Exercise 3.4.7 (Rational root of $au^2 + bu + c$)

Question

Let $a,b$, and $c$ be integers with $a 
e 0$. Show that if one root of the equation $au^2 + bu + c = 0$ is rational then the other one is, and that $b^2 - 4ac$ is a perfect square, possibly $0$. Show also that if $b^2 - 4ac$ is a perfect square, possibly $0$, then the roots of the equation $au^2 + bu + c = 0$ are rational.

Richard Ganaye · Accepted Answer

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

(Definitely $0 = 0^2$ is a perfect square.)
\begin{proof} 
Let $u_1, u_2$ be the (complex) roots of the polynomial $au^2 + bu +c$, where $u$ is a variable (indeterminate), and $u_1$ is rational. Then $au^2 + b u +c = a(u-u_1)(u-u_2) = au^2 - a(u_1 +u_2) + au_1u_2$.

Therefore
$b = - a(u_1+u_2)$, so $u_2 = (-b/a) - u_1$ is rational. Moreover, since there exists a real root $u_1$, $d = b^2 - 4ac \geq 0$, and the root
$$\frac{-b + \sqrt{d}}{2a} = \lambda \qquad (\lambda = u_1 	ext{ or } \lambda = u_2)$$
is rational.

Thus $\sqrt{d} = 2 a\lambda + b\in \Q$, so that 
$$\sqrt{d} = \frac{r}{s},$$
where $r,s$ are integers such that $r\geq 0, s>0, r\wedge s = 1$. If $p$ is a prime factor of $s$, then  $p \mid d s^2 = r^2$, thus $p \mid r$. This is impossible since $r\wedge s = 1$. This contradiction shows that $s>0$ has no prime factor, so $s = 1$, and $d = r^2$ is a perfect square.

\bigskip Conversely, suppose that $d = b^2 -4ac = m^2\ (m\in \N)$ is a perfect square. Then the roots of $au^2 + bu +c = 0$ are
$$\frac{-b \pm \sqrt{d}}{2a} =\frac{-b \pm m}{2a} ,$$
which are both rational numbers.
\end{proof}

Exercise 3.4.7 (Rational root of $au^2 + bu + c$)

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Exercise 3.4.7 (Rational root of $au^2 + bu + c$)

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