The Student Room Group

Roots of Polynomials

46BA3FFF-DE4C-41BE-890B-7C899C9CBAD2.jpg.jpeg
Doing number 4
I set equal to -the product roots times together

Giving( a=alpha)
a+13/a+46=52

The multiplied by a to give a quadratic.
However, this quadratic solved to 3+2i or 3-2i,
Yet none of other roots would become conjugates?
Not sure if working wrong but have checked a few tines because this seems impossible to have only one complex root?
Original post by Canary84
46BA3FFF-DE4C-41BE-890B-7C899C9CBAD2.jpg.jpeg
Doing number 4
I set equal to -the product roots times together

Giving( a=alpha)
a+13/a+46=52

The multiplied by a to give a quadratic.
However, this quadratic solved to 3+2i or 3-2i,
Yet none of other roots would become conjugates?
Not sure if working wrong but have checked a few tines because this seems impossible to have only one complex root?


Why?? Your working is fine. You might be thinking of the fact that if the coefficients of the polynomial are real then the roots must come in complex conjugate pairs. The thing here is that we don't know if m,nm,n are real, so we cannot rely on this property.
i got that alpha is 3+2i or 3-2i, factored it into aby = -m and got that M = -718/13 - (24/13) i , sounds wrong though tbh lol
Original post by RDKGames
Why?? Your working is fine. You might be thinking of the fact that if the coefficients of the polynomial are real then the roots must come in complex conjugate pairs. The thing here is that we don't know if m,nm,n are real, so we cannot rely on this property.


if we find that alpha is a conjucate pair, factoring that in to the other roots doesnt that mean theres a possibility of 6 roots?
Original post by Gent2324
if we find that alpha is a conjucate pair, factoring that in to the other roots doesnt that mean theres a possibility of 6 roots?


α\alpha takes on either 3+2i3+2i or 32i3-2i.

Considering one OR the other, we deduce different values of m,nm,n.

So we end up with two cubics which satisfy the given properties, each with 3 roots.

The question isn't very specific for us to narrow down to one cubic.
Reply 5
Original post by Gent2324
i got that alpha is 3+2i or 3-2i, factored it into aby = -m and got that M = -718/13 - (24/13) i , sounds wrong though tbh lol


Yeah thats what I got..... Agree it sounds wrong
Reply 6
Original post by RDKGames
Why?? Your working is fine. You might be thinking of the fact that if the coefficients of the polynomial are real then the roots must come in complex conjugate pairs. The thing here is that we don't know if m,nm,n are real, so we cannot rely on this property.


just gave me some very strange values for second part... perhaps its okay though
Thank you for you help all the same

Quick Reply