# Roots of a third degree polynomial

1. Nov 5, 2012

### V0ODO0CH1LD

1. The problem statement, all variables and given/known data

Knowing that the equation:
$$X^n-px^2=q^m$$
has three positive real roots a, b and c. Then what is
$$log_q[abc(a^2+b^2+c^2)^{a+b+c}]$$
equal to?

2. Relevant equations

$$a + b + c = -(coefficient \ of \ second \ highest \ degree \ term) = -k_2$$

$$abc = -(constant \ coefficient) = k_4$$

$$a^2 + b^2 + c^2 = (coefficient \ of \ second \ highest \ degree \ term)^2-(coefficient \ of \ third \ highest \ degree \ term) = k^2_2-2k_3$$

3. The attempt at a solution

First I assumed X^3 is equal to k_1x^3 and so x^3 = X^3/k_1.

$$X^n-px^2=q^m\\ x^3-\frac{p}{k_1}x^2-\frac{q^m}{k_1}=0\\ k_2=-\frac{p}{k_1}\\ k_3=0\\ k_4=-\frac{q^m}{k_1}\\ a+b+c=\frac{p}{k_1}\\ abc=\frac{q^m}{k_1}\\ a^2+b^2+c^2=\frac{p^2}{k^2_1}\\$$
$$log_q[abc(a^2+b^2+c^2)^{a+b+c}]=\\ \frac{q^m}{k_1}(\frac{p^2}{k^2_1})^{\frac{p}{k_1}}=\\ log_q\frac{q^mp^{\frac{2p}{k_1}}}{k_1^{1+\frac{2p}{k_1}}}=\\ log_qq^mp^{\frac{2p}{k_1}}-log_qk_1^{1+\frac{2p}{k_1}}=\\ log_qq^m+log_qp^{\frac{2p}{k_1}}-log_qk_1^{1+\frac{2p}{k_1}}=\\ m+\frac{2p}{k_1}log_qp-(1+\frac{2p}{k_1})log_qk_1=\\$$

If I substitute (X^3)/(x^3) for k_1 I have an equation that only resembles the possible answers if X = x or X = -x, which makes me think that either X was a typo and they meant x. Or that somewhere in the problem statement they tell you about something that according to some property of polynomials makes X = x or X = -x implied.

Is there some weird property of polynomial that I don't know that would help me solve this problem? What is it?

EDIT: Sorry for the latex error..

Last edited: Nov 5, 2012
2. Nov 5, 2012

### ehild

I think x and X mean the same, otherwise it would be explained.

ehild

3. Nov 5, 2012

### V0ODO0CH1LD

Doesn't the fact that the coefficient on the x^1 term is 0 mean that the coefficient on the x^3 term has to be zero or something of that nature?

EDIT: Doesn't the fact that the coefficient on the x^1 term is 0 mean that the coefficient on the x^3 term has to be ONE or something of that nature?

4. Nov 5, 2012

### Staff: Mentor

Since there is no x term, but there is an x2 term, it is reasonable to assume that the coefficient of x is 0. However, you can't assume that xn means x3. You are given this statement:
xn - px2 = qm has three positive real roots a, b and c. That doesn't necessarily imply that the equation is a cubic. What this says to me is that of all roots, three of them are real and positive. Any other roots could be negative or nonreal.

5. Nov 5, 2012

### V0ODO0CH1LD

I haven't thought of it like that, but you have a point. How would I proceed then? Do I actually have to find the real roots? I have no clue as to how I am supposed to do this now..

EDIT: No, I am sorry! I re-red the thread and realized something! There is a typo in the problem statement X^n is actually X^3! But couldn't X^3 mean x^(anything) if X ≠ x anyway?

6. Nov 6, 2012

### ehild

It is said that the roots of the equation are a, b, c. If X is not x it should be specified. Do not worry, it must be just a typo. Word starts a line with a capital letter, that can be the reason.

ehild

7. Nov 6, 2012

### V0ODO0CH1LD

Okay, but doesn't the fact that the coefficient on the x1 term is zero mean anything at all? Assuming for a second that X ≠ x; then $k_1(x-a)(x–b)(x–c)=k_1x^3-k_1(a+b+c)x^2+k_1(ab+ac+bc)x-k_1(abc)=X^3-px^2-q^m$
Doesn't that mean that:

$$k_1x^3 = X^3\\ k_1(a+b+c)=p\\ k_1(ab+ac+bc)=0\\ k_1(abc)=q^m$$

k_1 can't be zero otherwise the whole equation would be zero. So isn't there a way to prove that X = x maybe using the fact that a, b and c are positive? And how can "k_1 (ab + ac + bc) = 0" if a,b and c are positive and k_1 ≠ 0?

8. Nov 6, 2012

### Staff: Mentor

I believe that X is a typo, and should be x, which makes this the equation:
x3 -px2 = qm, with a, b, anc c being positive real roots.

Then
a3 -pa2 = qm
b3 -pb2 = qm
c3 -pc2 = qm

So

a2 = $\frac{q^m}{a - p}$
b2 = $\frac{q^m}{b - p}$
c2 = $\frac{q^m}{c - p}$

Substitute these into your log expression and use the properties of logs to simplify it as much as possible.

9. Nov 6, 2012

### V0ODO0CH1LD

The only thing keeping me from accepting that X is a typo is that x3 -px2 = qm does not have three real positive roots. Maybe I could find a condition on X for X3 -px2 = qm to have three positive real roots. But I don't know how to..

10. Nov 6, 2012

### ehild

You are right, the roots of x3 -px2 = qm can not be all positive. If X is some function of x, X3-px2 - qm=f(x)(x-a)(x-b)(x-c) where a, b, c are positive and f(x)=0 does not have positive roots. That means X3=f(x)(x-a)(x-b)(x-c) +px2 + qm, f(x) can be anything which does not have positive roots. But then you can not solve the problem.

I think it is just a badly worded problem. Instead of "all roots are positive" it should be "the product of the three roots is positive".

ehild

11. Nov 6, 2012

### Staff: Mentor

V0ODO0CH1LD, I think that you should ask your instructor for clarification on this problem. We are 11 posts into this thread, and still can't tell if X is the same as or different from x, whether it's X3 or Xn (or x3 or xn), and whether the three roots are positive or their product is positive.

Please check with you instructor as to exactly what is given in this problem. Until then, I'm done with it.

12. Nov 6, 2012

### V0ODO0CH1LD

I just sent an email to my professor, I will post his answer as soon as I receive it.

13. Nov 6, 2012

### uart

I'm pretty sure that the capitol X is just a typo, otherwise the question is not well defined.

More concerning to me is the "n". As Mark44 has pointed out, just because we're given that there are 3 real roots doesn't mean that there are only three roots and that therefore n=3.

However, if we are not able to assume that n=3, and if there may be other (non positive real) roots, then I don't think that this is a do-able problem. So I think that there is yet another typo (or poorly a worded part of question). Where it says "... has three positive real roots a, b and c", I believe that what the questioner *intended* to say was, "... has three roots a,b and c, all of which are positive real".