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Solving for a 'base' (eg binary) in a quadratic equation

  1. Sep 21, 2010 #1
    1. The problem statement, all variables and given/known data
    The solution of the quadratic equation x2 - 11x + 22 = 0 are x = 3 or x = 6. What is the base of the numbers?


    2. Relevant equations
    Knowledge of how to convert from a generic base to decimal?


    3. The attempt at a solution

    I tried to just place r in where I would have a value of s*r1

    (x-3)(x-6) = x2 - 11x + 22

    x2 - 6x - 3x + [1*r + 8] = x2 - [x( r + 1)] + 2r + 1

    r + 8 - 9x = x + 2r - xr + 1

    7 - 10x = r(1-x)

    r = (7-10x)/(1-x)

    when I try to put either of the values of x in I get r as either 10.6 or 11.5

    Please help me.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 21, 2010 #2

    Mark44

    Staff: Mentor

    The base should be an integer.

    Since 3 and 6 are roots of the equation, it's safe to assume that the base is at least 6.
    Also, since 3 and 6 are roots, x - 3 and x - 6 are factors of the quadratic.

    On the one hand you have (x - 3)(x - 6) = x2 - 9x + 18 (in base-10).
    On the other hand, you have x2 - 11x + 22 (in unknown base).

    Comparing the coefficients of the first expression with the second, you must have
    110 = 1b
    -910 = -11b
    1810 = 22b

    What does b need to be so that all three equations are true statements?
    Note that d1d2 in base b = d1 * b + d2 in base 10.
     
  4. Sep 21, 2010 #3
    It's base 8 right?
     
  5. Sep 21, 2010 #4

    Mark44

    Staff: Mentor

    Right. Notice that 118 means 1*8 + 1*1 = 910, and 228 means 2*8 + 2*1 = 1810.
     
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