Prove that √(11-6√2) + √(19-6√2) - 2√2 = 2?

Question

A calculator will show that the expression on the left seems to equal to 2. But is it exact?

Answer 1

use the notation rt for square root

rt(11-6rt2) + rt(19-6rt2) = 2 + 2rt2 ===> square both sides
(11-6rt2) + 2rt[ (11-6rt2)(19-6rt2) ] + (19-6rt2) = 4 + 8rt2 + 8 ==>simplify
30 -12rt2 + 2rt [209 - 180rt2+72 ] = 12+8rt2
2rt [209 - 180rt2+72] = -18 +20rt2 ===> square both sides again
4(281 - 180rt2) = 18^2 -720rt2 + 800
1124 - 720rt2 = 1124 -720rt2

:) both sides equal!

:)

Answer 2

Yes, it's exact.

√(11-6√2) = 1.585786438
√(19-6√2) = 3.242640687
So 1.585786438 + 3.242640687 = 4.828427125


2√2 = 2.828427125

4.828427125 - 2.828427125 = 2

Answer 3

sqrt(11-6sqrt(2)) = 3 - sqrt(2) ....1

this can be derived as

let 11-6sqrt(2) = (x+ysqrt(2))^2

= x^2+2y^2 + 2xy sqrt(2)
so x^2+2 y^2 = 11
2xy = -6
slving this we get
x = 3 y = -1

simliarly sqrt(19-6sqrt(2)) = 3sqrt(2)-1 ...2
( we need to take positive value)
add 1 and 2 and rearange to get the result