Find multiple roots (repeated roots) of a polynomial depending on the real parameter "a"

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vilgeforc5

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Hello everyone! I have got this task:
Find multiple roots (repeated roots) of a polynomial [math]?3−(1−2?)?^2−(?^2−2?)?+?^2[/math] depending on the real parameter a.
My idea was to factorize the polynomial using the Euclidean algorithm, but I was unable to do it , because it's becoming larger and larger every single step and I don't know what to do..Could anyone help me or give me a hint?
 
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For a cubic to have 3 distinct roots, it must satisfy these conditions:
1) Two real critical points points, [imath]x_1,x_2[/imath]
2) [imath]f(x_1)\times f(x_2)<0[/imath].
Visual aid:
Screen Shot 2022-06-14 at 8.14.33 PM.png
Find the critical points.
What values of a will sastify this condition [imath]f(x_1)\times f(x_2)<0?[/imath]
 
vilgeforc5 said:
Hello everyone! I have got this task:
Find multiple roots (repeated roots) of a polynomial [math]?3−(1−2?)?^2−(?^2−2?)?+?^2[/math] depending on the real parameter a.
My idea was to factorize the polynomial using the Euclidean algorithm, but I was unable to do it , because it's becoming larger and larger every single step and I don't know what to do..Could anyone help me or give me a hint?
Click to expand...
I believe your polynomial is divisible by [imath]t-1[/imath].
 
vilgeforc5 said:
Hello everyone! I have got this task:
Find multiple roots (repeated roots) of a polynomial [math]?3−(1−2?)?^2−(?^2−2?)?+?^2[/math] depending on the real parameter a.
My idea was to factorize the polynomial using the Euclidean algorithm, but I was unable to do it , because it's becoming larger and larger every single step and I don't know what to do..Could anyone help me or give me a hint?
Click to expand...

If you want to have a repeated root then working out the derivative might be helpful, can you see why?

Steven G said:
t=1 is not a root.
The corner is getting crowded tonight.
Click to expand...
Well spotted! I was in complete agreement with (t-1) being a factor :oops:
 
The interesting thing about t - 1, though, is if you divide by it you get a remainder which suggests that you set a = 0, which then leads to a double root. I think it's a very interesting approach to the problem so I think blamocur should at least be allowed to sit while he's in the corner.

-Dan
 
Somehow I misread "repeated roots" for "distinct roots" in the OP.
For a cubic to have repeated roots, it must satisfy these conditions:
1) Two real critical points: [imath]x_1,x_2[/imath]
2) [imath]f(x_1)\times f(x_2)=0 \implies f(x_1)=0[/imath] or [imath]f(x_2)=0[/imath]
Visual aid:

Screen Shot 2022-06-16 at 2.59.54 PM.png

I got a=0 as one of the three solutions, so I think @blamocur statement is justified.
 
BigBeachBanana said:
Somehow I misread "repeated roots" for "distinct roots" in the OP.
For a cubic to have repeated roots, it must satisfy these conditions:
1) Two real critical points: [imath]x_1,x_2[/imath]
2) [imath]f(x_1)\times f(x_2)=0 \implies f(x_1)=0[/imath] or [imath]f(x_2)=0[/imath]
Visual aid:

View attachment 33116

I got a=0 as one of the three solutions, so I think @blamocur statement is justified.
Click to expand...
I might get a permission to nap in the corner after all.
 
topsquark said:
Interesting. That does force a value of a that will work. How did you find it?

-Dan
Click to expand...
We have a specific cubic defined by a parameter a

[math]f(t) = t^3 + (1 - 2a)t^2 + (a^2 - 2a)t + a^2, \text { where } a \in \mathbb R.[/math]
Generally, a cubic with a leading coefficient of one can be factored into three linear terms

[math]f(t) = (t - p)(t - q)(t - r) = t^3 - (p + q + r)t^2 + (pq + pr + qr)t - pqr[/math].
Now if p is real and p = q = r, this general form reduces to

[math]f(t) = t^3 - 3pt^2 + 3p^2t - p^3 \implies p = \dfrac{1 - 2a}{3}, \ p^2 = \dfrac{a^2 - 2a}{3}, \text { and } p^3 = a^2.\\ \therefore \left ( \dfrac{1 - 2a}{3} \right )^2 = \dfrac{a^2 - 2a}{3} \implies 1 - 4a + 4a^2 = \dfrac{9}{3} * (a^2 - 2a) = 3a^2 - 6a \implies \\ a^2 + 2a + 1 = 0 \implies a = \dfrac{-1 \pm \sqrt{4 - 4}}{2} = - \dfrac{1}{2} \implies \\ p = \dfrac{1 - 2(-0.5)}{3} = \dfrac{2}{3}.\\ \text {However, there is an additional constraint, namely } p^3 = a^2. \text {But } p^3 = \left (\dfrac{2}{3} \right)^3 = \dfrac{8}{27} \ne \dfrac{1}{4} = \left ( - \dfrac{1}{2} \right )^2 = a^2. [/math]There is no real a that gives a real root of multiplicity 3.

We can use the same type of analysis to find what (if any) are the real values of a that result in a real root of multiplicity a.
 
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