Question

if x=root 3+root2 divided by root3-root2 and y=root3-root2 divided by root3+root2 then find the value of x square+ysquare​

Answer 1

$\underline\bold{\huge{SOLUTION \: :}}$

FIRST OF ALL, WE HAVE TO RATIONALISE "x" AND "y".

WE GET,

$x = \frac{( \sqrt{3 } + \sqrt{2}) ^{2} }{( \sqrt{3} - \sqrt{2} )( \sqrt{3} + \sqrt{2} ) }$

$= 3 \: + \: 2 \sqrt{6 \: } \: + \: 2$

SIMILARLY,

$y = \frac{ {( \sqrt{3} - \sqrt{2} ) }^{2} }{( \sqrt{3} + \sqrt{2)}( \sqrt{3 } - \sqrt{2)} }$

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

NOW, WE CAN PROCEED AS UNDER :

x² + y²

= (3+2√6+2)² + (3-2√6+2)²

= (5+2√6)² + (5-2√6)²

= 25 + 20√6 + 24 + 25 - 20√6 + 24

= 98 [REQUIRED ANSWER]

Answer 2

FIRST OF ALL, WE HAVE TO RATIONALISE "x" AND "y".

WE GET,

x = \frac{( \sqrt{3 } + \sqrt{2}) ^{2} }{( \sqrt{3} - \sqrt{2} )( \sqrt{3} + \sqrt{2} ) }x=

(

3

−

2

)(

3

+

2

)

(

3

+

2

)

2

= 3 \: + \: 2 \sqrt{6 \: } \: + \: 2=3+2

6

+2

SIMILARLY,

y = \frac{ {( \sqrt{3} - \sqrt{2} ) }^{2} }{( \sqrt{3} + \sqrt{2)}( \sqrt{3 } - \sqrt{2)} }y=

(

3

+

2)

(

3

−

2)

(

3

−

2

)

2

= 3 \: - \: 2 \sqrt{6 \: } + \: 2=3−2

6

+2

NOW, WE CAN PROCEED AS UNDER :

x² + y²

= (3+2√6+2)² + (3-2√6+2)²

= (5+2√6)² + (5-2√6)²

= 25 + 20√6 + 24 + 25 - 20√6 + 24

= 98 [REQUIRED ANSWER]