Suppose P(x) is a quadratic whose coefficients are all odd integers. Prove that P(x)=0 has no rational roots.?

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  • nbsale
    Lv 6
    1 month ago

    Let P(x) = ax^2 + bx + c, a,b,c all odd.

    To have rational roots, the discriminant, b^2 - 4ac must be a perfect square, call it k^2.

    b^2 is odd, and 4ac is even, so k^2 must also be odd.

    b^2 - 4ac = k^2

    b^2 - k^2 = 4ac is a multiple of 4, but not a multiple of 8 since a and c are odd.

    Consider b^2 - k^2 = (b-k)(b+k)

    Since b and k are odd, they must = 1 or 3 mod 4. Look at the 4 cases:

    b=k=1 mod 4. Then b-k = 0 mod 4, and b+k = 2 mod 4. That makes their product a multiple of 8.

    b=1, k=3 mod 4. Then b-k = 2 mod 4, and b+k = 0 mod 4. Again the product is a multiple of 8.

    b=3, k=1 mod 4. Then b-k = 2 mod 4, and b+k = 0 mod 4. Same as prior case.

    b=3, k=3 mod 4. Then b-k = 0 mod 4, and b+k = 2 mod 4. Again the product is a multiple of 8.

    Therefore we have a contradiction that b^2 - k^2 can't be a multiple of 8, but must be, so the assumption that b^2 - 4ac is a perfect square is wrong, and P can't have rational roots.

  • 1 month ago

    The general form of a quadratic is ax² + bx + c.

    Let's assume that a, b and c are all odd integers.

    In order to have rational roots, you must be able to factor that into a pair of binomials with integer coefficients.

    ax² + bx + c = (Ax + B)(Cx + D)

    Expanding that we have:

    ACx² + (AD + BC)x + BD

    Matching coefficients, we have:

    a = AC

    b = AD + BC

    c = BD

    The only way a can be odd is if A and C are both odd.

    The only way c can be odd is if B and D are both odd.

    So basically this will only work if A, B, C and D are odd.

    However, AD is odd and BC is odd but their sum will be *even*. Now we have a contradiction because b was assumed to be odd, but in order to factor it, b must be even. So it is not possible to factor it and hence the quadratic has no rational roots.

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