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Homework Help: Polynomial with 3 unknown variables

  1. Aug 31, 2009 #1
    1. The problem statement, all variables and given/known data

    (4m + 3n)x2 – 5nx + (m – 2) = 0

    The sum of the roots is 5/8
    The product of the roots is 3/32

    What is M + N?

    2. Relevant equations

    3. The attempt at a solution


    This is the actual solution. I just don't understand it.


    Why is it that although it says the SUM is equal to 5/8 the solution above has has things being divided? How do I find the roots of this equation?
  2. jcsd
  3. Aug 31, 2009 #2


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    For ax2+bx+c=0 (or x2+(b/a)x+c/a=0) if the roots are α and β, then x2+(b/a)x+c/a≡ (x-α)(x-β)

    The right side works out to be x2-(α+β)x+ (αβ). So equating coefficients we get

    α+β=-b/a and αβ=c/a

    or the sum of the roots= -b/a
    the product of the roots=c/a
  4. Aug 31, 2009 #3


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    If you want some further reading, you can have a look here. It's called http://en.wikipedia.org/wiki/Vi%C3%A8te%27s_formulas" [Broken].
    Last edited by a moderator: May 4, 2017
  5. Sep 1, 2009 #4
    Is that just a rule that I should memorize then? I couldn't really work this out using the actual numbers, could I?
  6. Sep 1, 2009 #5


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    Yes, this formula is pretty well-known, and you should memorize it. It's not very hard to memorize, yes?

    This formula comes very handy, when you need to work with the sum and/or product of the root of a polynomial, without the need to know what the roots are (since it's calculated based on the coefficients of the polynomial itself).


    For example, you are ask to evaluate the sum of the 2 roots of the following equation:

    x2 + x - 2 = 0

    The first way is to find the 2 roots of this equation, which turn out to be -2, and 1. And take the sum: -2 + 1 = -1.

    The second way is to apply the formula:
    x1 + x2 = -b/a = -1/1 = -1.

    Err, I don't really get what you mean. What do you mean by 'actual numbers'?
  7. Sep 1, 2009 #6


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    If you don't like memorizing, think it out: If [itex]\alpha_1[/itex] and [itex]\alpha_2[/itex] are roots of [itex]x^2+ bx+ c= 0[/itex], then we must have [itex](x- \alpha_1)(x- \alpha_2)= x^2+ bx+ c= 0[/itex] for all x. Multiplying that first product, [itex]x^2- (\alpha_1+ \alpha_2)x+ \alpha_1\alpha_2)= x^2+ bx+ c[/itex] and, since that is true for all x, taking x= 0 gives [itex]\alpha-1\alpha_2= c[/itex] and then, taking x= 1, [itex]1^2- (\alpha_1+ \alpha_2)(1)+ c= 1^2+ b(1)+ c[/itex] so [itex]\alpha_1+ \alpha_2= -b[/itex].
  8. Sep 2, 2009 #7
    I meant that if I tried solve this without knowing those equations it would be near impossible...Anyway, thanks again guys. Learned a little more. :)
  9. Sep 9, 2009 #8
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