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wu :: forums    riddles    easy (Moderators: Icarus, Eigenray, SMQ, Grimbal, ThudnBlunder, towr, william wu)    Nested radical equations « Previous topic | Next topic » Pages: 1  Reply Notify of replies Send Topic Print    Author  Topic: Nested radical equations  (Read 599 times) NickH Senior Riddler     Gender: Posts: 341 Nested radical equations   « on: Jul 27th, 2004, 2:01pm » Quote Modify Let x be a real number.  Find x if:   1) x = [sqrt](4 + [sqrt](4 + x)) 2) x = [sqrt](4 + [sqrt](4 - x)) 3) x = [sqrt](4 - [sqrt](4 + x)) 4) x = [sqrt](4 - [sqrt](4 - x))   (Four questions.) « Last Edit: Jul 27th, 2004, 2:18pm by NickH » IP Logged Nick's Mathematical Puzzles Grimbal wu::riddles Moderator Uberpuzzler     Gender: Posts: 7527 Re: Nested radical equations   « Reply #1 on: Jul 28th, 2004, 6:05am » Quote Modify :: 1) x = sqrt(4+x) x2 - x - 4 = 0 and x>=0 x1 = (1 + sqrt(17))/2   2) x = sqrt(4+sqrt(4-x)) note: x must be in [2,4] because 4-x>0 and x=sqrt(4+...) x2 = 4+sqrt(4-x) x2-4 = sqrt(4-x) x4-8x2+16 = 4-x x4-8x2+x+12 = 0 (x2-x-3)(x2+x-4) = 0 x = (1 [pm] sqrt(13))/2 or  x = (-1 [pm] sqrt(17))/2 The only valid solution to is x2 = (1+sqrt(13))/2 The other solution work if you compute negative roots.   3) x = sqrt(4-sqrt(4+x)) the solution is x3 is such that   x2 = sqrt(4+x3) or x3 = x22-4 ((1+sqrt(13))/2)2-4 = (14+2sqrt(13)-16)/4 = (-1+sqrt(13))/2 x3 = (-1+sqrt(13))/2   4) Just like 1) x = sqrt(4-x) x2 + x - 4 = 0 and 0<=x<=4 x4 = (-1 + sqrt(17))/2 :: IP Logged NickH Senior Riddler     Gender: Posts: 341 Re: Nested radical equations   « Reply #2 on: Jul 29th, 2004, 3:05pm » Quote Modify ::For 1, how do you know that x = sqrt(4+x) yields the only solution to the original equation?  Similarly for 4.:: IP Logged Nick's Mathematical Puzzles Grimbal wu::riddles Moderator Uberpuzzler     Gender: Posts: 7527 Re: Nested radical equations   « Reply #3 on: Jul 29th, 2004, 4:08pm » Quote Modify Hm.. you are right.  The shortcut was not very correct. :: But if it works the way it worked for 2 and 3, there should be 4 solutions, but the 4 solutions solve the variations where you choose the negative value of the one or the other square root.  So only one should work. But now that you mention it, with chaos theory in the back of my head, I suspect it might not work for other values of 4.   Anyway, nobody came up with another solution, so it should be the only one. :: IP Logged Pages: 1  Reply Notify of replies Send Topic Print Forum Jump: ----------------------------- riddles -----------------------------=> easy   - medium   - hard   - what am i   - what happened   - microsoft   - cs   - putnam exam (pure math)   - suggestions, help, and FAQ   - general problem-solving / chatting / whatever ----------------------------- general -----------------------------  - guestbook   - truth   - complex analysis   - wanted   - psychology   - chinese « Previous topic | Next topic »

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