Question

Consider the quadratic equation
$(a+c-b)x^2+2cx+(b+c-a)=0$, where $a,b,c$ are distinct real numbers and $a+c-b\ne 0$. 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)

$(a+c-b)x^2+2cx+(b+c-a)=0;a,b,c\in R$ and
distinct and $(a+c-b)\ne 0$, (given both roots $\alpha$ and $\beta$ rational).
$\Rightarrow \alpha +\beta =\frac{2c}{b-c-a}\in Q$ 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 $\frac{c}{a-b}=\frac{3}{2}\in Q,\frac{b}{c-a}=\frac{2+\sqrt{3}}{3-\sqrt{3}}\notin Q$
Now, given equation becomes $x^2+6x+(5)=0$
$\Rightarrow (x+1)(x+5)=0$
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.