• Congratulations to the Class of 2024 on your results!
    Let us know how you went here
    Got a question about your uni preferences? Ask us here

Polynomials Question (1 Viewer)

KFunk

Psychic refugee
Joined
Sep 19, 2004
Messages
3,323
Location
Sydney
Gender
Male
HSC
2005
This question (no.10 from further questions in the arnold & arnold) has been gving me grief so I figured I'd post it here to let you guys have a crack at it. I'm sure I must have overlooked something. Here it is:

The equation x<sup>3</sup> + 3px<sup>2</sup> + 3qx + r = 0, where p<sup>2</sup>&ne;q, has a double root.

Show that (pq - r)<sup>2</sup> = 4(p<sup>2</sup> - q)(q<sup>2</sup> - pr).
 

KFunk

Psychic refugee
Joined
Sep 19, 2004
Messages
3,323
Location
Sydney
Gender
Male
HSC
2005
Maybe this is just me being hopeful but... Bump.
 

Slidey

But pieces of what?
Joined
Jun 12, 2004
Messages
6,600
Gender
Male
HSC
2005
I'll be honest with you - I tried a large variety of algabraic manipulations, substitutions, and anything else you care to name, last night. In a few cases I almost got somewhere.

I bet it involves a very obscure algabraic manipulation such that if you didn't get it the first time, you wouldn't be likely TO get it.

I doubt they'd ask anything that hard in an exam.
 

Ogden_Nash

Member
Joined
Apr 27, 2004
Messages
35
Gender
Male
HSC
2005
Yeah, I've seen the solution to this question and it involves coming up with a lot of identities and manipulating them to get the answer.

Slide Rule said:
I doubt they'd ask anything that hard in an exam.
Agreed.
 

Estel

Tutor
Joined
Nov 12, 2003
Messages
1,261
Gender
Male
HSC
2005
I just had a look, got the paper out, and managed it first try :p... slice of luck :D
It's actually under a page, and I haven't studied polynomials, so I believe the art is to break down the question into parts you can manage:
p = -(2a+b)/3
q = a(a+2b)/3
r = -a^2b
and the rest really just follows...
 

KFunk

Psychic refugee
Joined
Sep 19, 2004
Messages
3,323
Location
Sydney
Gender
Male
HSC
2005
Thanks for trying :),
I had a go at a whole bunch of things too and the thing which seemed most on track was taking the first derivative ('cause there's a double root) which is a quadratic and using the quadratic equation. Under the square root you end up with (p<sup>2</sup> - q) which incurs that statement which was at the begining of p<sup>2</sup>&ne;q. Plus-minuses always scare me and I didn't get very far with it :(.
 

Estel

Tutor
Joined
Nov 12, 2003
Messages
1,261
Gender
Male
HSC
2005
Don't get fancy:
LHS = 4a/81 (a-b)^4
RHS = 4a/81 (a-b)^4 = LHS

I am, however, on the watch for a more elegant solution. :p
 

KFunk

Psychic refugee
Joined
Sep 19, 2004
Messages
3,323
Location
Sydney
Gender
Male
HSC
2005
Estel said:
Don't get fancy:
LHS = 4a/81 (a-b)^4
RHS = 4a/81 (a-b)^4 = LHS

I am, however, on the watch for a more elegant solution. :p
Bravo, very neatly done.
 

Estel

Tutor
Joined
Nov 12, 2003
Messages
1,261
Gender
Male
HSC
2005
I should add, btw, that I set a, a and b to be the roots.
Otherwise my statement is quite useless.
In fact, it's completely useless :p
 

Estel

Tutor
Joined
Nov 12, 2003
Messages
1,261
Gender
Male
HSC
2005
Kfunk: yup, that's what I got but for some reason I didn't type the squared... hmmmmmm

I am quite bored.
 

Slidey

But pieces of what?
Joined
Jun 12, 2004
Messages
6,600
Gender
Male
HSC
2005
Root-coefficient.. you're kidding me. I saw double root and just ignored everything that didn't involve the first derivative. Bah humbug.

EDIT: First derivative, not second.
 
Last edited:

Slidey

But pieces of what?
Joined
Jun 12, 2004
Messages
6,600
Gender
Male
HSC
2005
Of course. I thought this question was far harder than it actually is.
 

KFunk

Psychic refugee
Joined
Sep 19, 2004
Messages
3,323
Location
Sydney
Gender
Male
HSC
2005
haha
(spam spam spam spam, spam spam spam spam)
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top