Posers and Puzzles

Posers and Puzzles

  1. Joined
    21 Apr '05
    Moves
    54
    23 May '05 09:23
    For what values of a there are exactly two real numbers that satify the equation:

    x^3 + (a-1)x^3 + (9-a)x - 9 = 0 ?
  2. Joined
    12 Mar '03
    Moves
    37905
    23 May '05 11:01
    I assume the second term was x^2 and not x^3.

    Exactly 2 real solutions means two overlapping at a minimum or maximum, and one other. In order to have a minimum and maximum, the first derivative must have two real solutions, that means a non-negative discriminant. If my calc is correct then this is possible with a < -5.62 and a > 4.62.
  3. Standard memberPBE6
    Bananarama
    False berry
    Joined
    14 Feb '04
    Moves
    28719
    23 May '05 16:16
    By observation, x=1 is a solution of the polynomial. When you divide out (x-1), you are left with:

    x^2+a*x+9 = 0

    Solving this equation using the quadratic formula, you get:

    x = [-a +/- SQRT(a^2-36)]/2

    Since we already have one root (x=1), we find the second (repeated) root by setting the argument under the square root sign equal to zero:

    a^2 - 36 = 0
    a = +/- 6

    Neither solution is equal to 1, so we are assured we have only 2 roots. When we set a=6, the polynomial becomes:

    (x-1)(x+3)^2 = 0
    (x-1)(x^2+6x+9) = 0
    x^3+5x^2+3x-9 = 0

    And when we set a=(-6), the polynomial becomes:

    (x-1)(x-3)^2 = 0
    (x-1)(x^2-6x+9) = 0
    x^3-7x^2+15x-9 = 0

    So the solutions are a=6 or a=(-6).
  4. Joined
    12 Mar '03
    Moves
    37905
    24 May '05 07:48
    Originally posted by PBE6
    By observation, x=1 is a solution of the polynomial. When you divide out (x-1), you are left with:

    x^2+a*x+9 = 0

    Solving this equation using the quadratic formula, you get:

    x = [-a +/- SQRT(a^2-36)]/2

    Since we already have one root (x=1), we find the second (repeated) root by setting the argument under the square root sign equal to zero:

    a^2 - 3 ...[text shortened]... (x-1)(x-3)^2 = 0
    (x-1)(x^2-6x+9) = 0
    x^3-7x^2+15x-9 = 0

    So the solutions are a=6 or a=(-6).
    Nice. I forgot to check the peculiarities like dividing by x-1 or x+1, and therefor didn't expect only a few solutions.

    How about a= -10, leaving the double solution x=1 and the single one x=-9 ?
  5. Non-Sub Recs: 0
    Joined
    08 Apr '05
    Moves
    455
    24 May '05 08:331 edit
    Originally posted by Mephisto2
    I assume the second term was x^2 and not x^3.

    Exactly 2 real solutions means two overlapping at a minimum or maximum, and one other. In order to have a minimum and maximum, the first derivative must have two real solutions, that means ...[text shortened]... calc is correct then this is possible with a < -5.62 and a > 4.62.
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