Challenge Polynomials problem for current students (2 Viewers)

[CM_Tutor]

Here is a problem for current Extension 2 students to think about - it is not intended for those of use who are post-school to prove we can do it, because I'm sure we can.

The polynomial ax^4 + bx^3 + cx + d = 0 has a triple root.

(a) By proving that this root occurs at x = -b / 2a, or otherwise, prove that 4ad = bc.

(b) Is it true to say that any polynomial of this form whose coefficients are related by 4ad = bc must have a triple root? Explain.

 

[AGB]

doesnt the triple root occur at x = -b/4a??

 

[CM_Tutor]

No. I just checked it, and it's definitely at -b / 2a.

 

[Calculon]

Most 4u people haven't done polynomials yet...

 

[grimreaper]

Originally posted by Calculon
Most 4u people haven't done polynomials yet...

Well my class have. That question CM just posted is certainly harder than ANYTHING in the polynomials chapter of the Cambridge book (the book I use). Anyway I can do the first part of the first question:

f(x) = ax^4 + bx^3 + cx + d
f'(x) = 4ax^3 + 3bx^2 + c
f''(x) = 12ax^2 + 6bx

Since its a triple root, f''(x) = 0 so
12ax^2 + 6bx = 0
subbing x=0 into f(x) obviously doesnt work so
12ax = -6b
x = -b / 2a

However doing the next bit doesnt work for me... I sub x=-b/2a in for x in f(x) and I get a messy algebra thing (I think I made a stupid error somewhere)

 

[AGB]

i keep gettin 8ad = 3bc so close, but still no cigar

i think ive made a mistake in my working somewhere....

 

[grimreaper]

Originally posted by AGB
i keep gettin 8ad = 3bc so close, but still no cigar

i think ive made a mistake in my working somewhere....

lol I get b^4 + cba^2 - 16da^3 = 0, so youre way closer than me...

 

[Grey Council]

will have a go tommorow morning.

 

[AGB]

i got it!...just wait a sec and ill type it up

 

[AGB]

P(x) = ax^4 + bx^3 + cx + d

P'(x) = 4ax^3 + 3bx^2 + c

P"(x) = 6x(2ax + b)

:. x = -b/2a

P(-b/2a) = a(-b/2a)^4 + b(-b/2a) + c(-b/2a) + d

simplify.... and it = - b^4 - 8a^2bc + 16a^3d

P'(-b/2a) = 4a(-b/2a)^3 + 3b(-b/2a)^2 + c

simplify.....and it = b^3 + 4a^2c = 0 ...... multiply through by b

= b^4 + 4a^2bc = 0

:. b^4 = -4a^2bc

:. P(-b/2a) = - (-4a^2bc) - 8a^2bc + 16a^3d = 0

-4a^2bc + 16a^3d = 0

-4a^2 (bc - 4ad) = 0

:. bc = 4ad

 

[grimreaper]

doh I forgot it was true for f'(x) as well =\

 

[CM_Tutor]

That's interesting, it's not the way I did it. Now, what about the second part of the question? Also, can you show that the fourth root is b / 2a?

 

[AGB]

how did you do it?? im thinking about the 2nd question now...

 

[Grey Council]

aha! got the first part

 

[Grey Council]

second part I think requires you to use sum of roots theorems. doing it now, lets see if i can get it

bc/a = 4d
let triple root = @ and other root = $
4@³$ = (-3@/$ * c/a)/ a

 

[CM_Tutor]

AGB - How are you going with the second question? Any progress? - I will answer your question, but not until the second part is sorted out.

In the meantime, here's another couple for everyone to think about...

1. If P(x) = Ax⁴ + Bx³ + Cx² + Dx + E is an even polynomial, and one of the roots of P(x) = 0 is twice another of the roots, prove that 4C² = 25AE

2. The equation x³ + 3x - 5 = 0 has roots at alpha, beta and gamma (which, for clarity, I'm going to write as a, b, and c.). Find an equation with roots ab / c, bc / a and ca / b.

 

[AGB]

is the answer to the second question f(x) = 3x^3 + 25x^2 + 9x + 9?? if it is ill post up the working

im just having a study break from modern history (i have the half yearly in a couple of hours)....ill hav a go at the first question this arvo...it looks quite hard tho

 

[AGB]

oh...and i havent got the second question....yet

 

[CM_Tutor]

Originally posted by AGB
is the answer to the second question f(x) = 3x^3 + 25x^2 + 9x + 9?? if it is ill post up the working

Sorry, no.

 

[Xayma]

Originally posted by CM_Tutor
2. The equation x³ + 3x - 5 = 0 has roots at alpha, beta and gamma (which, for clarity, I'm going to write as a, b, and c.). Find an equation with roots ab / c, bc / a and ca / b.

Is it x³-9/5 x²-6x-5? If Im wrong Ill recheck my working but otherwise I will post it up.