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Arithmetic Progression + System of equations + binomial

  1. Aug 13, 2009 #1
    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. jcsd
  3. Aug 13, 2009 #2

    Hurkyl

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    (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? :confused:


    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.
     
  4. Aug 13, 2009 #3

    VietDao29

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

    The second equation:
    [tex](x + r) ^ 2 - (x - r) ^ 2 = \frac{24}{5}[/tex]. 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:

    [tex]\left( \frac{3}{5} + r \right) ^ 2 - \left( \frac{3}{5} - r \right) ^ 2 = \frac{24}{5}[/tex]

    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:

    [tex]f(x) = \alpha (x - x_1) (x - x_2) (x - x_3)[/tex], where [tex]\alpha[/tex] is some constant.

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

    Hope you can go from here. :)
     
  5. Aug 13, 2009 #4
    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. :/
     
  6. Aug 13, 2009 #5

    VietDao29

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    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: [tex](a \pm b) ^ 2 = a ^ 2 \pm 2ab + b ^ 2[/tex].
     
    Last edited by a moderator: May 4, 2017
  7. Aug 14, 2009 #6
    [tex]x^2+2xr+r^2-x^2-2xr+r^2[/tex]

    All I'm left with is

    [tex]2r^2=\frac{24}{5}[/tex]

    Which leads me to:

    [tex]r^2=\frac{24}{10}[/tex]


    [tex]r=\sqrt\frac{24}{10}[/tex]

    Which I can simplify to

    [tex]r=2\frac{\sqrt6}{10}[/tex]

    Which I'm pretty sure does not equal 2.
     
  8. Aug 14, 2009 #7

    HallsofIvy

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

     
  9. Aug 15, 2009 #8
    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 [tex]
    f(x) = \alpha (x - x_1) (x - x_2) (x - x_3)
    [/tex] but I'm confused about what to do with it.

    From what I know so far the polynomial looks like this [tex]f(x)=5x^3+bx^2+cx+d[/tex]
     
  10. Aug 15, 2009 #9
    [tex]x_1, x_2[/tex] and [tex]x_3[/tex] are the roots of the equation. Then it becomes clear i guess?
     
  11. Aug 15, 2009 #10

    VietDao29

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

    So, your third degree polynomial is:

    [tex]f(x) = \alpha \left( x + \frac{7}{5} \right) \left( x - \frac{3}{5} \right) \left( x - \frac{13}{5} \right)[/tex]

    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 [tex]\alpha[/tex]. And your problem is solved.
     
  12. Aug 15, 2009 #11

    HallsofIvy

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    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
    [tex] f(x)= \alpha(x-(-7/5))(x- 3/5)(x- 13/5)= \alpha(x+ 7/5)(x- 3/5)(x- 13/5)[/tex]
    and knowing that "the coefficient of the highest degree term is 5" tells you that [itex]\alpha= 5[/itex]. 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".
     
  13. Aug 21, 2009 #12
    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.
     
  14. Aug 23, 2009 #13
    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:

     
  15. Aug 26, 2009 #14
    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.
     
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