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Stumped on this one

  1. Apr 10, 2005 #1
    The quadratic equation x^2 + mx + n = 0 has roots that are twice those of x^2 + px + m = 0, and none of m, n, p is zero. What is the value of n/p?



    I'm stuck and don't know where to begin.
     
  2. jcsd
  3. Apr 10, 2005 #2

    dextercioby

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    Set the ratio of the roots to 2 and see whether u can isolate the ratio n/p

    Daniel.
     
  4. Apr 10, 2005 #3

    Curious3141

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    In a quadratic equation [tex]ax^2 + bx + c = 0[/tex], there are two roots (either both real or both complex). The sum of the two roots is [itex]-\frac{b}{a}[/itex] and the product of the two roots is given by [itex]\frac{c}{a}[/itex].

    Use this to come up with four simultaneous equations for the 2 pairs of roots for the given equations. The sum of the roots of the first equation is twice that of the sum of the roots of the second equation. The product of the roots of the first equation is 4 times the product of the roots of the second equation. Do a few algebraic manipulations to get the required ratio. Specifically, the property for the product of the roots will allow you to express n in terms of m, and the property for the sum of the roots will allow you to express m in terms of p, the rest is easy.
     
    Last edited: Apr 10, 2005
  5. Apr 10, 2005 #4
    this was an amc 12B problem?
    Anyway, the negative of roots add up to equal m, and mutliply to give n.
    2(a+b)=m
    (2a)(2b)=n

    in the 2nd equation
    (a)(b)=m
    (a+b)=p

    ab=2(a+b)..........(a+b)=(ab)/2=p.
    n=4ab
    p=0.5ab
    n/p=8
     
    Last edited: Apr 10, 2005
  6. Apr 10, 2005 #5
    Yeah this was an amc question and I couldn't do it. Did you take the test? Did you advance to the next round?


    p.s. Thanks for the help everyone
     
  7. Apr 10, 2005 #6
    yep, i went to the AIME (got a score of a 8/15), i didnt make the USAMO though. Did you make the AIME?
     
  8. Apr 10, 2005 #7
    yeah, i got a score of 5/15.
     
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