \: \: \: \: \: = \dfrac{ 2+ \sqrt{6} }{ 1 + \sqrt{6} } \times \dfrac{1 - \sqrt{6} }{1 - \sqrt{6} } }} \: \:$

$\\ \: { \bold{ \: \: \: \: \: \: \: \: = \frac{( 2+ \sqrt{6}) (1 - \sqrt{6} ) }{ (1 + \sqrt{6}) (1 - \sqrt{6}) } }} \: \:$

$\\ \: { \bold{ \: \: \: \: \: \: \: \: = \frac{2 - 2 \sqrt{6} + \sqrt{6} - 6 }{ 1 - 6 } }} \: \:$

$\\ \: { \bold{ \: \: \: \: \: \: \: \: = \frac{ - 4 - \sqrt{6} }{ - 5 } }} \: \:$

$\\ \: { \bold{ \: \: \: \: \: \: \: \: = \frac{ 4 + \sqrt{6} }{ 5 } }} \: \:$

▪︎Hence , $\: \large { \bold{ \: \: \dfrac{x - y}{x - 3y} = \dfrac{ 4 + \sqrt{6} }{ 5 } }} \: \:$

Answer 2

$\purple{\mathbb{ANSWER:-}}$

Given :–

${ \sf{x = \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \: and \: y \: = 1 }}$

To Find :–

${ \sf{ \dfrac{x - y}{x - 3y} }}$

Rationalising x :–

$\sf x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ \sf x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ \sf x = \frac{( \sqrt{3} + \sqrt{2}) {}^{2} }{( { \sqrt{3} }^{2} - \sqrt{2} ^{2} )} \\ \\ \sf x = \frac{3 + 2 \sqrt{6} + 2 }{3 - 2} \\ \\ \sf x = \frac{5 + 2 \sqrt{6} }{1} = 5 + 2\sqrt{6}$

To Find:–

$\sf \: \frac{x - y}{x - 3y}$

Solution:–

$\sf \: Putting \: values \: of \: x \: and \: y \: in \: \frac{x - y}{x - 3y}:- \\ \\ \sf \: \frac{5 + 2\sqrt{6} - 1 }{5 + 2\sqrt{6} - 3 \times (1) } \\ \\ \sf \: \frac{4 + 2\sqrt{6} }{5 + 2\sqrt{6} - 3 }$

$\sf \frac{4 + 2\sqrt{6} }{2 + 2\sqrt{6} } \\ \\ \sf \frac{2(2 + \sqrt{6} }{2(1 + \sqrt{6})}$

$\sf Rationalising \: by ~multiplying \: \frac{1 - \sqrt{6} }{1 - \sqrt{6} } :- \\ \\ \sf \frac{2 + \sqrt{6} }{1 + \sqrt{6}} \times \frac{1 - \sqrt{6} }{1 - \sqrt{6} } \\ \\ \frac{2 - 2 \sqrt{6} + \sqrt{6} - 6 }{1 - 6} \\ \\ \sf \: \frac{ - 4 - \sqrt{6} }{-5} \\ \\ \sf \: \frac{ 4 + \sqrt{6} }{5}$

$\red{\sf { Answer= \frac{ 4 + \sqrt{6} }{5}}}$