a polynomials question (1 Viewer)

[spice girl]

Because "OLDMAN" is complaining about the lack of problems, I've decided to cause one myself...

Prove that x^3 + 3px^2 + 3qx + r has a double root if and only if:

(pq - r)^2 = 4(p^2 - q)(q^2 - pr)

Enjoy

 

[OLDMAN]

I think this is the sort of slanging match that frightens a lot of students from posting their genuine problems. This problem is going to be messy because of the X^2 term.

3x^2+6px+3q=0 if double root.... p'(x)=0
therefore x^2=q-2px (1)
sub. into p(x), (q-2px)x+3p(q-2px)+3qx+r=0
qx-2px^2+3pq-6p^2x+3qx+r=0
which gives 2px^2=x(4q-6p^2)+3pq+r (2)
using (1) for x^2 , 2pq-4p^2x=x(4q-6p^2)+3pq+r
x(2p^2-4q)=pq+r giving x=(pq+r)/(2p^2-4q)

substituting this x into (2) gives the required identity, might post this first before I lose it.

 

[OLDMAN]

continuation.....

((pq+r)/(2p^2-4q))^2=q-2p(pq+r)(2p^2-4q)
(pq+r)^2=q(2p^2-4q)^2-2p(pq+r)(2p^2-4q)
=2(p^2-2q)(q(2p^2-q)-2p(pq+r))
=2(p^2-2q)(-q^2-2pr)
=2(2q-p^2)(q^2+2pr)

which gets a different result, but the methodology is true.

 

[spice girl]

Originally posted by OLDMAN
I think this is the sort of slanging match that frightens a lot of students from posting their genuine problems.

[to OLDMAN] slanging match?

umm...yeh oldman's soln is the usual approach and should always work unless the question is really a bitch. In summary the cookbook is as follows:
1) Differentiate
2) Sub in
3) repeat (2) until the solution comes out.


but this one has an alternative solution which is really fun to read so that's prolly why i posted it...

involves heaps of elegant elimination that ends up with a linear equation in x. Lots of algebra though.

 

[OLDMAN]

That's right, Spice Girl, its a standard approach. And what may be standard approach to you, might still not be standard approach for a struggling YR12. It is sad to see this Ext2 forum degenerate into "who can post the smartest question" rather than a sincere forum for helping young YR12s.

 

[OLDMAN]

Befor SpiceGirl puts up his alternative sol. Here is the standard method cleared up of numerical errors.

x^3+3px^2+3qx+r=0 (1)
3x^2+6px+3q=0 (2) p'x=0
from (2) x^2=-q-2px (3)
sub. into 1
-qx-2px^2-3pq-6p^2x+3qx+r=0
2px^2=r-3pq+(2q-6p^2)x (4)
2px^2=-2pq-4p^2x being (3)*2 (5)
(4)=(5) gives x=(r-pq)/2(p^2-q)
sub. into (3) (pq-r)^2/4(p^2-q)^2=-q-2p(r-pq)/2(p^2-q)
(pq-r)^2=-4(p^2-q)^2-4p(p^2-q)(r-pq)
=4(p^2-q)(-p^q+q^2-pr+p^2q)
=4(p^2-q)(q^2-pr)

 

[spice girl]

I think I'll let people have a shot at the question, so I'll make up a set of follow-up questions so it's easier:

NB: on second thoughts I don't think this is ~elegant~ but it's always nice to have alternatives.

Consider x^3 + 3px^2 + 3qx + r = 0 ...[1] has a double root, called alpha.

i) Explain why alpha is also a root of x^2 + 2px + q = 0 ...[2]
ii) Using [1] and [2] show that px^2 + 2qx + r = 0 ...[3]
iii) Using [2] and [3] show that (p^2 - q)x^2 + (pr - q^2) = 0 ...[4]
iv) Using [2] and [3] show that 2(p^2 - q)x + (pq - r) = 0 ...[5]
v) Hence show that (pq - r)^2 = 4(p^2 - q)(q^2 - pr)

 

[ND]

i)
If alpha is a double root for P(x), it is also a root at P'(x)=x^2 + 2px + q = 0 ...[2]
ii)
x^2 + 2px + q = 0
x^2 = -q -2px
sub into [1] (because i'm lazy, i'm not going to write every line)
-2px^2 + 2px -6p^2x -3pq + r = 0, now add [2] * -3p
px^2 + 2qx + r = 0 ...[3]
iii)
[2] * q - [3] * p = (p^2 - q)x^2 + (pr - q^2) = 0 ...[4]
iv)
[2] * p - [3] = 2(P^2 - q)x + px - r = 0 ...[5]
v)
from [5]:
(pq - r)^2 = 4(p^2 - q)^2*x^2 ...[6]
from [4]:
x^2 = (q^2 - pr)/(p^2 - q)
sub into [6]:
(pq - r)^2 = 4(p^2 - q)(q^2 - pr)

I wouldn't have been able to do that without all the separate parts.

 

[spice girl]

Sorry...sorry...
the sub-questions should be reworded to say prove alpha is also a root of [3], [4], [5]....

but close enough...

 

[OLDMAN]

Just a note on the standard approach. The steps are pretty straightforward and it will always deliver -for a 3rd degree polynomial. Just to recap, find x^2 in linear term ie mx+b;sub into p(x), find x^2; equate the two expressions, to solve for x. Now you can mention to the examiner that this x is the double root, so no need to mention alpha, just use x.
If the polynomial is 4th degree, the corresponding steps will produce a quadratic equation, one of the roots of which will be the repeated root.