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Math Help - Finding integer roots of quadratic

  1. #1
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    Finding integer roots of quadratic

    Hi all,

    I have a bit of a unique problem that I'm hoping someone amongst you may have some suggestions on how to obtain a short cut to the answer.

    Essentially, I have a list of quadratics that were generated from an initial quadratic equation by using some simple transformation rules.
    This family of quadratics (I like to call it a 'family' as they're related to each other because of the transformation rules), will always contain only ONE quadratic equation that has integer roots ... the remaining quadratics in the family have either complex roots or non-integer roots.

    Let me use an example to make this clearer ....

    lets start with the original quadratic say, x^2 - 4x + 4894

    Now, lets generate further members of this family by simply adding 24 to the 2nd term and subtracting 1 from the 3rd term ... so now we have the following:

    (1) x^2 - 4x + 4894 Roots: Complex
    (2) x^2 - 28x + 4893 Roots: Complex


    Essentially, we're using the following: x^2 - (4x + 24nx) + (4894 - n) to generate a family of 16 quadratics using n=0 to 15 iterations.

    n
    ---
    (0) x^2 - 4x + 4894 Roots: Complex
    (1) x^2 - 28x + 4893 Roots: Complex
    (2) x^2 - 52x + 4892 Roots: Complex
    (3) x^2 - 76x + 4891 Roots: Complex
    (4) x^2 - 100x + 4890 Roots: Complex
    (5) x^2 - 124x + 4889 Roots: Complex
    (6) x^2 - 148x + 4888 Roots: 49.75 and 98.25
    (7) x^2 - 172x + 4887 Roots: 35.91 and 136.09
    (8) x^2 - 196x + 4886 Roots: 29.31 and 166.69
    (9) x^2 - 220x + 4885 Roots: 25.06 and 194.94
    (10) x^2 - 244x + 4884 Roots: 22 and 222
    (11) x^2 - 268x + 4883 Roots: 19.66 and 248.34
    (12) x^2 - 292x + 4882 Roots: 17.80 and 274.20
    (13) x^2 - 316x + 4881 Roots: 16.29 and 299.71
    (14) x^2 - 340x + 4880 Roots: 15.02 and 324.98
    (15) x^2 - 364x + 4879 Roots: 13.94 and 350.06

    If we examine the above family of quadratics, we see that the ONLY integer roots of 22 and 222 occur at iteration 10.

    Please note that for every starting quadratic that I will be using, that quadratic WILL be guaranteed to only generate a single subsequent quadratic family member that has integer roots.

    Now what I would like, is for someone to suggest a way of QUICKLY finding that quadratic family member with the integer roots. I really don't want to have to plow my way through potentially millions of family members to find that single quadratic that has the only integer roots.

    Hopefully I've been reasonably clear in my request and explanation.

    Thanks for any assistance ...

    stevek
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  2. #2
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    Quote Originally Posted by stevek View Post
    Hi all,

    I have a bit of a unique problem that I'm hoping someone amongst you may have some suggestions on how to obtain a short cut to the answer.

    Essentially, I have a list of quadratics that were generated from an initial quadratic equation by using some simple transformation rules.
    This family of quadratics (I like to call it a 'family' as they're related to each other because of the transformation rules), will always contain only ONE quadratic equation that has integer roots ... the remaining quadratics in the family have either complex roots or non-integer roots.

