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Math Help - collector's choice

  1. #1
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    collector's choice

    I collect old school books. They have the advantage that you don't need calculators or computers to solve the problems published in those books.

    In an arithmetic and algebra exercise book from 1883 I found a way to solve this equation exactly:

    x^4+2x^3-13x^2+2x+1=0

    Give it a try.


    EB
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by earboth View Post
    I collect old school books. They have the advantage that you don't need calculators or computers to solve the problems published in those books.

    In an arithmetic and algebra exercise book from 1883 I found a way to solve this equation exactly:

    x^4+2x^3-13x^2+2x+1=0

    Give it a try.


    EB
    Perhaps this isn't what you had in mind, but note that the leading coefficient and constant terms are 1. So if this factors it must factor as:
    (x^2 + ax + 1)(x^2 + bx + 1) = x^4 + (a+b)x^3 + (ab + 2)x^2 + (a+b)x + 1

    So
    a + b = 2
    and
    ab + 2 = -13

    So b = 2 - a

    a(2 - a) = -15

    a^2 - 2a - 15 = 0

    (a - 5)(a + 3) = 0

    So a = -3 or a = 5.

    The a = 5, b = -3 solution works. (So does a = -3, b = 5)

    So
    x^4+2x^3-13x^2+2x+1 = (x^2 + 5x + 1)(x^2 - 3x + 1) = 0

    which we can now solve by setting the (irreducible) quadratic factors to 0.

    -Dan
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  3. #3
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    Quote Originally Posted by topsquark View Post
    Perhaps this isn't what you had in mind...

    -Dan
    Hello topsquark,

    Thanks for your reply!

    first: It wasn't me who "invented" this method.

    second: The following method only works if:
    - the coefficients are arranged(?) symmetrically in the equation and
    - x \neq 0

    Then you can divide the equation by x:

    x^4+2x^3-13x+2x+1=0 Divide by x
    x^2+2x-13+\frac{2}{x}+\frac{1}{x^2}=0 Re-arrange:
    x^2+\frac{1}{x^2}+2x+\frac{2}{x}-13=0 Factor:
    x^2+\frac{1}{x^2}+2 \left( x+\frac{1}{x} \right)-13=0

    Now set z = x+\frac{1}{x}. Then

    z^2 = x^2+2+ \frac{1}{x^2}. That means

    z^2 -2= x^2+ \frac{1}{x^2}

    And now you can use substitution. The equation becomes:

    z^2-2+2 \cdot z-13=0 which will give z = -5 or z = 3.

    Re-substituate to solve for x

    The actual solution for x doesn't matter. I only wanted to show a very special method to solve those equations.

    EB
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by earboth View Post
    Hello topsquark,

    Thanks for your reply!

    first: It wasn't me who "invented" this method.

    second: The following method only works if:
    - the coefficients are arranged(?) symmetrically in the equation and
    - x \neq 0

    Then you can divide the equation by x:

    x^4+2x^3-13x+2x+1=0 Divide by x
    x^2+2x-13+\frac{2}{x}+\frac{1}{x^2}=0 Re-arrange:
    x^2+\frac{1}{x^2}+2x+\frac{2}{x}-13=0 Factor:
    x^2+\frac{1}{x^2}+2 \left( x+\frac{1}{x} \right)-13=0

    Now set z = x+\frac{1}{x}. Then

    z^2 = x^2+2+ \frac{1}{x^2}. That means

    z^2 -2= x^2+ \frac{1}{x^2}

    And now you can use substitution. The equation becomes:

    z^2-2+2 \cdot z-13=0 which will give z = -5 or z = 3.

    Re-substituate to solve for x

    The actual solution for x doesn't matter. I only wanted to show a very special method to solve those equations.

    EB
    Ah yes! I've seen that kind of trick pulled occasionally. I've got an older advanced algebra (HS level) algebra book that works with that kind of stuff. I had totally forgotten about that method.

    -Dan
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