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| thyristor |
Posted: Tue Jun 24, 2008 3:03 am Post subject: Third-grade equations |
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 Forum Sophomore

Joined: 11 Feb 2008 Posts: 184 Location: Sweden
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How do you solve third-grade equations algebraicly?
For example y=(ax^3)+(bx^2)+d _________________ 373 13231-mbm-13231 373 |
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| JaneBennet |
Posted: Tue Jun 24, 2008 4:37 am Post subject: |
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 Forum Ph.D.

Joined: 06 Apr 2008 Posts: 874
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Cardano’s method
The working can be very complicated and tedious – best if you have a computer program for it.
PS: Such an equation is called a third-degree equation, not third-grade equation.  _________________
Did You Know?
Random Fact: Pablo Neruda
Last edited by JaneBennet on Wed Jun 25, 2008 7:42 am; edited 3 times in total |
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| Harold14370 |
Posted: Tue Jun 24, 2008 5:47 am Post subject: |
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 Forum Professor

Joined: 13 Apr 2007 Posts: 1665 Location: Pennsylvania
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Dang. I was still working on addition and subtraction when I was in the third grade.  |
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| sunshinewarrior |
Posted: Tue Jun 24, 2008 5:48 am Post subject: |
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 Forum Ph.D.

Joined: 26 Sep 2007 Posts: 1018 Location: London
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| Harold14370 wrote: |
Dang. I was still working on addition and subtraction when I was in the third grade.  |
Heh!
And I thought the answer would involve Fermat's Last Theorem! Or is that for equations of degree 4 and greater? |
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| serpicojr |
Posted: Tue Jun 24, 2008 7:14 am Post subject: |
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 Forum Professor

Joined: 17 Jul 2007 Posts: 1128 Location: JRZ
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| sunshinewarrior wrote: |
| And I thought the answer would involve Fermat's Last Theorem! Or is that for equations of degree 4 and greater? |
If you're just interested in finding roots and not integer roots, then FLT won't help you at all. Furthermore, FLT only talks about a very specific set of equations.
For equations of degree less than or equal to 4, you can always find the roots via arithmetic operations and root extractions (i.e., square root, cube root, etc.). However, beginning with degree 5 and continuing on forever, there exist polynomials whose roots cannot be expressed as such. For example, the family of polynomials xn-x-1, n ≥ 5, are such polynomials. However, this only says that there are no elementary algebraic methods of solving these polynomials. So, for example, there are analytic methods of finding roots, so not all hope is lost! |
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| sunshinewarrior |
Posted: Tue Jun 24, 2008 8:13 am Post subject: |
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 Forum Ph.D.

Joined: 26 Sep 2007 Posts: 1018 Location: London
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| serpicojr wrote: |
| sunshinewarrior wrote: |
| And I thought the answer would involve Fermat's Last Theorem! Or is that for equations of degree 4 and greater? |
If you're just interested in finding roots and not integer roots, then FLT won't help you at all. Furthermore, FLT only talks about a very specific set of equations.
For equations of degree less than or equal to 4, you can always find the roots via arithmetic operations and root extractions (i.e., square root, cube root, etc.). However, beginning with degree 5 and continuing on forever, there exist polynomials whose roots cannot be expressed as such. For example, the family of polynomials xn-x-1, n ≥ 5, are such polynomials. However, this only says that there are no elementary algebraic methods of solving these polynomials. So, for example, there are analytic methods of finding roots, so not all hope is lost! |
Had forgotten about integer roots, though my post was, in any case, made more light-heartedly. Thanks for that. |
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| serpicojr |
Posted: Tue Jun 24, 2008 1:00 pm Post subject: |
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 Forum Professor

Joined: 17 Jul 2007 Posts: 1128 Location: JRZ
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| sunshinewarrior wrote: |
| Had forgotten about integer roots, though my post was, in any case, made more light-heartedly. Thanks for that. |
My light-heartedness detectors turn off when math is involved. It's just a mathematician's instinct! |
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| JaneBennet |
Posted: Wed Jun 25, 2008 7:59 am Post subject: |
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 Forum Ph.D.

Joined: 06 Apr 2008 Posts: 874
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| serpicojr wrote: |
| sunshinewarrior wrote: |
| Had forgotten about integer roots, though my post was, in any case, made more light-heartedly. Thanks for that. |
My light-heartedness detectors turn off when math is involved. It's just a mathematician's instinct! |
I could tell SunshineWarrior was being light-hearted; I just couldn’t think of a light-hearted reply to make.
I know, it can be difficult to be light-hearted when trying to explain certain things in maths.  _________________
Did You Know?
Random Fact: Pablo Neruda |
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| sunshinewarrior |
Posted: Thu Jun 26, 2008 7:43 am Post subject: |
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 Forum Ph.D.

Joined: 26 Sep 2007 Posts: 1018 Location: London
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| serpicojr wrote: |
| sunshinewarrior wrote: |
| Had forgotten about integer roots, though my post was, in any case, made more light-heartedly. Thanks for that. |
My light-heartedness detectors turn off when math is involved. It's just a mathematician's instinct! |
Don't worry about it - we love you just the way you are. And it's just wonderful to have maths-expert guys like you and Lizzie's elder sister and Ratty here! |
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| NeptuneCircle |
Posted: Thu Jun 26, 2008 12:41 pm Post subject: |
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Forum Freshman

Joined: 22 Oct 2007 Posts: 45
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| couldn't you just plug the equation into your graphing calculator and just find when where the graph intercepts the x axis? |
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| serpicojr |
Posted: Thu Jun 26, 2008 3:00 pm Post subject: |
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 Forum Professor

Joined: 17 Jul 2007 Posts: 1128 Location: JRZ
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| If you're going to use a calculator, you may as well use it the right way. Most graphing calculators come with a polynomial solver. Use this instead of the graphing function. |
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| thyristor |
Posted: Fri Jun 27, 2008 12:50 am Post subject: |
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 Forum Sophomore

Joined: 11 Feb 2008 Posts: 184 Location: Sweden
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Thansk for all the posts but could anybody please tell me, if I have the function y=ax3+bx2+cx+d what solutions does it have? I know it can be solved algebraicly and this I want to learn. Please don't answer that you can factorize it in (x-a)(x-b)(x-c) where a,b & c are roots. _________________ 373 13231-mbm-13231 373 |
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| serpicojr |
Posted: Fri Jun 27, 2008 8:07 am Post subject: |
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Joined: 17 Jul 2007 Posts: 1128 Location: JRZ
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