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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)))))

See The Solution Submitted by Jer    
Rating: 1.4286 (7 votes)

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re: it is two easy | Comment 3 of 15 |
(In reply to it is two easy by Ady TZIDON)

When combining E*F, under its radical, m=2 and n=sqrt(2+sqrt(2+sqrt(2+sqrt(2)))), so m^2-n^2 = 4 - 2 - sqrt(2+sqrt(2+sqrt(2))).  Restoring the radical and simplifying gives sqrt(2-sqrt(2+sqrt(2+sqrt(2)))).

Now when combining with D, we have a situation of (m-n)(m-n) or just the square, under the radical, cancelling that radical, so D*E*F = 2-sqrt(2+sqrt(2+sqrt(2))).

But now we want to multiply this by C=sqrt(2-sqrt(2+sqrt(2))), and it no longer falls into the category of either (m+n)(m-n) or the square root of a square. I don't see where the answer can be obtained in 33 seconds.

A calculator shows the answer as

.0162298468649678...

but of course that's not very helpful in understanding the radical form.


  Posted by Charlie on 2005-05-14 16:32:41
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