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Roots of equation

  • Thread starter Saitama
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  • #1
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Homework Statement


Find the number and position of real roots of x4+4x3-4x-13=0.


Homework Equations





The Attempt at a Solution


I found the number of real roots using the Descarte's rule of signs. One is positive and other is negative. Now about the position of roots, i don't have any idea. Substituting the values of x and checking if f(x)=0 won't be a good idea. Also i can't factorize the given expression. I can't think of any other way for finding the roots.

Is it possible to find the roots using calculus? If yes, please tell me, i would be happy to know about that.

Thanks! :smile:
 

Answers and Replies

  • #2
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Use the rational root test.
 
  • #3
3,812
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Use the rational root test.
I don't think it will work here.
 
  • #4
I like Serena
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Hey Pranav!

There are a couple of algorithms to find non-nice roots in a quartic polynomial.
You can find a selection of them on wiki:
http://en.wikipedia.org/wiki/Quartic_function

However, to calculate them is quite involved.
And in the end they will give you the solutions that wolframalpha also gives.
If you take a look at wolframalpha, you'll see that the exact solutions are quite complex.

Alternatively there are other ways, like Newton-Raphson to approximate the roots.
Or else you can "bracket" the roots.
That is, find a (small) interval in which each of the roots must be.
 
  • #5
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Hey Pranav!

There are a couple of algorithms to find non-nice roots in a quartic polynomial.
You can find a selection of them on wiki:
http://en.wikipedia.org/wiki/Quartic_function

However, to calculate them is quite involved.
And in the end they will give you the solutions that wolframalpha also gives.
If you take a look at wolframalpha, you'll see that the exact solutions are quite complex.

Alternatively there are other ways, like Newton-Raphson to approximate the roots.
Or else you can "bracket" the roots.
That is, find a (small) interval in which each of the roots must be.
Thank you ILS for the reply but is there any calculus based way to find the roots?
 
  • #6
Ray Vickson
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Thank you ILS for the reply but is there any calculus based way to find the roots?
No. However, you could regard Newton's method (which is an iterative algorithm to find successively better approximations to a root) as calculus-based.

RGV
 
  • #7
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No. However, you could regard Newton's method (which is an iterative algorithm to find successively better approximations to a root) as calculus-based.

RGV
Thanks for the clarification, i will look into that. :smile:
 
  • #8
Curious3141
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The question asked for "number" and "position" of real roots. You might be able to simplify your work a teensy tiny bit by doing a rough sketch of the curve. Even this is tough, when you differentiate to find the turning points, you get a cubic with a discriminant greater than zero, meaning there are 3 real distinct turning points, none of which can be found "easily" (e.g. with RRT). You'll definitely need to use Newton's method here. Or you can just roll up your sleeves and go through the tedious algebraic process for an exact solution (which, after you're done, will likely look ugly enough that you'll be wondering why you even bothered!). :biggrin:
 
  • #9
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The question asked for "number" and "position" of real roots. You might be able to simplify your work a teensy tiny bit by doing a rough sketch of the curve. Even this is tough, when you differentiate to find the turning points, you get a cubic with a discriminant greater than zero, meaning there are 3 real distinct turning points, none of which can be found "easily" (e.g. with RRT). You'll definitely need to use Newton's method here. Or you can just roll up your sleeves and go through the tedious algebraic process for an exact solution (which, after you're done, will likely look ugly enough that you'll be wondering why you even bothered!). :biggrin:
Graphing? I am always poor at that. :biggrin:
I still need to look into the Newton's method on Wikipedia, there is a huge information about that on Wikipedia. I guess i won't look into the algebraic process as you have already made me aware of the results. :P

I am lucky enough as i have found a way of solving equations of degree 4, i found these in one of my elder brother's engineering books so i won't have to refer Wikipedia for those. :tongue2:
The methods given in the book are Ferrari's method and Descarte's method.
 
  • #10
Ray Vickson
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The question asked for "number" and "position" of real roots. You might be able to simplify your work a teensy tiny bit by doing a rough sketch of the curve. Even this is tough, when you differentiate to find the turning points, you get a cubic with a discriminant greater than zero, meaning there are 3 real distinct turning points, none of which can be found "easily" (e.g. with RRT). You'll definitely need to use Newton's method here. Or you can just roll up your sleeves and go through the tedious algebraic process for an exact solution (which, after you're done, will likely look ugly enough that you'll be wondering why you even bothered!). :biggrin:
There might only be one turning point, since it is possible that two of the roots of the cubic are complex.

