Question

Which one of the `sqrt17-sqrt12` and `sqrt11-sqrt6` is greater?

Answer 1

`sqrt17-sqrt12 < sqrt11-sqrt6` and hence `sqrt11-sqrt6` is greater.

Explanation:

Let us consider `sqrt17+sqrt6` and `sqrt11+sqrt12` for comparison.

Well we just square both of them and we get

`(sqrt17+sqrt6)^2=(sqrt17)^2+2xxsqrt17xxsqrt6+(sqrt6)^2`

= `17+6+2sqrt102=23+sqrt102`

and `(sqrt11+sqrt12)^2=(sqrt11)^2+2xxsqrt11xxsqrt12+(sqrt12)^2`

= `11+12+2sqrt132=23+2sqrt132`

As `(sqrt17+sqrt6)^2 < (sqrt11+sqrt12)^2`,

we have `sqrt17+sqrt6 < sqrt11+sqrt12`

and hence transposing `sqrt6` to right and `sqrt12` to left

`sqrt17-sqrt12 < sqrt11-sqrt6` and hence `sqrt11-sqrt6` is greater.