If X=(√3+√2÷√3-√2) and Y=(√3-√2÷√3+√2) Then find a)

1.	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» Given :- $\rm :\longmapsto\:x = \dfrac{ \sqrt{3} + \sqrt{<klux>2</klux>} }{ \sqrt{3} - \sqrt{2} }$ and $\rm :\longmapsto\:y = \dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} }$ To FIND :- $\rm :\longmapsto\:(i) \: {x}^{2} + {y}^{2}$ $\rm :\longmapsto\:(<klux>II</klux>) \: {x}^{2}y + {y}^{2}x$ 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)

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»

Given :-

$\rm :\longmapsto\:x = \dfrac{ \sqrt{3} + \sqrt{<klux>2</klux>} }{ \sqrt{3} - \sqrt{2} }$

and

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

To FIND :-

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

$\rm :\longmapsto\:(<klux>II</klux>) \: {x}^{2}y + {y}^{2}x$

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)