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Math Help - Equivalence problem

  1. #1
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    Equivalence problem

    I have to find with what value ''m'' those two equations are equivalent?

    x^2 + m^2x + (6-5m)x = 0

    and x^2 +2(m-3)x + m^2 -7m +12 = 0

    I found that with m=3 ... I did this: x^2 + m^2x + (6-5m)x = x^2 +2(m-3)x + m^2 -7m +12

    Not sure if this is the right way?
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  2. #2
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    Quote Originally Posted by Ellla View Post
    I have to find with what value ''m'' those two equations are equivalent?

    x^2 + m^2x + (6-5m)x = 0

    and x^2 +2(m-3)x + m^2 -7m +12 = 0

    I found that with m=3 ... I did this: x^2 + m^2x + (6-5m)x = x^2 +2(m-3)x + m^2 -7m +12

    Not sure if this is the right way?
    What if m=4 ?

    You should write something like

    x^2+\left(m^2-5m+6\right)x+0=0

    x^2+2(m-3)x+\left(m^2-7m+12)=0

    and comparing, we solve for

    m^2-7m+12=0
    Last edited by Archie Meade; May 29th 2011 at 01:46 PM.
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  3. #3
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    Quote Originally Posted by Archie Meade View Post
    What if m=4 ?

    You should write something like

    x^2+\left(m^2-5m+6\right)x+0=0

    x^2+2(m-3)x+\left(m^2-7m+12)=0

    and comparing, we solve for

    m^2-7m+12=0
    So we get m = 2,3 and 4. If I got it right.
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  4. #4
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    m=2 doesn't work.

    m^2-5m+6=2(m-3)

    as the multipliers of "x" must be equal.

    m^2-5m+6=2m-6

    m^2-5m-2m+6+6=2m-2m-6+6

    m^2-7m+12=0

    which is the exact same as the two constants being equal
    (the constant in one equation is zero).

    Then to find the 2 solutions for "m",
    find the factors of 12 that sum to -7
    in order to factor the quadratic in "m".
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