Prove $ax^2+bx+c=0$ has no rational roots if $a,b,c$ are odd

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.

Consider a quadratic equation of the form $a\cdot x^2 + b\cdot x + c = 0$. The only way, it can have rational roots IFF there exist two integers $\alpha$ and $\beta$ such that

$$\alpha \cdot \beta = a\cdot c\tag1$$ $$\alpha + \beta = b\tag2$$ $$ Explanation\left\{ \begin{align} if\,\alpha\cdot \beta &= a\cdot c,\\ \frac{\alpha}{a} &= \frac{c}{\beta}\\ a\cdot x^2 + b\cdot x + c& = a\cdot x^2 + (\alpha + \beta)\cdot x + c\\ & = a\cdot x^2 + \alpha\cdot x + \beta\cdot x + c\\ & =a\cdot x(x + \frac{\alpha}{a}) + \beta\cdot (x + \frac{c}{\beta}))\\ & =a\cdot x(x + \frac{\alpha}{a}) + \beta\cdot (x + \frac{\alpha}{a}))\\ & =(x+\frac{\alpha}{a})\cdot (a\cdot x + \beta)\\ &\text {As a Quadratic equation has only two roots,}\\ &\text {there would be no other way to factorize the equation} \end{align}\right. $$ $$\text{Reason }\alpha,\beta\in\mathbb{Z}\begin{cases} \text{Given a Ring R, with two operations }\left\{⋅,+\right\},\text{ on }\mathbb{Q}\\ \text{and if } \alpha,\beta \in \mathbb{Q},\alpha\cdot \beta \in \mathbb{Z},\alpha+\beta\in\mathbb{Z}\\ \Rightarrow\alpha,\beta \in \mathbb{Z}\\ \text{ where } \mathbb{Z}\subset\mathbb{Q} \end{cases} $$ If $(a,b,c)$ are odd then $\alpha \cdot \beta$ is odd and $\alpha + \beta$ is odd, but you cannot have two integers whose product and sum are odd.

So by contradiction we prove that the equation cannot have rational roots

Suppose that $a,b,c$ are odd. $ax^2+bx+c=0$ has rational roots iff the discriminant is the square of an integer. That is, there is an integer $d$ so that $d^2=b^2-4ac$. Since $a,b,c$ are odd, $d$ must also be odd.

Note that the right hand side of $$ (b-d)(b+d)=4ac\tag{1} $$ has exactly two factors of $2$. However, since $b$ and $d$ are both odd, $2d\equiv2\pmod{4}$ and so one of $b-d$ and $b+d$ is $0\bmod{4}$ and the other is $2\bmod{4}$. Thus, the left hand side of $(1)$ has at least three factors of $2$. Contradiction.