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

### 2 Answers

- 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.