Challenge Polynomials problem for current students (2 Viewers)

[AGB]

i just did it again and got 5x^3 + 9x^2 + 15x - 25 = 0....i think it might be wrong....it doesnt match xaymas answer and he is hardly ever wrong

 

[grimreaper]

I dont have time to put working up, but is the answer to question 2: x^3 + 9x^2 + 30x - 5??

 

[CM_Tutor]

Originally posted by Xayma
Is it x³+19/5 x² -6x +5? If Im wrong Ill recheck my working but otherwise I will post it up.

Originally posted by AGB
i just did it again and got 5x^3 + 9x^2 + 15x - 25 = 0....i think it might be wrong....it doesnt match xaymas answer and he is hardly ever wrong

Originally posted by grimreaper
I dont have time to put working up, but is the answer to question 2: x^3 + 9x^2 + 30x - 5??

Sorry, but no to all of you.

 

[Xayma]

Hmm I was rushing through it Ill check the expansion again. Yeah I stuffed up the expansion Ill go edit it above.

Ok heres my working:

If they are the roots.

f(x)=(x-ab/c)(x-ac/b)(x-bc/a)
=x³ -x²([a²b²+a²c²+b²c²]/abc) +(a²+b²+c²)-a²b²c²/abc
=x³ -x²([a²b²+a²c²+b²c²]/abc) +x(a²+b²+c²)-abc
=x³-x²({[ab+ac+bc]²-2[a+b+c][ab+ac+bc]}/abc)+x([a+b+c]²-2[ab+ac+bc])-abc

But a+b+c=0
ab+ac+bc=3
abc=5

Subbing in gets x³-9/5x-6x-5

Originally posted by AGB
doesnt match xaymas answer and he is hardly ever wrong

Why do people keep saying that, I dont even do 4 unit.

 

[CM_Tutor]

Originally posted by Xayma
x³-9/5x-6x-5

is the correct answer but for a typo in the last line - it should be x³ - 9x² / 5 - 6x - 5 = 0, or alternately,
5x³ - 9x² - 30x - 25 = 0

Extn 2 people - can you find an easier way to do this?

 

[Xayma]

Originally posted by CM_Tutor
is the correct answer but for a typo in the last line - it should be x³ - 9x² / 5 - 6x - 5 = 0, or alternately,
5x³ - 9x² - 30x - 25 = 0

Forgot that, I had it on the paper I worked it out on .

 

[DcM]

how do u do the second part of the question of the first page question?

 

[CM_Tutor]

Originally posted by DcM
how do u do the second part of the question of the first page question?

I'm not going to say anything until someone offers an attempt. AGB is working on it, as (I guess) are others, and no one benefits from me just stating the answer.

 

[DcM]

kkz..

working on it..

 

[DcM]

ahhh just had dinner....which was great..

anyway another way to do q1 on page on is:

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

= - b^4 - 8a^2bc + 16a^3d

sum of roots: -3b/2a + @ = -b/a

:. @ = b/2a

product of roots: (-b/2a)^3 * b/2a = d/a

:. -b^4 = 16a^3d

sub back into previous eqn: - b^4 - 8a^2bc + 16a^3d = 0

which = 32a^3d - 8a^2bc = 0

factorise.. 8a^2 (4ad - bc) = 0

:. 4ad = bc


is that the way u did it?

 

[CM_Tutor]

Originally posted by DcM
ahhh just had dinner....which was great..

Good idea, I think I'll go do that pretty soon myself

is that the way u did it?

Yes. (AGB, take note)

 

[Xayma]

Bah Im stuck on the second one on the first page.

Heres where Im stuck at (probably gone by the completly wrong way)

Let w=alpha
x=beta
y=gamma
z=delta

w²xy+w²xz+w²yz... the rest is similar just the squared jumps across to the x then y then z etc... =0

Im trying to prove that for that to equal 0 w=x=y=-z

 

[Grey Council]

aarrrgggh!

can people not say blah question on x page? cause I've set my setting so that it shows 40 posts per page. and i get confused. Just quote the question please. thanks

and out of curiousity, why don't you do 4u xayma? i'd think you'd be good enough. mmm

 

[Xayma]

Long story, the teachers reackon I wouldn't be good enough to do it via correspondance (due to various issues) and there was staff issues so they couldn't run a class with just me in it. They said if there were multiple people doing it I could do it but...

 

[CM_Tutor]

Xayma, here's a hint. Consider for a moment the question you are trying to answer. It was:

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

You are effectively trying to answer the question "prove that it is true to say that any polynomial of this form whose coefficients are related by 4ad = bc must have a triple root". These are not the same thing.

 

[Xayma]

Im trying to do the explain bit, if I can disprove it, it isnt true to say (or am I stuffed up on the wording again).

 

[CM_Tutor]

In order to disprove it, you only need provide a single counter-example... And i'm getting dinner, back in 1/2 to 3/4 of an hour.

 

[Grey Council]

Originally posted by CM_Tutor
You are effectively trying to answer the question "prove that it is true to say that any polynomial of this form whose coefficients are related by 4ad = bc must have a triple root". These are not the same thing.

hrm, I couldn't think of a way to answer the other question, so i also tried to answer "prove that it is true to say that any polynomial of this form whose coefficients are related by 4ad = bc must have a triple root". And i got nowhere.

 

[Xayma]

Originally posted by CM_Tutor
In order to disprove it, you only need provide a single counter-example...

And thats what Im trying to show. But I cant find any Wait if b=c=d=0 ax^4=0 only has one root, So it isnt true to say...

 

[Grey Council]

lol, nice.
wd