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#1 2014-12-09 22:52:44

thedarktiger
Member
Registered: 2014-01-10
Posts: 91

vieta's formulas

For math class homework, I have several problems like this and I don't really know how to do these:

Let g(x) = x^4-5x^3+2x^2+7x-11, and let the roots of g(x) be p, q, r, and s.

(a) Compute pqr + pqs + prs + qrs.
(b) Compute


(c) Compute p^2 + q^2 + r^2 + s^2.
(d) Compute p^2qrs + pq^2rs + pqr^2s + pqrs^2.

Can you help me understand the method, not just do them for me?

Thanks!


Good. You can read.

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#2 2014-12-09 22:57:47

Agnishom
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From: Riemann Sphere
Registered: 2011-01-29
Posts: 24,974
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Re: vieta's formulas

Are you aware of Vieta?


'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
I'm not crazy, my mother had me tested.

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#3 2014-12-10 00:20:47

Agnishom
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From: Riemann Sphere
Registered: 2011-01-29
Posts: 24,974
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Re: vieta's formulas

Silly me. of course you are.

(a) The answer is The coefficient of x divided by that of x^4.

(b) Try taking the LCM of the denominator and writing it out as a single large fraction. You'll get a Vieta form.

(c) p^2 + q^2 + r^2 + s^2 = (p+q+r+s)^2 - 2(pq+qr+rs+...) Apply Vieta from here.

(d) I do not see it yet but look for some similar algebraic manipulation

Last edited by Agnishom (2014-12-10 01:37:53)


'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
I'm not crazy, my mother had me tested.

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#4 2014-12-10 00:58:48

Bob
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Registered: 2010-06-20
Posts: 10,011

Re: vieta's formulas

hi the darktiger,

Because I can never be bothered to learn formulas, I went back to first principles:

See below for help with getting these expressions.**

So the first is now easy, I hope.

Agnishom has given you the way to do the second and third.

The fourth:

Bob

** When you multiply the four brackets together you must take one term from each bracket, so:

x^4 ... take x from every bracket; only one way to do this.

x^3 ... take x from three brackets and a letter from the other bracket.  There are four ways of doing that.

x^2 ... take x from two brackets and letters from the other two.  There are six ways of pairing up the letters to achieve this. 4C2 in combinations.

x     ... take x from a bracket and letters from the others.  It's easiest to think about choosing all four letters and then leaving one out.  Four ways of leaving one out.

no x ... take a letter every time; one way.


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#5 2014-12-10 01:38:33

Agnishom
Real Member
From: Riemann Sphere
Registered: 2011-01-29
Posts: 24,974
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Re: vieta's formulas

Oh, I made a silly mistake in my last post. I corrected it.

Thank you bob.


'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
I'm not crazy, my mother had me tested.

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#6 2014-12-10 19:58:15

thedarktiger
Member
Registered: 2014-01-10
Posts: 91

Re: vieta's formulas

@Agnishom, on your before last post, you said the answer to (a) is the coefficient of x divided by that of x^4. Thats seven, and its incorrect... I think the answer is -7

Thanks both of you


Good. You can read.

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#7 2014-12-10 20:45:39

Bob
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Registered: 2010-06-20
Posts: 10,011

Re: vieta's formulas

I agree with -7

http://en.wikipedia.org/wiki/Vieta's_formulas

The sign alternates between + and - for each coefficient.

Using my earlier post, you can see it depends on the number of letters multiplied together.

eg.  The term with pqr is negative because -p times -q times -r yields a negative result.

Bob

ps.  For purists only.  In my posts on this thread the set of letters = {p,q,r,s}  I am not considering x as a letter.  smile


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#8 2014-12-11 15:11:13

thedarktiger
Member
Registered: 2014-01-10
Posts: 91

Re: vieta's formulas

I just got another similar problem I thought was really cool:

Let r, s, and t be roots of the equation

.

Compute

rs/t+st/r+tr/s = ((rs+st+tr)^2-2rst(r+s+t))/rst = (6^2-2*9*(-5))/9 = -6

but strangly, -6 = -(rs+st+rt)...

I dont see how that works smile

I just thought that was a really cool factorization.

so that can also be

(b^2+2ac)/c = a

maybe I discovered something new. Duno smile

Last edited by thedarktiger (2014-12-11 15:19:25)


Good. You can read.

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#9 2014-12-11 20:17:46

Bob
Administrator
Registered: 2010-06-20
Posts: 10,011

Re: vieta's formulas

hi thedarktiger,

just checking this ...............

Could that result be a  coincidence.

Try this:

Make up a new cubic and compute the values this time.  Do you get the same result ?

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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