12th March 2004, 12:55  #1 
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tripled equation Noob
image says it all
(please help me) 
12th March 2004, 13:02  #2 
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Try the AVS forums

12th March 2004, 13:08  #3 
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ha ha, very funny...easter bunny

12th March 2004, 13:09  #4 
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Here goes...
(x^2  9)/(x+3) + x = (x18)/(x3) (x^2  9) > is a difference of squares, so you can replace it with this (x+3)(x3) b/c if you expand it you end up with (x^2  9) so: ((x+3)(x3))/(x+3) + x = (x18)/(x3) x3 + x = (x18)/(x3) 2x  3 = (x18)/(x3) (2x  3)(x3) = (x18) 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 
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gah... nuffin' here
Phi = (1+sqrt(5))/2 
12th March 2004, 13:29  #6 
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gah, found 1 mistake... in the 1st line, i multiplied x^218 with (x3)... 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 
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well, what is the bloody equation and ill solve it, your attachment is messy (ill try if its correct):
(x^29)/(x+3)+x=x18; (x^29)/(x+3)=x18x; (x^29)/(x+3)=18; 18*(x+3)=(x^29); 18x+54=x^29; x^218x63=0; x1 + x2 = 18; x1 * x2 =63; x1 = 3; x2 = 21; I think Phi = (1+sqrt(5))/2 
12th March 2004, 13:41  #8 
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(x^29)/(x+3)+x=x^218 <correct one

12th March 2004, 14:05  #9 
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aaargh:
(x^29)/(x+3)+x=x^218; [(x^29)+x*(x+3)]/x+3=(x^218)/1; 2x^2+3x9=(x+3)*(x^218); 2x^2+3x9=x^3+3x^218x54; x^3+x^221x45=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=yb/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=2/4766; y=pow[(q/2+sqrt(q^2/4+p^3/27)),1/3]+pow[(q/2sqrt(q^2/4+q^3/27)),1/3]; now solve the y in terms of p and q, then x=yb/(3*a); and you have the x [edit] 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 
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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 
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well, its thurd power equation
[edit] 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 
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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 
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Is nice to see people doing math in here
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13th March 2004, 00:56  #14 
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x cannot equal 3; the original equation had a fraction with x+3 as the base, therefore x cannot equal 3
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13th March 2004, 01:40  #15 
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Using Jaak's working
(x^29)/(x+3)+x=x^218; [(x^29)+x*(x+3)]/x+3=(x^218)/1; .............. (i) 2x^2+3x9=(x+3)*(x^218); ...................... (ii) 2x^2+3x9=x^3+3x^218x54; x^3+x^221x45=0; This factorizes to (x5)(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 
13th March 2004, 08:38  #16  
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Quote:
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 
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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 
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(x^29)/(x+3)+x=x^218;
if x=3; ((3)^29)/((3)+3)+(3)=(3)^218; (99)/(33)=918; 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  
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13th March 2004, 13:11  #20 
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13th March 2004, 13:56  #21 
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X^225=0 /+25 X^2=25 /sqrt x=5 maybe the same way you can solve x^3x^2+21x+45=0 can you? 
13th March 2004, 15:42  #22 
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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 
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umm, if the original equation was (x^29)/(x+3) + x = x^2  18, then
((x+3)(x3))/(x+3) + x = x^2  18 2x  3 = x^2  18 x^2  2x  15 = 0 (x5)(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 
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here's the work i did.

13th March 2004, 21:15  #25 
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i allways go longer way :P
anyways, 3 is not solution you get division by zero with it (x^29)/(x+3) + x = x^2  18; Phi = (1+sqrt(5))/2 
14th March 2004, 00:16  #26  
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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. Never underestimate the bandwidth of a station wagon full of tapes hurtling down the highway. 

14th March 2004, 07:09  #27 
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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  
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Quote:
http://www.sosmath.com/algebra/factor/fac11/fac11.html Phi = (1+sqrt(5))/2 

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


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