RGV
 
  • #11
Ray Vickson
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Homework Statement


Find the number and position of real roots of x4+4x3-4x-13=0.


Homework Equations





The Attempt at a Solution


I found the number of real roots using the Descarte's rule of signs. One is positive and other is negative. Now about the position of roots, i don't have any idea. Substituting the values of x and checking if f(x)=0 won't be a good idea. Also i can't factorize the given expression. I can't think of any other way for finding the roots.

Is it possible to find the roots using calculus? If yes, please tell me, i would be happy to know about that.

Thanks! :smile:
Nowadays there are easier ways. For example, you can use Wolfram Alpha http://www.wolframalpha.com . Just type the formula x^4 +4*x^3 - 4*x - 13 and press 'enter'---see what you get! Note: there is no cost other than internet connection charges.

RGV
 
  • #12
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There might only be one turning point, since it is possible that two of the roots of the cubic are complex.

RGV
But there are two real roots here in this case.
 
  • #13
Curious3141
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  • #14
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  • #15
NascentOxygen
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A general iteration that converges nicely comes from the re-arrangement:

4x³ = 13 + 4x - x⁴

i.e., x = ∛( (13 + 4x - x⁴)/4 )

1.0 → 1.588 → 1.48 → .....

I understand you wanting a method with a better guarantee than the hit and miss of a general iteration re-arrangement, but the lure of its "lucky dip" outcome always beckons me to give it a try. :smile:

If you need certainty: http://m.wolframalpha.com/input/?i=x^4++4*x^3+-+4*x+-+13=0&x=6&y=4&js=off
 
  • #16
Curious3141
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Erm..instead of going in such an algebraic mess, can't we use Descarte's rule of signs?
Yes, but how much does that really help you? You can tell that there's exactly one positive root, but there may be either 3 or 1 negative root(s).
 
  • #17
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Yes, but how much does that really help you? You can tell that there's exactly one positive root, but there may be either 3 or 1 negative root(s).
We can find the number of negative roots by the same rule. Just plugin -x at the place of x and see how many times the sign changes from negative to positive. ;)
 
  • #18
Curious3141
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We can find the number of negative roots by the same rule. Just plugin -x at the place of x and see how many times the sign changes from negative to positive. ;)
Yes, but the rule does not fix the number of negative roots down in this case. There are 3 sign changes - that means there can be *either* 3 *or* 1 negative root. In this case, it's the latter (if you graph the curve), but there's no way to exclude the former possibility (3 negative roots) without graphing the curve.
 
Last edited:
  • #19
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There are 3 sign changes - that means there can be *either* 3 *or* 1 negative root.
How?? :confused:
I only see one sign change.
Substituting -x i get x4-4x3+4x-13=0.
I see only one sign change, from negative sign to positive sign. There is only one negative root.
 
  • #20
Curious3141
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How?? :confused:
I only see one sign change.
Substituting -x i get x4-4x3+4x-13=0.
I see only one sign change, from negative sign to positive sign. There is only one negative root.
+x^4 -> -4x^3 : one sign change
-4x^3 -> +4x: one sign change
+4x -> -13: one sign change

3 sign changes, implying 3 or 1 negative root(s).
 
  • #21
3,812
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+x^4 -> -4x^3 : one sign change
-4x^3 -> +4x: one sign change
+4x -> -13: one sign change

3 sign changes, implying 3 or 1 negative root(s).
I am sorry Curious, you are right, i just rechecked the rule! :smile:

Can there be 2 negative roots?
 
  • #22
Curious3141
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I am sorry Curious, you are right, i just rechecked the rule! :smile:

Can there be 2 negative roots?
No. It's either the number of sign changes, or less than it by a multiple of 2.
 
  • #23
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No. It's either the number of sign changes, or less than it by a multiple of 2.
Thanks, there's nothing like that mentioned in the book i am referring, so i will remember it. :smile:
 

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