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 $\frac pq$.
 A: 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 
A: 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.
A: By Gauss's lemma It suffices to consider the domain of $x$ as the integers. If $x$ is even or odd then $ax^{2}+bx+c=x(ax+b)+c$ is odd hence not equal to $0$.
A: You can actually prove this in quite an elementary way without even knowing anything about the roots of the quadratic equation. Suppose, on the contrary, that we have rational root $\dfrac {p}{q}$. Then your equation is equivalent to $ap^2+bpq+cq^2=0$. You have that $a,b,c$ are all odd. I will denote odd as an $O$ and even as an $E$.
This breaks into four cases.
1) if $p=O$ and $q=E$ then you have $O+E+E=O=0$, which is impossible
2) if $p=E$ and $q=O$ then you have $E+E+O=O=0$, which is impossible
3) if $p=O$ and $q=O$ then you have $O+O+O=O=0$, which is impossible
4) if $p=E$ and $q=E$ then you have$E+E+E=E=0$. which could be possible ($0$ is an even number), so we treat this case below
Suppose $p=2k$ and $q=2l$ is a solution, then you have that $(p,q)$ is a solution of the equation if and only if $(k,l)$ is a solution, if $k=2e$ and $l=2f$ then you have that $(p,q)$ is a solution if and only if $(e,f)$ is a solution, proceeding in this way you can see that solution, if it exists, must be of the form $(p,q)=(g2^r,h2^s)$ and this also cannot be a solution beacuse this reduces to one of the first 3 cases.
A: Hint $\ $ By the Rational Root Test, any rational root is integral, hence it follows by
Theorem Parity Root Test $\ $ A polynomial $\rm\:f(x)\:$ with integer coefficients 
has no integer roots if its constant coefficient and coefficient sum are both odd. 
Proof $\ $ The test  verifies that $\rm\  f(0) \equiv  1\equiv f(1)\ \ (mod\ 2)\:,\ $ i.e. 
that $\rm\:f(x)\:$ has no roots modulo $2$, hence it has no integer roots. $\ $ QED
This test extends to many other rings which have a "sense of parity", i.e. an image $\cong \Bbb Z/2,\:$ for example, various algebraic number rings such as the Gaussian integers.
A: If $\frac{p}{q}$ is a root then $a\frac{p^2}{q^2}+b\frac{p}{q}+c=0\Rightarrow ap^2+bpq+cq^2=0$. Now we may assume that $p,q$ are not both even; $a,b,c$ are odd, whence contradiction ($p(ap+bq)=-cq^2$ with $p$ even $c,q$ odd or $q(bp+cq)=-ap^2$ with $q$ even $a,p$ odd or ...). 
A: 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.
A: $x= \dfrac{-b+ \sqrt{b^2-4ac}}{2a} $ or $\dfrac{-b- \sqrt{b^2-4ac}}{2a}$
If $x=\dfrac{p}{q} \implies b^2-4ac=k^2$, and $k$ is odd. ($odd-even=odd$).
Considering $a,b,c$ odd. 
$k^2 \equiv 1 \mod 8$
$b^2 \equiv 1 \mod 8$
$4ac \equiv 4 \mod 8$
$b^2-4ac=-3 \mod 8 \implies 5 \mod 8$, a contradiction.(Either of $a$ or $c$ has to be even), since $k^2 \equiv 1 \mod 8$
$\therefore$ You don't get rational roots when all are odd.
A: Here is a proof that is elementary, i.e. requires no knowledge of modular arithmetic. First we use the AC method to reduce to a monic quadratic, i.e. one with leading coefficient $= 1.$
$$\rm\begin{eqnarray} 0\: =\ f(x)\, = &&\rm\ \,  a\,x^2+\,b\,x+c\\ \rm \Rightarrow\ \ 0 = a f(x)\, = &&\rm\! (ax)^2\! + b (ax) + ac\\ \rm \Rightarrow\ \  0 =\, F(X) = &&\rm\ \ \, X^2 \,+\ \color{#C00}b\,\ X\ \,+ \color{#0A0}{ac},\quad X = ax\end{eqnarray}$$
Suppose $\rm\,f(x_i)=0\,$ for $\rm\:\color{brown}{x_i\in\Bbb Q}.\:$ Then $\rm\,F(X_i)=0\,$ for $\rm\: X_i = a\, x_i\in\Bbb Q.\:$ By the Rational Root Test, the rational roots $\rm\,X_i$ are integers. Since their product $\rm = \color{#0A0}{ac}\:$ is odd, the roots are both odd, so their sum is even. This  contradicts: by Vieta, the root sum $\rm\, = -\color{#C00}b\,$ is odd, by hypothesis. Hence $\rm\:\color{brown}{x_i \not\in \Bbb Q}.$
Remark $\ $ This yields a conceptual view of the calculations in Abhijit's answer (which, alas, is incomplete, since it does not justify the reduction from rational to integer roots).
The AC-method generalizes to higher degree polynomials (see the above-linked answer). It is intimately connected to various refinement-based views of unique factorization.
A: Recall that the square of an odd number is always equivalent to $1$ modulo $8$.
If $a,b,c$ are all odd, then
$$ b^2 - 4 a c \equiv 5 \pmod 8 $$
and thus, $b^2 - 4ac$ cannot be the square of an integer (and thus cannot be the square of a rational number). Therefore the roots are irrational.
A: 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.
enter image description here
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.
