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Topic: Not AbSURDly Easy (Read 684 times) |
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ThudnBlunder
wu::riddles Moderator Uberpuzzler
The dewdrop slides into the shining Sea
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Not AbSURDly Easy
« on: Dec 22nd, 2008, 11:36am » |
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2 + x 2 - x Solve --------------- + -------------- = 2 2 + (2 + x) 2 - (2 - x)
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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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towr
wu::riddles Moderator Uberpuzzler
Some people are average, some are just mean.
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Re: Not AbSURDly Easy
« Reply #1 on: Dec 23rd, 2008, 2:22am » |
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hidden: | (2 + x)/[2 + (2 + x)] + (2 - x)/[2 - (2 - x)] = 2 (2 + x)[2 - (2 + x)]/-x + (2 - x)[2 + (2 - x)]/x = 2 (2 - x)[2 + (2 - x)] - (2 + x)[2 - (2 + x)] = 2 x (2 - x)(2 - x) + (2 + x)(2 + x) - 22 x = 2 x (2 - x)(2 - x) + (2 + x)(2 + x) = 32 x (2 - x)(2 - x) = 3 2 x - (2 + x)(2 + x) (2 - x)3 = 18 x2 + (2 + x)3 - 6 2 x(2 + x)(2 + x) 18 x2 + (2 + x)3 - (2 - x)3 = 6 2 x(2 + x)(2 + x) (18 x2 + (2 + x)3 - (2 - x)3)2 = 72 x2(2 + x)3 4x2(x2 + 9x + 12)2 = 72 x2(2 + x)3 (x2 + 9x + 12)2 = 18(2 + x)3 x2 (x2 - 3) = 0 x = 0 v x = +/- 3 x=0: 2/[2 + 2] + (2)/[2 - 2] = 2 => division by 0. x=-3: (2 - 3)/[2 + (2 - 3)] + (2 + 3)/[2 - (2 + 3)] = 2 (2 - 3)/[2 + (6-2)/2] + (2 + 3)/[2 - (6+2)/2] = 2 -5 2 = 2 => false (duh!) x=3: (2 + 3)/[2 + (2 + 3)] + (2 - 3)/[2 - (2 - 3)] = 2 (2 + 3)/[2 + (6+2)/2] + (2 - 3)/[2 - (6-2)/2] = 2 2 = 2 ==> x = 3 |
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Wikipedia, Google, Mathworld, Integer sequence DB
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Barukh
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Re: Not AbSURDly Easy
« Reply #2 on: Dec 24th, 2008, 4:48am » |
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Hmm... I wonder if the following substitution x = 2cos(y) may lead to a more succinct solution?
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