Suppose P(x) is a quadratic whose coefficients are all odd integers. Prove that P(x)=0 has no rational roots.?
- nbsaleLv 61 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.
- PuzzlingLv 71 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.