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Radically confusing (Posted on 2005-05-14) Difficulty: 2 of 5
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)

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 forwardphi2005-06-20 16:33:27
a question to Jerpcbouhid2005-05-24 17:00:24
it is 2 easyAdy TZIDON2005-05-17 15:51:43
re(2): Fixing my Problem (Solution)Bruno2005-05-17 01:55:08
re: Fixing my Problem (Solution)ajosin2005-05-16 22:23:16
re: Fixing my Problem:Charlie2005-05-16 20:43:19
Fixing my Problem:Jer2005-05-16 16:51:09
Questionre(2): it is two easyCharlie2005-05-16 03:39:04
Some ThoughtsI'm be continued !pcbouhid2005-05-15 23:07:40
my previous comment is wrongpcbouhid2005-05-14 23:25:53
No Subjectpcbouhid2005-05-14 23:09:30
The most I can simplify...ajosin2005-05-14 17:36:23
re: it is two easyCharlie2005-05-14 16:32:41
Questionre: it is two easyAmon2005-05-14 14:27:30
Solutionit is two easyAdy TZIDON2005-05-14 13:46:11
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