Arithmetic Progression + System of equations + binomial

[Paulo Serrano]

Homework Statement

• 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)

Homework Equations

None that I can think of.

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

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.

 

[Hurkyl]

(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)

(since the r's cancel themselves out)

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.

 

[VietDao29]

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.

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. :)

 

[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. :/

 

[VietDao29]

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. :/

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" (Read the part 1, 2, and 3).

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

 

[Paulo Serrano]

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

All I'm left with is

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

Which leads me to:

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.

 

[HallsofIvy]

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

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.

All I'm left with is

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

Which leads me to:

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.

 

[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 <br /> f(x) = \alpha (x - x_1) (x - x_2) (x - x_3)<br /> 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

 

[Fightfish]

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

 

[VietDao29]

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

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)

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.

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.

 

[HallsofIvy]

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

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".

I found the value of the three roots, which are the following:

7/5 , 3/5 , 13/5

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

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 <br /> f(x) = \alpha (x - x_1) (x - x_2) (x - x_3)<br /> 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

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".

 

[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.

 

[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:

Given that the coefficient of the highest degree term is 5...
A third degree polynomial...

 

[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.