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Thread: Sum and Product of roots ><

  1. #1
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    Sum and Product of roots ><

    Ive got a problem and my maths test is 2mw! gaaaaaa

    The roots of the equation X^2 + (P-10)X = 10P are a and a + 4 . Find the possible values of P


    a is actually the aplha sign thingy but i dunno how to type it
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  2. #2
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    Hello,

    You have the equation : X^2 + (P-10)X = 10P

    You know that $\displaystyle \alpha$ (click on it to see how i wrote it) and $\displaystyle \alpha+4$ are solutions.

    This means that :

    $\displaystyle \alpha^2+(P-10) \alpha=10P$
    $\displaystyle (\alpha+4)^2+(P-10)(\alpha+4)=10P$
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  3. #3
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    I dont get how you derive to this:

    $\displaystyle \alpha^2+(P-10) \alpha=10P$
    $\displaystyle (\alpha+4)^2+(P-10)(\alpha+4)=10P$

    Why is X replaced by $\displaystyle \alpha$

    Sorry, but im really dumb at roots. have a phobia of it somehow... Never manage to understand it ><

    And thanks for showing me the way to do e alpha thing! ^_^
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  4. #4
    Moo
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    Well, as a and a+4 are solutions, they verify the equation. The equation has x as the unknown, the variable.
    So if "a takes the place of x", the equation is true.

    The same goes for a+4
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  5. #5
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    So does it mean that i dont need to do the aplha + Beta = -b/a way?
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  6. #6
    Moo
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    You can have a go with that

    There are multiple ways of solving it, your suggestion is more direct

    You may also know alpha*beta

    ($\displaystyle \alpha = \alpha \text{ \ and \ } \beta = \alpha + 4$)
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  7. #7
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    and how do i do that way then? :P
    Im really very confused
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  8. #8
    Moo
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    Well, if you have a trinomial :

    ax˛+bx+c=0, $\displaystyle \alpha$ and $\displaystyle \beta$ two roots of the equation, you have :

    $\displaystyle \alpha+\beta=\frac{-b}{a}$
    and
    $\displaystyle \alpha \beta=\frac{c}{a}$

    Here, a=1, b=P-10, c=-10P, $\displaystyle \beta=\alpha+4$

    And just replace ^^

    The system to solve will be :

    $\displaystyle \alpha + \alpha +4=\frac{-(P-10)}{1}=10-P$

    $\displaystyle \alpha (\alpha+4)=\frac{-10P}{1}=-10P$

    The first equation will give : $\displaystyle 2 \alpha = 6-P$

    Then replace it in the second equation to get P

    Sorry, i didn't understand the title as i should have
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  9. #9
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    okies...so i got:
    $\displaystyle 2 \alpha + 4 = 10 - P$ ----- (1)
    $\displaystyle \alpha^2 + 4 \alpha = 10P$ -----(2)
    from (2) $\displaystyle P = (- \alpha^2 - 4 \alpha )/10 $
    Is this correct?
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  10. #10
    Moo
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    It's -10P, try to see why :P

    (i was editing my post >.<)
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  11. #11
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    hahas, i was also editing my post XD
    i managed to get the answer now after trying it out

    Thanks alot!! ^_^ ^_^
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  12. #12
    Moo
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    So what are the solutions for P ? :-)
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  13. #13
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    -6 and -14 ^_^
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  14. #14
    Moo
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    Oh yeaaah
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  15. #15
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    Thanks again! =)
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