(In reply to

re: Long answer - very interesting by Erik O.)

I got a similar result this way: I supposed there existed some identity like √(a±√b)= X±Y. Squaring to remove a square root sign, you get a±√b=X²+Y²±2XY.

Since this is ONE equation with TWO unknowns, and there is still another square root to remove, in order to simplify a little I decided to try out letting a=X²+Y² so some terms would go away.

That identity implies that X²=a/2+z and Y²=a/2-z. Substituting, ±√b=±2√(a²/4-z²). Squaring again to remove both square roots at the same time, we finally get z=(1/2)√(a²-b).

From now on, operating I reached the same √(14) result as e.g.

*Edited on ***June 4, 2004, 2:59 pm**