    Let me use an example to make this clearer ....

    lets start with the original quadratic say, x^2 - 4x + 4894

    Now, lets generate further members of this family by simply adding 24 to the 2nd term and subtracting 1 from the 3rd term ... so now we have the following:

    (1) x^2 - 4x + 4894 Roots: Complex
    (2) x^2 - 28x + 4893 Roots: Complex


    Essentially, we're using the following: x^2 - (4x + 24nx) + (4894 - n) to generate a family of 16 quadratics using n=0 to 15 iterations.

    n
    ---
    (0) x^2 - 4x + 4894 Roots: Complex
    (1) x^2 - 28x + 4893 Roots: Complex
    (2) x^2 - 52x + 4892 Roots: Complex
    (3) x^2 - 76x + 4891 Roots: Complex
    (4) x^2 - 100x + 4890 Roots: Complex
    (5) x^2 - 124x + 4889 Roots: Complex
    (6) x^2 - 148x + 4888 Roots: 49.75 and 98.25
    (7) x^2 - 172x + 4887 Roots: 35.91 and 136.09
    (8) x^2 - 196x + 4886 Roots: 29.31 and 166.69
    (9) x^2 - 220x + 4885 Roots: 25.06 and 194.94
    (10) x^2 - 244x + 4884 Roots: 22 and 222
    (11) x^2 - 268x + 4883 Roots: 19.66 and 248.34
    (12) x^2 - 292x + 4882 Roots: 17.80 and 274.20
    (13) x^2 - 316x + 4881 Roots: 16.29 and 299.71
    (14) x^2 - 340x + 4880 Roots: 15.02 and 324.98
    (15) x^2 - 364x + 4879 Roots: 13.94 and 350.06

    If we examine the above family of quadratics, we see that the ONLY integer roots of 22 and 222 occur at iteration 10.

    Please note that for every starting quadratic that I will be using, that quadratic WILL be guaranteed to only generate a single subsequent quadratic family member that has integer roots.

    Now what I would like, is for someone to suggest a way of QUICKLY finding that quadratic family member with the integer roots. I really don't want to have to plow my way through potentially millions of family members to find that single quadratic that has the only integer roots.

    Hopefully I've been reasonably clear in my request and explanation.

    Thanks for any assistance ...

    stevek
    You are looking for the member whose discriminant is a perfect square (or for the square root of the discriminant to be an integer). The discriminant of Q(x)=ax^2+bx+c is \text{disc}(Q)=b^2-4ac.

    CB
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    You are looking for the member whose discriminant is a perfect square (or for the square root of the discriminant to be an integer). The discriminant of Q(x)=ax^2+bx+c is \text{disc}(Q)=b^2-4ac.

    CB
    Yes, I'd already seen that but many thanks anyway for responding.

    I actually need somekind of a "shortcut" to allow me to go from the original quadratic at (n=0) and find the quadratic at (n=10) without having to manually examine all the other quadratics inbetween (n=1 to n=9).

    Imagine if the quadratic with the integer roots was at n=2,000,000 !!!
    It'd take forever to search all the quadratics for n < 2,000,000.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by stevek View Post
    Yes, I'd already seen that but many thanks anyway for responding.

    I actually need somekind of a "shortcut" to allow me to go from the original quadratic at (n=0) and find the quadratic at (n=10) without having to manually examine all the other quadratics inbetween (n=1 to n=9).

    Imagine if the quadratic with the integer roots was at n=2,000,000 !!!
    It'd take forever to search all the quadratics for n < 2,000,000.
    Write the discriminant in terms of the coeficients of the n-th quadratic and find the value of n that gives a perfect square.

    CB
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  5. #5
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    Ottawa, Canada
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    Well, as far as I can see, something has to be added for clarification;
    like a definition of "family" or something...
    In your example, n=10 works, BUT also n=67:
    X^2 - 1612x + 4827 : roots 3 and 1609

    Is there a "purpose" for this?

    Starting point: ax^2 + bx + c = 0
    a,b and c each increase or decrease n times, by a constant (possibly 0)
    Let u = change to a, v = change to b, w = change to c
    Then:
    (a + nu)x^2 + (b + nv)x + c + nw = 0

    A = a + nu, B = b + nv, C = c + nw
    Ax^2 + Bx + C = 0
    So:
    sqrt(B^2 - 4AC) = integer ?

    I see no unique solutions or quick ways...but it's Sunday
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