Hello!
I have two questions about two cases of an integration. The integration is given.
Fa(x) = 4/3 and a = 0 is the first case in which I should find out the value of x. That is my calculation:
in the end I got the value of x by trials. It is x = 0,732 (ca.). My question in this case is, what I have to do to solve my calculation to x? What should I do to complete my calculation?
In the second case x is 2 and the integration is a function of zero. The value of a is ask for. After an integration rule, namely I suppose that the value of a must be 2 too. Here is my calculation:
And when I put a = 2 in the equation, I get the function of zero for the integration, but my question is the same: what I have to do to solve the equation to a? I would love to know the rest of the solution processes. I guess it is possible to use the solution process of a cubic function in the second case to get the value of a. And in the first case? The solution process of a cubic function makes not sense in my opinion, as the integration is not a zero of a function.
Thanks in advance!

Kallisto
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 22032013 23:52
Last edited by Kallisto; 30032013 at 23:31. 
WhiteGroupMaths
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 23032013 01:20
Why are the variables x and t, as well the integration operators dt and dx randomly scattered all over the place?
Peace. 
ghostwalker
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 23032013 09:27
(Original post by Kallisto)
in the end I got the value of x by trials. It is x = 0,732 (ca.). My question in this case is, what I have to do to solve my calculation to x? What should I do to complete my calculation?
Rearrange your equations into the standard form of a polynomial. Multiply through by 3/2 to simplify it a bit and get integer coefficients.
Now try factors (positive and negative) of the constant term as possible roots, and you'll find an easy one.
Using polynomial division you get a linear factor and a quadratic one, and you can use the standard formula on the quadratic one to get the exact answer.
Which is:
I presume x is >=0, otherwise you'll have problems eliminating the other solutions.
Similar method to the first one. Except in this case you know one of the roots  it's 2. So, by the factor theorem, (a2) is a factor. If you take that out you're left with a quadratic, which in this case has no roots. 
Kallisto
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 23032013 10:33
(Original post by WhiteGroupMaths)
Why are the variables x and t, as well the integration operators dt and dx randomly scattered all over the place?
Peace.
(Original post by ghostwalker)
Rearrange your equations into the standard form of a polynomial. Multiply through by 3/2 to simplify it a bit and get integer coefficients.
(...)
(Original post by ghostwalker)
Yes, that is one possible value of "a". And the remainder of your working will show that there is no other possible value.
Similar method to the first one. Except in this case you know one of the roots  it's 2. So, by the factor theorem, (a2) is a factor. If you take that out you're left with a quadratic, which in this case has no roots.Last edited by Kallisto; 23032013 at 10:49. 
ghostwalker
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 23032013 10:59
(Original post by Kallisto)
Why I should multiply by 3/2? or did you mean 2/3? as the term is I would multiply by three and then I get and now I subtract the equation by 4 and I get
and then it is possible to use a polynomial division which you told, I would get an quadratic function and I am able to solve the variable by pqformula. But then I apply these steps, I would calculate a value of a zero of a function. That's why it makes no sense to use a polynomial division in my opinion.
Once you have a factor, you can then use polynomial division.
You are trying to find the zeros (roots) of the equation, to find the values of x that satisfy it. Polynomial division is the way to go.
Could you be more precise, please? I don't still know what should I do exactly in both of them. 
Kallisto
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 23032013 17:23
(Original post by ghostwalker)
(...) Which part don't you understand, and need more details for?
First I divide by 2/3, after that I substrate by 2 and then I got the equation which you posted above: . And then I used a polynomial division by x+1: .
To get the value of x, I used a pqfromula. I got x = 0,732. Then I put the value in the term of integration and I got 4/3!
I have found out a solution process at last! I'm so happy that it works in the first case.
I did the same procedure in the second one, but it doesn't work! I used a polynomial division by x+2... perhaps it would work, if I used an other term as divisor?Last edited by Kallisto; 23032013 at 17:29. 
aznkid66
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 23032013 17:42
I have found out that there is no difference whether I multiply by 3/2 or divide by 2/3
I used a polynomial division by x+2... 
Kallisto
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 23032013 18:34
When I divide by 2/3, I get . And when I use a polynomial division by x2 I get: . I see at the first glance that a = 2 is not a zero of a function in the term. What is wrong again?
Last edited by Kallisto; 24032013 at 09:23. 
ghostwalker
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 23032013 18:54
(Original post by Kallisto)
When I divide by 2/3, I get . And when I use a polynomial division by x2 I get: . I see at the first glance that a = 2 is not a zero of a function in the term. What is wrong again?
When you take out the factor (a2), then a=2 will not be a zero of what remains
The quadratic has no real roots, and hence the only zero is a=2. 
Kallisto
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 23032013 22:40
(Original post by ghostwalker)
(...)
The quadratic has no real roots, and hence the only zero is a=2. 
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 24032013 02:26
(Original post by Kallisto)
In other words: it is useless to use the polynomial division. That is to say I have to solve the cubic function to get the variable. Am I right? then I have to use the solution process of a cubic function. That is the only possibility which makes sense in my opinion.
You know x=2 is a solution to Fa(x)=0
So, (x2) is a factor.
After polynomial division, the remaining quadratic has no roots, no zeros.
So x=2 is the only solution.
Done. 
aznkid66
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 24032013 08:04
Maybe this image will help you understand?

Kallisto
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 24032013 09:46
(Original post by ghostwalker)
No.
You know x=2 is a solution to Fa(x)=0
So, (x2) is a factor.
After polynomial division, the remaining quadratic has no roots, no zeros.
So x=2 is the only solution.
Done.
and I'm done at last. Why I should use a polynomial division, if get a term which has no roots and no zeros?

aznkid66
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 24032013 10:08
Because you want to check if there are any remaining roots, aside from a=2. What you were getting tripped up on was that there were not any remaining roots, so the whole process seemed pointless to you. However, if you recall the first exercise, it is a quick and easy way to find the other 2 roots without testing more integrals, as we have the quadratic formula. Your original question was "what else do I have to do to solve the equation for a?" and the answer is to find the other roots, in which case none exist and the final answer is "nothing else".

ghostwalker
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 24032013 10:11
(Original post by Kallisto)
Then I use the integration rule which I have posted above to reason my answer.
But it does not show you that it is the only answer.
Hence the work I outlined above.
You seem very keen to solve a general cubic equation, but it is not necessary.
I suspect your command of English is making this a lot more difficult than it needs to be.Last edited by ghostwalker; 24032013 at 10:12. 
Kallisto
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 24032013 12:51
I guess it is enough to use a polynomial division to get the quadratic function from the cubic one. After that I use the pqformula to show up that there is no another solution. I realize that the solution process of a cubic function is too complicate, especially there is an easier way to solve it.

Kallisto
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 25032013 22:14
As there is no answer anymore and my problems seem to be solved, this thread is answered. Thanks for help!
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