# Arithmetic Progression + System of equations + binomial

1. Aug 13, 2009

### Paulo Serrano

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

• A third degree polynomial has 3 roots that, when arranged in ascending order, form an arithmetic progression in which the sum of the 3 roots equal 9/5.
• The difference between the square of the greatest root and the smallest root is 24/5
• Given that the coefficient of the highest degree term is 5, determine:

a) the arithmetic progression
b) the coefficient of the first degree term of the polynomial

(I hope I translated this right)

2. Relevant equations

None that I can think of.

3. The attempt at a solution

Using the first 2 pieces of given information I constructed the following:

http://img29.imageshack.us/img29/4939/systemofequations.jpg [Broken]

With just the first equation I was able to determine that x=3/5 (since the r's cancel themselves out)

After that, I'm completely lost. I don't know how to find the value of r. I can tell you guys right now that it's supposed to equal r=2, but I don't know how or why. If I substitute the value of x into the second equation they cancel themselves out anyway and I get a different answer for r.

I also don't know what to do after I've found the value of r.

Last edited by a moderator: May 4, 2017
2. Aug 13, 2009

### Hurkyl

Staff Emeritus
(Just to make sure you confused yourself -- as you're using them, "x" has nothing to do with the indeterminate variable of your polynomial, and "r" doesn't directly relate to the roots of the polynomial)

They do?

I note you haven't used the fact that (x-r), x, and (x+r) are roots of a particular polynomial, nor have you used the fact you were told its leading coefficient.

3. Aug 13, 2009

### VietDao29

So, you've used the first equation to solve for x. Now that you have: $$x = \frac{3}{5}$$

The second equation:
$$(x + r) ^ 2 - (x - r) ^ 2 = \frac{24}{5}$$. This equation contains x (which you've already found out), and r (the unknown). So, you can just simply plug x in, and solve for r. Like this:

$$\left( \frac{3}{5} + r \right) ^ 2 - \left( \frac{3}{5} - r \right) ^ 2 = \frac{24}{5}$$

You'll have 2 values for r, and in this problem, only one is valid. Can you see why?

Having x, and r, can you find out the progression (also the 3 roots of the polynomial)?

If a third degree polynomial has 3 roots, say x1, x2, and x3; then it is of the form:

$$f(x) = \alpha (x - x_1) (x - x_2) (x - x_3)$$, where $$\alpha$$ is some constant.

Now, can you find the coefficient of the first degree term of the polynomial?

Hope you can go from here. :)

4. Aug 13, 2009

### Paulo Serrano

It's just about midnight where I live now, so I'm going to try this tomorrow morning. Sorry if I seem oblivious, it's just that a lot of this stuff I never learned in school or if I did learn I've already forgotten it. I'm going to try this tomorrow morning with the advice you guys gave me but I'll likely need more help. :/

5. Aug 13, 2009

### VietDao29

Well, don't worry then. Just try, and see how far you can go, then post it here. So that we can know which part you are not sure about.

I think you might want to review some concepts about http://en.wikipedia.org/wiki/Quadratic_equation" [Broken] (Read the part 1, 2, and 3).

And as a reminder: $$(a \pm b) ^ 2 = a ^ 2 \pm 2ab + b ^ 2$$.

Last edited by a moderator: May 4, 2017
6. Aug 14, 2009

### Paulo Serrano

$$x^2+2xr+r^2-x^2-2xr+r^2$$

All I'm left with is

$$2r^2=\frac{24}{5}$$

$$r^2=\frac{24}{10}$$

$$r=\sqrt\frac{24}{10}$$

Which I can simplify to

$$r=2\frac{\sqrt6}{10}$$

Which I'm pretty sure does not equal 2.

7. Aug 14, 2009

### HallsofIvy

No. The second square is $(x- r)^2= x^2- 2xr+ r^2$ and so $-(x-r)^2= -x^2+ 2xr- r^2$, not $-x^2- 2x+ r^2$. In particular, the two "$r^2$" terms cancel so you have a linear equation for r.

8. Aug 15, 2009

### Paulo Serrano

This is why I'm glad I come to this forum for help. Although I am surprised that I could forget how to do such a basic operation. Thank you.

Now that I know that x=3/5 and y=2 I found the value of the three roots, which are the following:

7/5 , 3/5 , 13/5

That's the first part of the question done. Now I need a point in the right direction for the second part.

VietDao did give me this formula $$f(x) = \alpha (x - x_1) (x - x_2) (x - x_3)$$ but I'm confused about what to do with it.

From what I know so far the polynomial looks like this $$f(x)=5x^3+bx^2+cx+d$$

9. Aug 15, 2009

### Fightfish

$$x_1, x_2$$ and $$x_3$$ are the roots of the equation. Then it becomes clear i guess?

10. Aug 15, 2009

### VietDao29

This is wrong, the three roots should be:
-7/5, 3/5, and 13/5.

So, your third degree polynomial is:

$$f(x) = \alpha \left( x + \frac{7}{5} \right) \left( x - \frac{3}{5} \right) \left( x - \frac{13}{5} \right)$$

Memorize the formula is one thing, but to "understand" it is another. Understanding a formula helps one remember it longer, and more easily.

Let's see if you know how I got that formula.

-----------------------

There is one extra piece of information, which should helps you determine $$\alpha$$. And your problem is solved.

11. Aug 15, 2009

### HallsofIvy

You mean "r= 2" don't you? That's the problem with using "x" or "y" for parameters. I would have said "a- d, a, and a+ d".

3/5- 2= -7/5, not 7/5.

So VietDao's formula becomes
$$f(x)= \alpha(x-(-7/5))(x- 3/5)(x- 13/5)= \alpha(x+ 7/5)(x- 3/5)(x- 13/5)$$
and knowing that "the coefficient of the highest degree term is 5" tells you that $\alpha= 5$. All that's left to do is to multiply it out in order to get "the coefficient of the first degree term of the polynomial".

12. Aug 21, 2009

### Paulo Serrano

I haven't replied in so long because of a bit of embarrassment and a bit of not having time. Can you tell me what this type of calculation is called? I only know how to solve second degree polynomials. Why is the highest degree term a constant like that? I know I'm asking for a lot so if you could just link me to a wikipedia page or something I would appreciate it.

13. Aug 23, 2009

### DorianG

The highest degree term isn't constant.
It is a variable in the third degree, (i.e. its value depends on a variable, x, which is then cubed) which in this case is then multiplied by a constant, 5.
This was given in the initial question you posted:

14. Aug 26, 2009

### Paulo Serrano

Finally got the courage to do this again and found it to be pretty easy once I took my time. Thanks very much for the help guys. I'll be needing more soon. One thing, though...

In order for me to solve this I multiplies two of the expressions in parenthesis and then multiplied what I got from there with the other one. Is there a faster way of doing this? The way I did it would take a very long time if it was something like a 6th degree polynomial.