polynomials (1 Viewer)

[shkspeare]

hey ive done polynomial questions with roots where they are in arithmetic progression and geometric progression

i was wondering if anyone has questions involving limiting sums (if they can be applied to roots in polynomials?)...

i have a big test coming up ... and ive literally done every polynomial questoin i could get my hands on...

i just want one on limiting sums out of interest... thx

 

[nike33]

perhaps..the limiting sum of this series is a root of multiplicity x of a polynomial p(x), show this is true, hence factorise p(x) although i doubt something like this could be asked..or perhaps they could give polynomial of degree infinite, written as a sum using sigma err.. i giveup

 

[redslert]

i think you are going way beyond what is in the 3u course.....

 

[AGB]

Originally posted by nike33
perhaps..the limiting sum of this series is a root of multiplicity x of a polynomial p(x), show this is true, hence factorise p(x) although i doubt something like this could be asked..or perhaps they could give polynomial of degree infinite, written as a sum using sigma err.. i giveup

haha yeh is roots of multiplicity even in the 3u course?

 

[CM_Tutor]

Show that any degree three polynomial whose roots are in both arithmetic and geometric progression has a triple root at some value other than 0.

 

[CM_Tutor]

Originally posted by AGB
haha yeh is roots of multiplicity even in the 3u course?

The concept of multiple roots is Extn 1, the multiple root theorem (invloving the derivative) is Extn 2 only.

 

[AGB]

Originally posted by CM_Tutor
The concept of multiple roots is Extn 1, the multiple root theorem (invloving the derivative) is Extn 2 only.

i thought as much...

 

[shkspeare]

P(x) = ax^3+bx^2+cx+d

Let roots be ... crap i give up ... >.<"

 

[CM_Tutor]

Shkspeare, don't give up, it sounds harder than it is.

 

[shkspeare]

how would we "formulate" the roots ... =\

both AP and GP ...

both have common ratio... both have common difference

ARGHH >__<"
xayma wot did u get for the roots

 

[Xayma]

Originally posted by CM_Tutor
Shkspeare, don't give up, it sounds harder than it is.

Hmm I might have an answer, can you check it, I will put it as an attachment so Shkspeare can still do it.

Originally posted by shkspeare
xayma wot did u get for the roots

look in the zip for the html file if you want to know, you wont get the answer at all as it can be anything in the form (x-b)³ where b is a real number. but you will prove that all roots are equal

 

[CM_Tutor]

Xayma, right so far, but the question said that the root was at some value other than zero.

BTW, this can be massively shortened by starting with a better choice of roots. Personally, I'd go for a, ar and ar².

 

[Xayma]

Oh yeah,

I was thinking (and if I didnt have to reload and change the zip etc I would have) of putting it must be at some value other then 0 as N/M=O/N if M=0 then both sides become undefined therefore it is at some value other then 0.

 

[shkspeare]

a, ar, ar^2...

how is that an arithmetic progression... isnt that only a geometric one???

thx xayma

 

[shkspeare]

one more question =\

Prove that (p^2 + q^2)x^2 - 2(p + q)x + 2 = 0 has no roots unless p = q

thx

 

[nike33]

"Show that any degree three polynomial whose roots are in both arithmetic and geometric progression has a triple root at some value other than 0."

the polynomial has roots x,x,x ie a triple root as it is the only possible way to be in a GP (increasing by x1) and an AP (increasing/decreasing by +0)..also meaning x=! 0

 

[nike33]

"Prove that (p^2 + q^2)x^2 - 2(p + q)x + 2 = 0 has no roots unless p = q"

i dont have paper here atm but i can almost certainly say you will need to find the discriminant and find when it will >= 0 and at discriminant = 0 p = q...just a thought

 

[CM_Tutor]

Originally posted by shkspeare
a, ar, ar^2...

how is that an arithmetic progression... isnt that only a geometric one???

Having defined roots as a GP, you'd now require they also be an AP, hence ar - a = ar² - ar. This leaves two possibilities, a = 0 (impossible, since 0, 0, 0 is not a GP, as it has no common ratio) or a <> 0 and r = 1, in which case the roots are a, a, and a, for a <> 0.

Originally posted by shkspeare
one more question =\

Prove that (p^2 + q^2)x^2 - 2(p + q)x + 2 = 0 has no roots unless p = q

thx

As nike33 said, look at the discriminant. It turns out to be -4(p - q)², which is negative for all cases except p = q. Hence, p = q gives a double root, and p <> q gives no real roots

 

[shkspeare]

thx CM!