A Polynomials Question From Fitzpatrick (1 Viewer)

[.ben]

10. find the cubic eqn whose roots are twic htose of the eqn 3x^3-2x^2+1=0

thanks

ps: the answer is 3x³-4x²+24=0 bu t i keep getting 3x^3-4x^2+2=0 so is there a problem with the my answer or the books

 

[Dreamerish*~]

Re: polyenomialss form fitzpartric

Note: Fitzpatrick makes tons of mistakes.

 

[Riviet]

Re: polyenomialss form fitzpartric

It's more likely to be one of the many mistakes that the textbook has in the answers, basically, don't rely on the answers all the time. If you keep getting your answer and keep getting the same answer that's different to fitzpatrick's answers, and you're sure everything's right, just give yourself a pat on the back and move on to the next question.

P.S It's nice seeing you around the maths forums Dreamerish.

 

[.ben]

Re: polyenomialss form fitzpartric

yeh but in this case can you just confirm im right, cos i might be wrong?

 

[pLuvia]

Re: polyenomialss form fitzpartric

I got 3x³-4x²+2=0

Since
x+y+z=-b/a

xy+xz+yz=c/a

xyz=-d/a

So yeh, that answers are probably wrong

 

[.ben]

Re: polyenomialss form fitzpartric

ok cool thaks

 

[Riviet]

Slidey, I'm afraid I'm going to have to correct you.

let p, q and r be the roots of the original cubic polynomial, say p(x)=ax³+bx²+cx+d.

p+q+r=-b/a=2/3

.'. sum of roots that are double the roots of p(x) = 2p+2q+2r = 2(p+q+r) = 2. 2/3 = 4/3
Equating 4/3=-b/a,
a=3, b=-4

product of roots of p(x) = pqr = -d/a = -1/3
2p.2q.2r = 8pqr = 8.-d/a = 8.(-1/3) = -8/3
equating -d/a=-8/3
d=8

.'. new equation with roots 2p, 2r, 2q is 3x³-4x²+8=0

 

[Slidey]

Aye, that's what you get for misreading subscripts.

Anyway, a better and more systematic way to solve this problem is the following (although that method is quite neat):

3x^3-2x^2+1=0

Let y=2x, x=y/2. Sub in:

3y^3/8-y^2/2+1

3y^3-4y^2+8

Rewrite in terms of x:

3x^3-4x^2+8

Which means pLuvia and .ben still need to check their working.

 

[.ben]

but that answer is not the one in the boook

 

[Slidey]

Well spotted. That's because the book is wrong and multiplied the constant by 3 accidentally - not hard to believe considering the leading coefficient is 3.

Further, I'm presuming your mistake, and pLuvia's, lies in multiplying -b/a, c/a and -d/a all by 2 instead of pushing 2 lots of each root through the original identities, as Riviet did.

 

[.ben]

what's wrong with this method:

3x^3-2x^2+1=0
let a, b and c be roots

a+b+c=2/3
ab+ac+bc=0
abc=-1/3

new eqn:
x^3-2(a+b+c)x^2+2(ab+ac+bc)x-2abc=0

2(a+b+c)=4/3
2(ab+ac+bc)=0
2(abc)=-2/3

therefore
x^3-4/3x^2+2/3=0

and
3x^3-4x^2+2=0

 

[c0okies]

well to confirm two other people got riviet's answer.. in the back of my fitzpatrick someone changed it to says 3x^3-4x^2+8=0 and i got that answer as well

 

[Slidey]

what's wrong with this method:

3x^3-2x^2+1=0
let a, b and c be roots

a+b+c=2/3
ab+ac+bc=0
abc=-1/3

new eqn:
x^3-2(a+b+c)x^2+2(ab+ac+bc)x-2abc=0

2(a+b+c)=4/3
2(ab+ac+bc)=0
2(abc)=-2/3

therefore
x^3-4/3x^2+2/3=0

and
3x^3-4x^2+2=0

You can't simply multiply each sum/multiple of roots by 2 because that won't acheive what you want.

If the roots are double, then the roots are: 2a, 2b and 2c.

That means this:
a+b+c=2/3
ab+ac+bc=0
abc=-1/3

Becomes this:
(2a)+(2b)+(2c)=2(a+b+c)=4/3
(2a)(2b)+(2a)(2c)+(2b)(2c)=4(ab+ac+bc)=0
(2a)(2b)(2c)=8abc=-8/3

 

[.ben]

another question:

25. Show that th ecubic eqn 8x^3-6x+1=0 can be reduced to the form cos3(theta)=-1/2.

 

[.ben]

ok cool ^dont do q25

 

[.ben]

You can't simply multiply each sum/multiple of roots by 2 because that won't acheive what you want.

If the roots are double, then the roots are: 2a, 2b and 2c.

That means this:
a+b+c=2/3
ab+ac+bc=0
abc=-1/3

Becomes this:
(2a)+(2b)+(2c)=2(a+b+c)=4/3
(2a)(2b)+(2a)(2c)+(2b)(2c)=4(ab+ac+bc)=0
(2a)(2b)(2c)=8abc=-8/3

yeh thats what i did

 

[Slidey]

No it isn't. Take a careful look at Riviet's (or my) working.

You have multiplied each one by 2, when you should instead be substituting in twice each root, which will multiply by 2, 4 and 8 respectively.

 

[Riviet]

Aye, that's what you get for misreading subscripts.

Anyway, a better and more systematic way to solve this problem is the following (although that method is quite neat):

3x^3-2x^2+1=0

Let y=2x, x=y/2. Sub in:

3y^3/8-y^2/2+1

3y^3-4y^2+8

Rewrite in terms of x:

3x^3-4x^2+8

A very interesting and ingenious method Slidey, I would presume this works for similar type questions and is perfectly valid to use in the HSC exam?

 

[pLuvia]

Ar ok, damn lol careless thanks slide and Riviet

 

[pLuvia]

Aye, that's what you get for misreading subscripts.

Anyway, a better and more systematic way to solve this problem is the following (although that method is quite neat):

3x^3-2x^2+1=0

Let y=2x, x=y/2. Sub in:

3y^3/8-y^2/2+1

3y^3-4y^2+8

Rewrite in terms of x:

3x^3-4x^2+8

Which means pLuvia and .ben still need to check their working.

Interesting method Slide Rule, just wondering do we learn that in MX2 polynomials? because that is hell of lot faster than the conventional method