Re: Australian Maths Competition 2013
wait isn it root(x+y) + root(y+z) = root(x+z), it probably works your way as well because when i did it i got the same answer as well, what i did was this:
root(x+y) + root(y+z) = root(x+z)
(x+y) + (y+z) + 2root((x+y)(y+z)) = z + x
cancelling out...