Question

11. If x = (sqrt(3) + sqrt(2))/(sqrt(3) - sqrt(2)); y = (sqrt(3) - sqrt(2))/(sqrt(3) + sqrt(2)) & Find x ^ 2 + y ^ 2​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:x = \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$

On rationalizing the denominator, we get

$\rm :\longmapsto\:x = \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} }$

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

We know that,

$\underbrace{ \boxed{ \bf \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}$

and

$\underbrace{ \boxed{ \bf \: (x + y)(x - y) = {x}^{2} - {y}^{2}}}$

Using, these Identities, we get

$\rm :\longmapsto\:x = \dfrac{3 + 2 + 2 \times \sqrt{3} \times \sqrt{2} }{ {( \sqrt{3}) }^{2} - {( \sqrt{2} )}^{2} }$

$\rm :\longmapsto\:x = \dfrac{5 + 2 \sqrt{6} }{ 3 - 2 }$

$\rm :\longmapsto\:x = \dfrac{5 + 2 \sqrt{6} }{ 1 }$

$\bf\implies \:x = 5 + 2 \sqrt{6}$

Again, It is given that

$\rm :\longmapsto\:y = \dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} }$

On rationalizing the denominator, we get

$\rm :\longmapsto\:y = \dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \times \dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$

$\rm :\longmapsto\:y = \dfrac{(\sqrt{3} - \sqrt{2})^{2}}{(\sqrt{3} - \sqrt{2})( \sqrt{3} + \sqrt{2})}$

$\rm :\longmapsto\:y = \dfrac{3 + 2 - 2 \times \sqrt{3} \times \sqrt{2} }{ {( \sqrt{3}) }^{2} - {( \sqrt{2} )}^{2} }$

$\rm :\longmapsto\:y = \dfrac{5 - 2 \sqrt{6} }{ 3 - 2 }$

$\rm :\longmapsto\:y = \dfrac{5 - 2 \sqrt{6} }{1 }$

$\bf\implies \:y = 5 - 2 \sqrt{6}$

Now, Consider,

$\rm :\longmapsto\: {x}^{2} + {y}^{2}$

$\rm \: = \: {(5 + 2 \sqrt{6})}^{2} + {(5 - 2 \sqrt{6})}^{2}$

We know,

$\underbrace{ \boxed{ \bf \: {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}}$

On applying this identity, we get

$\rm \: = \: 2\bigg( {(5)}^{2} + {(2 \sqrt{6} }^{2}) \bigg)$

$\rm \: = \: 2(25 + 24)$

$\rm \: = \: 2 \times 49$

$\rm \: = \: 98$

Hence

$\bf :\longmapsto\: {x}^{2} + {y}^{2} = 98$