Winamp & Shoutcast Forums tripled equation Noob

12th March 2004, 12:55   #1
kingo'mountain
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tripled equation Noob

image says it all

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 12th March 2004, 13:02 #2 ryan not fucked, not quite. (Forum King)     Join Date: Feb 2002 Location: Tn Posts: 8,808 Try the AVS forums
 12th March 2004, 13:08 #3 kingo'mountain in need of banned aid     Join Date: Mar 2001 Location: over the rainb... uh... nevermind Posts: 2,716 ha ha, very funny...easter bunny
 12th March 2004, 13:09 #4 eleet-2k2 Forum King     Join Date: Aug 2001 Location: Mobil Ave. Posts: 5,381 Here goes... (x^2 - 9)/(x+3) + x = (x-18)/(x-3) (x^2 - 9) --> is a difference of squares, so you can replace it with this (x+3)(x-3) b/c if you expand it you end up with (x^2 - 9) so: ((x+3)(x-3))/(x+3) + x = (x-18)/(x-3) x-3 + x = (x-18)/(x-3) 2x - 3 = (x-18)/(x-3) (2x - 3)(x-3) = (x-18) 2x^2 - 6x + 9 = x - 18 2x^2 - 5x + 27 = 0 to solve for x now, use the quadratic formula: x = [-b +/- sqrt(b^2 - 4ac)]/(2a) x = 5 +/- sqrt(25 - 4(2)(27)) / (2(2)) x cannot be solved for unless you want imaginary roots: x = [5 +/- sqrt(-191)] / 4 And if x = 5 at the end, it doenst work because if you sub x = 5 into the original equation and solve, it doesnt work out. "Welcome to the Island of people who know too much."..."Did you really think balloons would stop him?!" See what I'm listening too.
 12th March 2004, 13:25 #5 Jaak Major Dude     Join Date: Jan 2003 Location: Estonia. Posts: 851 gah... nuffin' here Phi = (1+sqrt(5))/2
 12th March 2004, 13:29 #6 kingo'mountain in need of banned aid     Join Date: Mar 2001 Location: over the rainb... uh... nevermind Posts: 2,716 gah, found 1 mistake... in the 1st line, i multiplied x^2-18 with (x-3)... its supposed to be (x+3) because i wanna lose the fractions, right? how is it now? forgot to put ^2 after the X in the first line, and multiply it all with x+3
 12th March 2004, 13:35 #7 Jaak Major Dude     Join Date: Jan 2003 Location: Estonia. Posts: 851 well, what is the bloody equation and ill solve it, your attachment is messy (ill try if its correct): (x^2-9)/(x+3)+x=x-18; (x^2-9)/(x+3)=x-18-x; (x^2-9)/(x+3)=-18; 18*(x+3)=(x^2-9); 18x+54=x^2-9; x^2-18x-63=0; x1 + x2 = 18; x1 * x2 =-63; x1 = -3; x2 = 21; I think Phi = (1+sqrt(5))/2
 12th March 2004, 13:41 #8 kingo'mountain in need of banned aid     Join Date: Mar 2001 Location: over the rainb... uh... nevermind Posts: 2,716 (x^2-9)/(x+3)+x=x^2-18
 12th March 2004, 14:05 #9 Jaak Major Dude     Join Date: Jan 2003 Location: Estonia. Posts: 851 aaargh: (x^2-9)/(x+3)+x=x^2-18; [(x^2-9)+x*(x+3)]/x+3=(x^2-18)/1; 2x^2+3x-9=(x+3)*(x^2-18); 2x^2+3x-9=x^3+3x^2-18x-54; x^3+x^2-21x-45=0; (what is true if x=5) but solving quibics: we have: ax^3+bx^2+cx+d=0; wile: a=1; b=1; c=-21; d=-45; x=y-b/3a; y^3+y*[(c/a)-(b^2)/(3a^2)]+[(2b^2)/(27a^3)-(bc)/(3a^2)+d/a]=0; y^3+y*p+q=0; p=(c/a)-(b^2)/(3a^2)=-21.5; q=(2b^2)/(27a^3)-(bc)/(3a^2)+d/a=2/47-66; y=pow[(-q/2+sqrt(q^2/4+p^3/27)),1/3]+pow[(-q/2-sqrt(q^2/4+q^3/27)),1/3]; now solve the y in terms of p and q, then x=y-b/(3*a); and you have the x  I did it and got that x ~= 5 (5.1 coz i rounder a lot) Phi = (1+sqrt(5))/2
 12th March 2004, 14:12 #10 kingo'mountain in need of banned aid     Join Date: Mar 2001 Location: over the rainb... uh... nevermind Posts: 2,716 the formula should be ax^2+bx+c=0, no d, and cubics should go like eleet - 2k2 wrote, i should then get: x1=0 x2=5 (or the opposite) (P.S.: you are bored if you edited your post )
 12th March 2004, 14:22 #11 Jaak Major Dude     Join Date: Jan 2003 Location: Estonia. Posts: 851 well, its thurd power equation  you'll learn solving quartics in highschool, i just happen to know how to solve them... Maybe there is mistake in your math book or something, but the way I solved (correct way imo) it seems to be correct and leads to correct ansver aswell Phi = (1+sqrt(5))/2 Last edited by Jaak; 12th March 2004 at 17:06.
 12th March 2004, 17:22 #12 ujay Forum King     Join Date: Jul 2001 Location: London Posts: 6,072 I got x=5 and x=-3 with x=-3 as a repeated root.(turning point on the x axis) UJ
 13th March 2004, 00:32 #13 Joel Debian user(Forum King)     Join Date: Jan 2003 Location: Arch land Posts: 4,917 Is nice to see people doing math in here BTW: I think that X=-15 * PC: Intel Core 2 DUO E6550 @ 2.33 GHz with 2 GB RAM: Archlinux-i686 with MATE. * Laptop: Intel Core 2 DUO T6600 @ 2.20 GHz with 4 GB RAM: Archlinux-x86-64 with MATE.
 13th March 2004, 00:56 #14 Nimelennar Major Dude     Join Date: Mar 2003 Location: Canada Posts: 841 x cannot equal -3; the original equation had a fraction with x+3 as the base, therefore x cannot equal -3 The world is made of conflicts: good and evil, order and chaos, light and dark, hot and cold. All are essential to life. None can prevail for any length of time, or life will fail. In the end, the best any can hope for is balance.
13th March 2004, 01:40   #15
ujay
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Using Jaak's working

