Question

Consider the quadratic equation
$(a+c-b) x^{2}+2 c x+(b+c-a)=0$, where $a, b, c$ are distinct real numbers and $a+c-b \neq 0 $. Suppose that both the roots of the equation are rational, then

• A. $a, b$ and $c$ are rational
• B. $\frac{c}{(a-b)}$ is rational
• C. $\frac{b}{(c-a)}$ is rational
• D. None of these

Answer 1

Answer: (B)

$(a+c-b) x^{2}+2 c x+(b+c-a)=0 ; a, b, c \in ℝ$ and
distinct and $(a+c-b) \neq 0$, (given both roots $\alpha$ and $\beta$ rational).
$\Rightarrow \alpha+\beta=\frac{2 c}{b-c-a} \in ℚ $ and $ \alpha \beta=\frac{b+c-a}{a+c-b} \in ℚ $
$\Rightarrow \frac{2}{\frac{b-a}{c}-1} \in ℚ$
$ \Rightarrow \frac{b-a}{c} \in ℚ$
$ \Rightarrow \frac{c}{a-b} \in ℚ$
Now, consider the following counter example
Let $c=3, a=\sqrt{3}, b=2+\sqrt{3}$
Clearly, $D=4(2)^{2}=16$ i.e., perfect square of rational
and $\frac{c}{a-b}=\frac{3}{2} \inℚ, \frac{b}{c-a}=\frac{2+\sqrt{3}}{3-\sqrt{3}} \notin ℚ$
Now, given equation becomes $x^{2}+6 x+(5)=0$
$\Rightarrow (x+1)(x+5)=0$
Clearly, both roots are rational.
Thus if both roots are rational, then it is not necessary
that $a, b, c$ are rational and $\frac{b}{c-a}$ is rational.