TMUA practice: Polynomials, problem 3

Suppose, for contradiction, that $\;ax^{2} + bx + c\;$ has a rational root $\;p/q\;$ in lowest terms (so $\gcd(p, q) = 1$).

Substituting and clearing denominators: $\;a p^{2} + b p q + c q^{2} = 0$.

Analyse parities of $p, q$:

• Both $p$ and $q$ odd. Then $a p^{2}$, $b p q$, $c q^{2}$ are all (odd)(odd) = odd. Sum of three odds = odd. But $0$ is even — contradiction.
• $p$ even, $q$ odd. Then $a p^{2}$ is even, $b p q$ is even, $c q^{2}$ is odd (since $c$ odd and $q^{2}$ odd). Sum: $\text{even} + \text{even} + \text{odd} = \text{odd}$. Again odd, contradiction.
• $p$ odd, $q$ even. Symmetric: $a p^{2}$ is odd, $b p q$ is even, $c q^{2}$ is even. Sum: odd. Contradiction.
• Both $p$ and $q$ even is excluded by $\gcd(p, q) = 1$.

In every case the sum is odd, hence non-zero. So there is no rational root. The number of rational roots must be $0$. This is option A.

Why the other options fail.

• B, $1$: would require exactly one rational root, but every case yields zero.
• C, $2$: as above.
• D, ‘cannot be determined’: the answer is determined by the parity argument.

The lesson: parity arguments are powerful when coefficients are restricted to all odd. The key step is checking the three parity cases of $(p, q)$ (excluding both even by coprimality).