(x^2-9)/(x+3)+x=x^2-18;
[(x^2-9)+x*(x+3)]/x+3=(x^2-18)/1; .............. (i)
2x^2+3x-9=(x+3)*(x^2-18); ...................... (ii)
2x^2+3x-9=x^3+3x^2-18x-54;
x^3+x^2-21x-45=0;

This factorizes to (x-5)(x+3)(x+3)=0

So x=5 or x=-3(repeated)

As nimelennar points out, implicit in moving from (i) to (ii) above involves multiplying through by (x+3), but when x=-3 this is equivalent to multiplying through by zero. Strictly speaking this means that the function is not defined when x=-3.

It's always a good idea to do a quick sketch of the function in such cases, examining the turning points gives something like the attatched (sorry about the lack of l33tness, just got back from the pub ).

UJ
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 cubic.jpg (10.9 KB, 132 views)

13th March 2004, 08:38   #16
Jaak
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Quote:
 Originally posted by ujay As nimelennar points out, implicit in moving from (i) to (ii) above involves multiplying through by (x+3), but when x=-3 this is equivalent to multiplying through by zero. Strictly speaking this means that the function is not defined when x=-3.
Yea, finiding those multiplications with zero sucks hard :P
I didnt want to simply draw the graph (with avs its too easy ) so I tryed to solve this with pure math and I actulay succeeded (heh, yea... i was bored)

