Question

If x= \[\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}\] and y = \[\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}\] , then x² + y +y² =

विकल्प

• 101

• 99

• 98

• 102

Answer 1

Given that  `x= (sqrt3 - sqrt2) /(sqrt3 + sqrt2)` and `y = (sqrt3 + sqrt2) /(sqrt3 - sqrt2)`.

We need to find  `x^2 +xy +y^2`

Now we will rationalize x. We know that rationalization factor for   `sqrt3+sqrt2` is `sqrt3-sqrt2`   `sqrt3+sqrt2`. We will multiply numerator and denominator of the given expression `x= (sqrt3 - sqrt2) /(sqrt3 + sqrt2)` by `sqrt3 - sqrt2`, to get

`x =  `x= (sqrt3 - sqrt2) /(sqrt3 + sqrt2) xx (sqrt3 - sqrt2) /(sqrt3 - sqrt2)`

`= ((sqrt3)^2 +(sqrt2)^2 - 2 xx sqrt3 xx sqrt2)/((sqrt3)^2 - (sqrt2)^2)`

`= (3+2-2sqrt6)/(3-2)`

` = 5-2sqrt6`

Similarly, we can rationalize y. We know that rationalization factor for `sqrt3 - sqrt2` is  `sqrt3 +sqrt2`. We will multiply numerator and denominator of the given expression `(sqrt3 +sqrt2) /(sqrt3 - sqrt2)`by `sqrt3 + sqrt2`, to get

`y = (sqrt3 + sqrt2) /(sqrt3 - sqrt2) xx (sqrt3 + sqrt2) /(sqrt3 +sqrt2)`

`= ((sqrt3)^2 +(sqrt2)^2 +2 xx sqrt3 xx sqrt2)/((sqrt3)^2 - (sqrt2)^2)`

`= (3+2-2sqrt6)/(3-2)`

` = 5-2sqrt6`

Therefore,

`x^2 + xy + y ^ 2  = ( 5 - 2 sqrt6 )^2+ (5-2sqrt6) (5 + 2 sqrt6)+ (5+2sqrt6)^2`

` = 5^2 +(2sqrt6 )^2 - 2 xx 5 xx 2 sqrt6 +5^2 - (2sqrt6 )^2+ 5^2 + (2sqrt6)^2 + 2xx 5 xx 2sqrt6`

` = 25 +24 -20sqrt6 +25 - 24 +25 +24 +20sqrt6`

` = 49 + 1+ 49`

`  = 99`