MarkT Posted April 22, 2004 Report Share Posted April 22, 2004 It is soluble; there are two solutions.Go for it, Tim Quote Link to comment Share on other sites More sharing options...
Mystic-Al-Bob Posted April 22, 2004 Report Share Posted April 22, 2004 good luck at all!!complex number: a + b*i; i^2 = -1; a and b are real numbers Quote Link to comment Share on other sites More sharing options...
Argalius Posted April 22, 2004 Report Share Posted April 22, 2004 I hate working with I Quote Link to comment Share on other sites More sharing options...
Klaas Posted April 22, 2004 Report Share Posted April 22, 2004 a^2 + b^2 = -103a - b = 0(a^2 + b^2)/10 = -1=> (a^2 + b^2)/10 = i^23a - b = 0=> b = 3a; a = b/3=> (a^2 + 9a^2)/10 = i^2=> a^2 = i^2=> a = i => b = 3ior=> a = -i=> b = -3iNot sure if that's right, the only thing I remembered is that i^2 = -1, so I went further on that. Quote Link to comment Share on other sites More sharing options...
Mystic-Al-Bob Posted April 22, 2004 Report Share Posted April 22, 2004 that's right. good work. Quote Link to comment Share on other sites More sharing options...
Klaas Posted April 22, 2004 Report Share Posted April 22, 2004 Cool Well I'm not good at asking questions, so anyone else who likes to ask one, go ahead Quote Link to comment Share on other sites More sharing options...
DarkAngelBGE Posted April 22, 2004 Report Share Posted April 22, 2004 Nice one. Too bad I read the solution before I got my hand at finding the solution. Maybe you ask a similar question, Klaas ? Quote Link to comment Share on other sites More sharing options...
Mystic-Al-Bob Posted April 22, 2004 Report Share Posted April 22, 2004 I have another one:a^2 + b^2 = 0c^2 - a = 34*(c^2) - 3b = 16i Quote Link to comment Share on other sites More sharing options...
DarkAngelBGE Posted April 22, 2004 Report Share Posted April 22, 2004 Gah, I won't do this till the end, but here is the stuff that I have now:a^2 + b^2 = 0 (1)c^2 - a = 3 <=> a = c^2 / 3 (2)4*(c^2) - 3b = 16i <=> b = ( 4c^2 - 16i ) / 3 (3)(1) ^ (2) ^ (3) =>( c^2 / 3 )^2 + ( (4c^2 - 16i ) / 3 ) ^2 = 0<=> c^4 / 9 + ( 16c^4 - 128c^2 i^2 + 256i^2 ) / 9 = 0<=> 17c^4 + 128c^2 - 256 = 0<=> c^4 + 128/17 c^2 - 256/17 = 0<=> (c^2 + 64/17)^2 - 8448/17^2 = 0<=> (c^2 + 64/17)^2 = 8448/17^2Now do sqrt(), then find out what c is. Put the value that you get for c into (2) and (3) and you get a and b.Not enough time to complete, sorry. Quote Link to comment Share on other sites More sharing options...
Mystic-Al-Bob Posted April 22, 2004 Report Share Posted April 22, 2004 looks very difficult Quote Link to comment Share on other sites More sharing options...
DarkAngelBGE Posted April 22, 2004 Report Share Posted April 22, 2004 Actually it is not very difficult. Well okay, let me complete the thing then. ....<=> (c^2 + 64/17)^2 = 8448/17^2<=> c^2 + 64/17 = sqrt( 8448/17^2 ) V c^2 + 64/17 = -sqrt( 8448/17^2 )<=> c = sqrt( sqrt( 8448/17^2 ) -64/17 ) V c^2 = sqrt( -sqrt ( 8448/17^2 ) -64/17 )=> c = (rounded) 1.28 V c = sqrt( -9.17 )Well, next time I find some time I will post all the squareroots of -9.17. Quote Link to comment Share on other sites More sharing options...
Mystic-Al-Bob Posted April 22, 2004 Report Share Posted April 22, 2004 that looks wrongc=2+imaybe you get the other one (a and Quote Link to comment Share on other sites More sharing options...
Klaas Posted April 23, 2004 Report Share Posted April 23, 2004 I don't really find it either, I just end up with the same long thing as Tim Btw you made mistake in the beginning, so if you fix that your sollution could be right?c^2 - a = 3 <=> a = c^2 / 3 (2)it's:c^2 - a = 3 <=> a = c^2 - 3 Quote Link to comment Share on other sites More sharing options...
Mystic-Al-Bob Posted April 23, 2004 Report Share Posted April 23, 2004 so tim should try again. maybe he will do it Quote Link to comment Share on other sites More sharing options...
DarkAngelBGE Posted April 23, 2004 Report Share Posted April 23, 2004 Stupid mistake Yeah I will prolly try again. Quote Link to comment Share on other sites More sharing options...
Mystic-Al-Bob Posted April 27, 2004 Report Share Posted April 27, 2004 should I tell you the right solution? Quote Link to comment Share on other sites More sharing options...
chichigrande Posted April 28, 2004 Report Share Posted April 28, 2004 I guess you shall. It would be a good idea. Quote Link to comment Share on other sites More sharing options...
MarkT Posted April 28, 2004 Report Share Posted April 28, 2004 And yet:a^2 + b^2 = 0 => a^2 = -b^2; b = +/- i.ac^2 - a = 3=> c^2 = a + 34c^2 - 3b = 16i=> 4(a+3) - 3b = 16i=> 4a - 3b = 16i - 12=> 4a +/- 3i.a = 16i - 12=> a = ( 16i - 12 ) / ( 4 +/- 3i )Either: a = -3.84 + 1.12i, b = +/- (3.84.i + 1.12), c = +/- ( 0.529 + 1.058.i ) (approx).Or: a = 4i, b = +/- 4, c = +/- ( 2 + i )Eight solutions total.*chants* TeX, TeX, TeX, TeX, ....Next question, anyone? Quote Link to comment Share on other sites More sharing options...
chichigrande Posted April 28, 2004 Report Share Posted April 28, 2004 Sure, go ahead and ask a new question. Quote Link to comment Share on other sites More sharing options...
Mystic-Al-Bob Posted April 28, 2004 Report Share Posted April 28, 2004 very nice, mark Quote Link to comment Share on other sites More sharing options...
Mystic-Al-Bob Posted May 4, 2004 Report Share Posted May 4, 2004 NEW QUESTION:How many possibilities are there to write the number 2004 as a difference of two square numbers of natural numbers??QUESTION 2:0 = 4*x^3 + 9*x^2 - 1x=? Quote Link to comment Share on other sites More sharing options...
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