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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

    2agnpdt.jpg

    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

    rock.freak667

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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

    VietDao29

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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

    VietDao29

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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

    HallsofIvy

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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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