Prove $ax^2+bx+c=0$ has no rational roots if $a,b,c$ are odd If $a,b,c$ are odd, how can we prove that $ax^2+bx+c=0$ has no rational roots?

I was unable to proceed beyond this: Roots are $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
and rational numbers are of the form $\frac pq$.
You should think about factoring the polynomial rather than finding its roots.  If $ax^2 + bx + c$ has a rational root, then its other root must also be rational, and then it factors in some way like this:
$$ax^2 + bx + c = (Ax + B)(Cx + D) \quad A,B,C,D \in \mathbb{Z}$$
Then $a = AC$ and $c = BD$, so if both are odd, then all of $A,B,C,D$ are odd.  But then we also have $b = AD + BC$, which is the sum of odd integers, and therefore is even.
Assume the your root z is rational (= q/r where at at most one of q, r is even)

Re-write the  equation with z on the left hand size and 1/z * c/a  - b/a
on the other. Substitute q/r for z, and put the left hand side into a fraction.

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If neither q or r are even then numerators don't match 
(q != rc - qb because odd != odd + odd)
and if only one is even then the denominator doesn't match 
(r != qa because odd != even * odd and even != odd * odd)

Therefore, z is not rational.