how do I know which roots are repeated on the calculator?

Thread Starter

PG1995

Joined Apr 15, 2011
816
Hi

I was trying to find roots of some equations using the calculator TI-92 Plus (Voyage 200 and TI-89 are similar in functionality to 92 Plus). The calculator only gives me distinct roots and I can't know which roots are repeated. For an example, please have a look on the attachment. Thanks for the help.

Regards
PG
 

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Papabravo

Joined Feb 24, 2006
12,767
In this particular case the obvious answer is in the factorization. The factor of m to the third power is the source of the root with a multiplicity of three. You could also use the quadratic formula to find the roots of the quadratic factor and finally you could graph the equation to see where the multiple roots lie.
 

steveb

Joined Jul 3, 2008
2,436
Hi

I was trying to find roots of some equations using the calculator TI-92 Plus (Voyage 200 and TI-89 are similar in functionality to 92 Plus). The calculator only gives me distinct roots and I can't know which roots are repeated. For an example, please have a look on the attachment. Thanks for the help.

Regards
PG
That is a limitation of the calculator, it seems. Hopefully, you can figure it out by looking at the equation. In your example it's kind of obvious that m=0 is three of the required 5 solutions for a 5th order polynomial, but other examples may not be so obvious.

For reference, here is what I get when I use the "allroots" function in Maxima.

allroots(x^3*(x^2+3*x+2));

\([x=0.0,x=0.0,x=0.0,x=-1.0,x=-2.0]\)

I expect Mathematica and Maple will also give you the full list including repeating roots. If you really need a tool that can do this, any of the three options should be good, but Maxima is freely available. I use it to do various symbolic calculations, and it also has good numerical solution features.
 

Thread Starter

PG1995

Joined Apr 15, 2011
816
Thank you, Papabravo, Steve.

@Steve: I know about about Maxima and I even once downloaded it but then didn't use. I'm happy with the calculator because serves my purpose mostly. Anyway, thanks for the suggestion.

1: When a quadratic equation has equal repeated roots, then this means that the curve touches the x-axis on only one point. But what does repeated equal roots mean for higher degree equations. e.g. What would this mean when cubic equation has two equal roots? I'm asking this because Papabravo suggested to draw a graph to check which roots are repeated.

2: I think to solve this roots problem factorization function of the calculator could help. I was wondering if there is any connection between repeated roots and factorization of the expression.

Please help me with the queries above. Much grateful for all this help.

Regards
PG
 
Last edited:

steveb

Joined Jul 3, 2008
2,436
finally you could graph the equation to see where the multiple roots lie.
Plotting is a good idea in general, but it won't reveal multiple roots always. The degenerate root could be complex. For example, consider x^4+2*x^3+3*x^2+2*x+1, which has two degenerate roots with multiplicity of 2. (x=-0.5+i*sqrt(3)/2 and x=-0.5-i*sqrt(3)/2 ). As the plot shows, there are no zero crossings here.

Also, in the case of real degenerate roots, one must be adept at reading the plot to know there is a degeneracy and the order of the degeneracy. For example y=x^2 and y=x^4 are both similar looking, but can you tell the degeneracy in a simple way. It's clear there is a degeneracy, but you need to be good to know how to read the order.
 

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steveb

Joined Jul 3, 2008
2,436
2: I think to solve this roots problem factorization function of the calculator could help. I was wondering if there is any connection between repeated roots and factorization of the expression.
Most definitely. Consider the example I gave above.

y=x^4+2*x^3+3*x^2+2*x+1

Factor it out and you get ...

y=(x^2+x+1)^2=(x+0.5-i*sqrt(3)/2)(x+0.5-i*sqrt(3)/2)(x+0.5+i*sqrt(3)/2)(x+0.5+i*sqrt(3)/2)

It all is clear here. This is the preferred method, when you can do it.

By the way, I noticed that the Matlab "roots" function gives all roots, including the degeneracy. In this case the command would be "roots([1 2 3 2 1])".
 

1chance

Joined Nov 26, 2011
46
By looking at the graph you can tell a bit about the multiplicity of the roots by whether they touch or cross the x-axis. Even multiplicities "touch" and odd multiplicities "cross". Besides factoring, don't forget that that you can determine the multiplicity by repeatedly depressing the the polynomial using synthetic division. That's a really quick method since you don't have to hunt around using the Rational Zero Theorem--you already know the distinct real roots. Combine this with your knowledge of the degree equaling the total number of roots and Descartes Rule of Signs (gives possible number of positive/negative/imaginary roots), then you can even find the imaginary roots from the depressed polynomial as well. As a college algebra teacher, I encourage students to use the graphing calculator as a tool to help them find a root and then depress the polynomial using synthetic division. Obviously, if you can depress it down to a quadratic, then the quadratic equation always works for those last two.
 

Tobin

Joined May 5, 2018
1
Just use the roots obtained on the calculator to find the other root.
example- x^3 - 9.x^2 + 24.x - 20 = 0
the calculator gives roots 5,2.
then we know (x-5)(x-2)(x-a) = x^3 - 9.x^2 + 24.x - 20 = 0
checking the constant term -10a= -20 ..... a=2 is the third root.
 

WBahn

Joined Mar 31, 2012
25,062
It's unlikely that the TS is still seeking an answer to a question asked 6 years ago (there's a reason that there is warning and checkbox requirement associated with threads that are more than a year dead).

You essentially suggested the same thing as the prior post, but the mechanics you use are a bit different, so it might be useful to people reading the thread in the future.
 

MrAl

Joined Jun 17, 2014
6,940
Hi,

Yes this thread is a 'little' old :)

It could get tricky though too. For example:
x^4-14*x^3+69*x^2-140*x+100=0

and also:
x^4-11*x^3+42*x^2-68*x+40

They both have solutions 2 and 5, but each equation has different repeated roots and one has two pairs of repeated roots. Note however that the constant term must be the product of all four roots a*b*c*d so that might help. We know 2 and 5, and to get 100 from only using 2 and 5 as factors means only certain combinations of roots are possible, and to get 40 a different combination of roots is necessary. For example, we know 5 does not repeat 3 times because 5*5*5*2=250 and neither constant is 250.
It could of course get more complicated with higher order equations but the constant would always be the product of all roots.
 

WBahn

Joined Mar 31, 2012
25,062
Hi,

Yes this thread is a 'little' old :)

It could get tricky though too. For example:
x^4-14*x^3+69*x^2-140*x+100=0

and also:
x^4-11*x^3+42*x^2-68*x+40

They both have solutions 2 and 5, but each equation has different repeated roots and one has two pairs of repeated roots. Note however that the constant term must be the product of all four roots a*b*c*d so that might help. We know 2 and 5, and to get 100 from only using 2 and 5 as factors means only certain combinations of roots are possible, and to get 40 a different combination of roots is necessary. For example, we know 5 does not repeat 3 times because 5*5*5*2=250 and neither constant is 250.
It could of course get more complicated with higher order equations but the constant would always be the product of all roots.
You seem to be assuming that the roots are all integers. Was there something in the problem that imposed that constraint?
 

MrAl

Joined Jun 17, 2014
6,940
You seem to be assuming that the roots are all integers. Was there something in the problem that imposed that constraint?
Hi there,

Not really, but the example is more clear when we use integers. The product of roots still applies although it could be harder to see. If they are not integers, then that's the way it goes. We will certainly know the difference between 250.123 and 100.123 or 40.123 for example.

Also, i happen to know the OP and he mostly works with integer solutions, although i wont force that. What i dont know though is the highest degree he intends to work with.
 
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