If the coefficients of a quadratic equation $$ax^2+bx+c=0$$ are all odd numbers, show that the equation will not have rational solutions.
I am also not sure if I should consider $c$ as a coefficient of $x^0$, suppose if I take that $c$ is also odd,
then $$b^2-4ac $$ will be odd. But that $-b$(odd), in the quadratic formula will cancel out the oddness of $\sqrt{b^2-4ac}$, in case if it is a perfect square. If it's not a perfect square then the root is irrational.
If I take $c$ to be even, even then the same argument runs but we noticed that when we take $c$ odd, we get that when discriminant is perfect square, so that means the question asks for $c$ not to be an coefficient.
Final question: Is this right to take $c$ as one of the coefficients of the equation $ax^2+bx=c=0$?
Best Answer
If the quadratic has rational roots, it can be expressed in the form $$ ax^2+bx+c = (Ax+B)(Cx+D) $$ for integers A, B, C, and D. Expanding and matching, we see that $$ a=AC\qquad b=AD+BC\qquad c=BD $$ For $a$ to be odd, we require $A$ and $C$ to both be odd. Similarly, for $c$ to be odd, we require both $B$ and $D$ to be odd. However, if all of $A$, $B$, $C$, and $D$ are odd, then $AD+BC$ must be even, and thus $b$ must be even.
Thus, to have rational roots, all the coefficients cannot be odd at the same time.