Question

if X=(√3+√2÷√3-√2) and Y=(√3-√2÷√3+√2) Then find a) X^2+Y^2 b) X^2Y+Y^2X​

Answer 1

Given :-

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

and

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

To Find :-

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

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

Identities Used :-

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

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

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

Solution :-

Given that,

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

On multiply and divide by conjugate of denominator,

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

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

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

$\:  \: \rm  =  \:  \: 5 + 2 \sqrt{6}$

$\: \: \: \: \: \: \: \: \red{\bf :\implies\:x = \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2}} = 5 + 2 \sqrt{6} }$

Also, Given that,

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

On multiply and divide by conjugate of denominator,

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

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

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

$\:  \: \rm  =  \:  \: 5 - 2 \sqrt{6}$

$\: \: \: \: \: \: \: \: \green{\bf :\implies\:x = \dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2}} = 5 - 2 \sqrt{6} }$

Now,

Consider (i)

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

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

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

$\:  \: \rm  =  \:  \: 2(25 + 24)$

$\:  \: \rm  =  \:  \: 2 \times 49$

$\:  \: \rm  =  \:  \: 98$

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

Consider (ii)

$\pink{\bf :\longmapsto\: {x}^{2}y + {xy}^{2}}$

$\:  \: \rm  =  \:  \: xy(x + y)$

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

$\:  \: \rm  =  \:  \: (25 - 24)(10)$

$\:  \: \rm  =  \:  \: 10$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \pink{\bf :\implies\: {x}^{2}y + {xy}^{2} = 10}$

Additional Information :-

More Identities to know :-

• (a + b)² = a² + 2ab + b²

• (a - b)² = a² - 2ab + b²

• a² - b² = (a + b)(a - b)

• (a + b)² = (a - b)² + 4ab

• (a - b)² = (a + b)² - 4ab

• (a + b)² + (a - b)² = 2(a² + b²)

• (a + b)³ = a³ + b³ + 3ab(a + b)

• (a - b)³ = a³ - b³ - 3ab(a - b)