vieta's formulas

thedarktiger · 2014-12-09 22:52:44

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!

Agnishom · 2014-12-09 22:57:47

Are you aware of Vieta?

Agnishom · 2014-12-10 00:20:47

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)

Bob · 2014-12-10 00:58:48

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.

Agnishom · 2014-12-10 01:38:33

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

Thank you bob.

thedarktiger · 2014-12-10 19:58:15

@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

Bob · 2014-12-10 20:45:39

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.

thedarktiger · 2014-12-11 15:11:13

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

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

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

Bob · 2014-12-11 20:17:46

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

Math Is Fun Forum

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

vieta's formulas

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

Re: vieta's formulas

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

Re: vieta's formulas

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

Re: vieta's formulas

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

Re: vieta's formulas

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

Re: vieta's formulas

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

Re: vieta's formulas

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

Re: vieta's formulas

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

Re: vieta's formulas

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