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 Radically confusing (Posted on 2005-05-14)
Simplify the product A*B*C*D*E*F

A = (√2)
B = (√(2-√2))
C = (√(2-√(2+√2)))
D = (√(2-√(2+√(2+√2))))
E = (√(2-√(2+√(2+√(2+√2)))))
F = (√(2+√(2+√(2+√(2+√2)))))

 Submitted by Jer Rating: 1.4286 (7 votes) Solution: (Hide) 2 Beginning at the right, the radicands are conjugates. Keep simplifying. . Maybe some day I'll be a scholar and I'll remember this problem and fix it. Until then the wrongly posted problem gives some decimal solution and the correct problem gives the nice anwer everyone wants it to be (2). The reason the problem ended up wrong was that in the queue we tried fixing the radical symbols about 10 times and every time I changed something else they would all revert to gobbledy-gook. I changed around the negatives and positives so many times I got confused and by the time the problem was posted it came out wrong.

Comments: ( You must be logged in to post comments.)
 Subject Author Date straight forward phi 2005-06-20 16:33:27 a question to Jer pcbouhid 2005-05-24 17:00:24 it is 2 easy Ady TZIDON 2005-05-17 15:51:43 re(2): Fixing my Problem (Solution) Bruno 2005-05-17 01:55:08 re: Fixing my Problem (Solution) ajosin 2005-05-16 22:23:16 re: Fixing my Problem: Charlie 2005-05-16 20:43:19 Fixing my Problem: Jer 2005-05-16 16:51:09 re(2): it is two easy Charlie 2005-05-16 03:39:04 I'm tired...to be continued ! pcbouhid 2005-05-15 23:07:40 my previous comment is wrong pcbouhid 2005-05-14 23:25:53 No Subject pcbouhid 2005-05-14 23:09:30 The most I can simplify... ajosin 2005-05-14 17:36:23 re: it is two easy Charlie 2005-05-14 16:32:41 re: it is two easy Amon 2005-05-14 14:27:30 it is two easy Ady TZIDON 2005-05-14 13:46:11
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