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

image says it all

(please help me)
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Old 12th March 2004, 13:02   #2
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Old 12th March 2004, 13:08   #3
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ha ha, very funny...easter bunny
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Old 12th March 2004, 13:09   #4
eleet-2k2
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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.

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Old 12th March 2004, 13:25   #5
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gah... nuffin' here

Phi = (1+sqrt(5))/2
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Old 12th March 2004, 13:29   #6
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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
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Old 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^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
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Old 12th March 2004, 13:41   #8
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(x^2-9)/(x+3)+x=x^2-18 <correct one
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Old 12th March 2004, 14:05   #9
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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

[edit] I did it and got that x ~= 5
(5.1 coz i rounder a lot)

Phi = (1+sqrt(5))/2
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Old 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 )
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Old 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.
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Old 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
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Old 13th March 2004, 00:32   #13
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Is nice to see people doing math in here

BTW:
I think that X=-15


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Old 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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Old 13th March 2004, 01:40   #15
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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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Old 13th March 2004, 08:38   #16
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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
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Old 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
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Old 13th March 2004, 11:02   #18
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(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
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Old 13th March 2004, 12:42   #19
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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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Old 13th March 2004, 13:11   #20
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Old 13th March 2004, 13:56   #21
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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?
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Old 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
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Old 13th March 2004, 19:23   #23
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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
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Old 13th March 2004, 19:29   #24
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here's the work i did.
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Old 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^2-9)/(x+3) + x = x^2 - 18;

Phi = (1+sqrt(5))/2
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Old 14th March 2004, 00:16   #26
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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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Old 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.
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Old 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
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Old 14th March 2004, 18:20   #29
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heres how we solve it in israel
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