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Roots and coeffecients of polynomial equations

  1. Mar 2, 2010 #1
    1. The problem statement, all variables and given/known data
    One root of 2x2-kx+k=0 is twice the other. Find K. (assume k is not 0)


    2. Relevant equations
    not sure what this means.


    3. The attempt at a solution

    [tex]\alpha[/tex]+[tex]\beta[/tex]=[tex]\frac{-b}{a}[/tex]

    [tex]\alpha\beta[/tex]=[tex]\frac{c}{a}[/tex]

    then i tried

    [tex]\alpha[/tex]=2[tex]\beta[/tex] then substituted in..

    2x2-3[tex]\beta[/tex]+[tex]\alpha\beta[/tex]=0 and now im stuck

    im not sure if this is right way to do this
     
  2. jcsd
  3. Mar 2, 2010 #2

    Mark44

    Staff: Mentor

    Let r and 2r be the two roots. Substitute each one in your equation. That will give you two equations in two unknowns - r and k. Solve the two equations simulataneously for these variables.
     
  4. Mar 2, 2010 #3
    thanks for reply. which equation are you mentioning?
     
  5. Mar 2, 2010 #4

    Mark44

    Staff: Mentor

    The only equation you were given, and the one whose roots you are trying to find.
    2x2 - kx + k=0
     
  6. Mar 2, 2010 #5
    thanks for reply. but i still have two unknowns and i can't solve

    edit: and i still have the x at the start of that equation. all i know is 'a' (which is 2)
     
  7. Mar 2, 2010 #6

    Mark44

    Staff: Mentor

    You should have two equations in the two unknowns. Solve these equation simultaneously to find r and k.
     
  8. Mar 2, 2010 #7
    by substituting in, do you mean where the -k is (which is b) i substitute in 2r+4r? (to get this i had r+2r = -[tex]\frac{b}{2}[/tex])

    and then where the +k is (which is c) i substitute in 4r2? (to get this i rearranged 2r2=[tex]\frac{c}{2}[/tex]
     
  9. Mar 2, 2010 #8
    edit: was to remote the extra bracket
     
  10. Mar 2, 2010 #9

    Mark44

    Staff: Mentor

    What I'm saying is that you should ignore the equations you have with alpha and beta and a and b, and focus on the equation you're trying to solve, 2x2 - kx + k = 0.

    You are given that r and 2r are roots of this equation, so if you replace x by r you'll get one equation, and replace x by 2r, you'll get another equation. These two equations will be in terms of r and k.

    Solve those two equations simultaneously for r and k.
     
  11. Mar 2, 2010 #10

    epenguin

    User Avatar
    Homework Helper
    Gold Member

    You got the equation

    2x2-3[tex]\beta[/tex]+[tex]\alpha\beta[/tex]=0

    In the 2nd term here you have used the information you are given (one root is twice the other) and in the third term you haven't yet.

    Oh, and you must have meant

    2x2- 3[tex]\beta[/tex] x + [tex]\alpha\beta[/tex]=0
     
    Last edited: Mar 2, 2010
  12. Mar 2, 2010 #11
    thanks for reply. doing this way i made the two equations equal eachother (2r2-kr+k = 8r2-2kr+k
    ended up with r=[tex]\frac{6}{k}[/tex] and i can't solve that because of two unknown variables.





    thanks for reply. this method i ended up with 2x2-3[tex]\beta[/tex]x+2[tex]\beta[/tex]2
     
  13. Mar 2, 2010 #12

    Mark44

    Staff: Mentor

    I get r = k/6 (not 6/k) or r = 0.

    If r = 0, then 2r = 0, so what is k?
    If r = k/6, then 2r = k/3, so what is k?
     
  14. Mar 2, 2010 #13
    thanks for reply. yeah i got r=k/6 (i mistyped)

    If r = 0, then 2r = 0, so what is k? k=0
    If r = k/6, then 2r = k/3, so what is k? plugged r=k/6 back into original equation and got k = 9 or 0.. right?

    EDIT: made a mistake, recalculated and got 9 and 0
     
  15. Mar 2, 2010 #14

    epenguin

    User Avatar
    Homework Helper
    Gold Member

    Sorry -I made a typo.

    It is usually less confusing to work with an equation whose leading coefficient is 1. So better to divide your initial equation by 2. So use

    x2 - kx/2 + k/2

    A quadratic with one root twice the other is

    x2-3[tex]\beta[/tex]x+2[tex]\beta[/tex]2

    I think you had the idea of that but the leading coefficient caused you confusion. You should easily be able to see your last result must be wrong; do not accept any reult without checking it against your original problem.
     
    Last edited: Mar 2, 2010
  16. Mar 2, 2010 #15
    so how do i solve it from here? im really confused..
     
  17. Mar 2, 2010 #16

    epenguin

    User Avatar
    Homework Helper
    Gold Member

    Any, in other words every, quadratic with one root twice the other is of form

    [tex]x^2-3\beta x+2\beta^2[/tex] (1)

    (a result you got but for a little mistake.)

    You are told

    [tex]x^2 - \frac{kx}{2} + \frac{k}{2} [/tex] (2)

    is a quadratic with one root twice the other.

    So (2) must be of form (1). Or if you like identical to it for all x. So do you see how to compare them (remembering similar exercises you've probably done before for identical expressions)?
     
    Last edited: Mar 2, 2010
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