Phi = (1+sqrt(5))/2

 13th March 2004, 10:10 #17 kingo'mountain in need of banned aid     Join Date: Mar 2001 Location: over the rainb... uh... nevermind Posts: 2,716 so this is a parabula? cant be... cause in mah book it says that just X=5, no other x... odd thx for the solutions though, they helped
 13th March 2004, 11:02 #18 Jaak Major Dude     Join Date: Jan 2003 Location: Estonia. Posts: 851 (x^2-9)/(x+3)+x=x^2-18; if x=-3; ((-3)^2-9)/((-3)+3)+(-3)=(-3)^2-18; (9-9)/(3-3)=9-18; 0/0=-9; what is flase and cant be solution, your math book is correct Phi = (1+sqrt(5))/2
13th March 2004, 12:42   #19
eleet-2k2
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Quote:
 Originally posted by kingo'mountain the formula should be ax^2+bx+c=0, no d, and cubics should go like eleet - 2k2 wrote, i should then get: x1=0 x2=5 (or the opposite) (P.S.: you are bored if you edited your post )
Quadratic equation (what I did) only works for quadratics, that is when the highest degree of any term is 2. Cubics dont work the same way, and requires some calculus and/or factoring.

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 13th March 2004, 13:11 #20 Vie Forum King     Join Date: Dec 2001 Location: Thoron fields and Duranium shadows. Posts: Crap mostly Posts: 8,002 42 I dont know, Algibra was never my strong point. Member most in need of SpellCheck Lifetime Achievement Award I'm a Twitch Streamer these days, it's weird.
 13th March 2004, 13:56 #21 kingo'mountain in need of banned aid     Join Date: Mar 2001 Location: over the rainb... uh... nevermind Posts: 2,716 remember this action? X^2-25=0 /+25 X^2=25 /sqrt x=5 maybe the same way you can solve -x^3-x^2+21x+45=0 can you?
 13th March 2004, 15:42 #22 Jaak Major Dude     Join Date: Jan 2003 Location: Estonia. Posts: 851 no, exaclty like you cant solve ax^2+bx+c=0; like this way Phi = (1+sqrt(5))/2
 13th March 2004, 19:23 #23 intoxiChaos Junior Member     Join Date: Aug 2003 Location: Florida, USA Posts: 34 umm, if the original equation was (x^2-9)/(x+3) + x = x^2 - 18, then ((x+3)(x-3))/(x+3) + x = x^2 - 18 2x - 3 = x^2 - 18 x^2 - 2x - 15 = 0 (x-5)(x+3) = 0 x = 5, -3 edit: i'm scanning the work so its easier to see. i'll post the pic after its done
13th March 2004, 19:29   #24
intoxiChaos
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here's the work i did.
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 work.gif (61.0 KB, 128 views)

 13th March 2004, 21:15 #25 Jaak Major Dude     Join Date: Jan 2003 Location: Estonia. Posts: 851 i allways go longer way :P anyways, -3 is not solution you get division by zero with it (x^2-9)/(x+3) + x = x^2 - 18; Phi = (1+sqrt(5))/2
14th March 2004, 00:16   #26
k_rock923
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Quote:
 Originally posted by intoxiChaos here's the work i did.
wow. that makes waaaaaaay more sense to me than what jaak did. is that calculus? I hope not because i'm screwed next year if it is!

edit// i mean is what jaak did calculus. hopefully i can figure out what intoxiChaos did or calculus wouldn't be an option for me.

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 14th March 2004, 07:09 #27 intoxiChaos Junior Member     Join Date: Aug 2003 Location: Florida, USA Posts: 34 thats not calc at all. that's just basic algebra. jaak's right, btw, about the -3 not being a solution because it makes the point undefined. 5 is the only answer that works.
14th March 2004, 11:01   #28
Jaak
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Quote:
 Originally posted by k_rock923 edit// i mean is what jaak did calculus. hopefully i can figure out what intoxiChaos did or calculus wouldn't be an option for me.
I dont know how its called in usa or england or whatever, there are several way for solving cubics and they all are hard (if you dont have the formula in front of you). I used this way if your wondering how it works:
http://www.sosmath.com/algebra/factor/fac11/fac11.html

Phi = (1+sqrt(5))/2

14th March 2004, 18:20   #29
kingo'mountain
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heres how we solve it in israel
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