Question

The value of `1/"x"^2+1/"y"^2`, where x = `2 +sqrt3` and y `=2-sqrt3`, is

विकल्प

• 12

• 16

• 14

• 10

Answer 1

14

Explanation:

x = `2+sqrt3` ,  y =`2-sqrt3`

`1/"x"=1/(2+sqrt3)`   `1/"y"=1/(2-sqrt3)`

By Rationalizing

`1/"x"=1/((2+sqrt3))((2-sqrt3))/((2-sqrt3))1/"y"=((2+sqrt3))/((2-sqrt3)(2+sqrt3))`

`1/"x"=2-sqrt3`        `1/"y"=2sqrt3`

`1/"x"^2=(2-sqrt3)^2`   

`1/"y"^2=(2+sqrt3)^2`1/"x"^2+1/"y"^2=2^2+(sqrt3)^2-4sqrt3+2^2+(sqrt3)^2+4sqrt3`

`1/"x"^2+1/"y"^2=4+4+3+3+=14`