Question

X=root3 + root2/root3-rrot2 and y=root3-root2 /root3+root2 then find the value of xsqure + ysquare

Answer 1

It is given that the value of x is $\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ and the value of y is $\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$




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



Multiply and divide the fraction by ( √3 + √2 ),

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



From the properties of expansion, we know ; -

• ( a + b )( a - b ) = a² - b²
• ( a + b )² = a² + b² + 2ab



$\implies x =\dfrac{3 + 2 + 2 \sqrt{6} }{ {( \sqrt{3}) }^{2} - {( \sqrt{2} )}^{2} } \\ \\ \\ \implies x = \dfrac{5 + 2 \sqrt{3}}{3 - 2} = \dfrac{5 + 2 \sqrt{6} }{1} = 5 + 2 \sqrt{6}$



By the same process,

= > y = 5 - 2√6



Now,
= > x² + y²

= > ( 5 + 2√6 )² + ( 5 - 2√6 )²


= > 25 + 24 + 20√6 + 25 + 24 - 20√6


= > 98



Therefore the value of x² + y² is 98.
$\:$

Answer 2

$\huge{\bold{\underline{\underline{Heya!!}}}}$
___________________________________

Given:-
$x = \frac{ \sqrt{3 } + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \: \: \: ........1)\\ \\ \\ \\ y = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \: \: \: \: ........2)$

Now,

Rationalising 1) by multiplying √3+√2 to both numerator and denominator.

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

From here we get x = 5 +2√6

Now rationalising 2) by multiplying √3-√2 to both numerator and denominator.

$y = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ \\ \\ y = \frac{ ({ \sqrt{3} - \sqrt{2} )}^{2} }{ { \sqrt{3} }^{2} - { \sqrt{2} }^{2} } \\ \\ y = \frac{3 + 2 - 2. \sqrt{3}. \sqrt{2} }{1} \\ \\ y = 5 - 2 \sqrt{6} ..........4)$
From here we get y =5-2√6

By adding the squares of both x and y we get,

x^2+y^2=(5+2√6)^2. + (5-2√6)^2

x^2 + y^2 = 25 +24 + 20√6+25+24-20√6

x^2 + y^2= 98

$\bf{So, \:the\: value\: of \:x^2+y^2\: = 98}$

Note:-

Some Formulae are used here=>

•(a+b)^2 =a^2 + b^2 +2ab

•(a-b)^2=a^2+2ab+b^2

•a^2 - b^2 =(a+b)(a-b)

$\huge{\bold{\underline{Thanks}}}$

$\large{\bold{\mathfrak{Hope \:it\:helped\:uh!!}}}$