Existence of rational roots in a quadratic equation 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 is not equal to 0. Suppose that both the roots of the equation are rational . Then 
a) a,b, and c are rational 
b)$c/(a-b)$ is rational 
c)$b/(c-a)$ is rational 
d)$a/(b-c)$ is rational 
My attempt - I used the discriminant method to find out the possible roots of the given equation. 
Which for me came out to be $(-2c + a-b)/(2a -2b +2c)$ and $ ( -2c + b - a)/(2c + 2a - 2b )$
Hence according to me option a is the correct answer , while the correct answer which is given is b. Please tell me where am I going wrong , or what am I missing ?
 A: Notice that your polynomial $$(a+c-b)x^2 +2cx +(b+c-a) = (a+c-b) (x - \alpha)(x - \beta) =$$
$$=(a+c-b)x^2 - (a+c-b)(\alpha + \beta) x + (a+c-b)\alpha \beta$$
where $\alpha, \beta$ are the rational roots.
Now, the coefficient of $x$ must be equal
$$2c = -(a+c-b)(\alpha + \beta)$$
and with some algebra you get
$$\frac{c}{a-b} = -\frac{\alpha + \beta}{\alpha + \beta +2} \in \Bbb{Q}$$
A: Note that, when $x=-1$, we get
$$
(a+c-b)-2c+(b+c-a)=0
$$
so $-1$ is a root for every choice of the parameters $a$, $b$ and $c$. Even if $a+c-b=0$, that would give $c/(a-b)=-1$.
This leads into checking option b. So, suppose $a+c-b\ne0$. The factorization of the polynomial is
$$
(x+1)\bigl((a+c-b)x+c-a+b\bigr)
$$
so the second root is
$$
\frac{(a-b)-c}{(a-b)+c}=\frac{1-\dfrac{c}{a-b}}{1+\dfrac{c}{a-b}}
$$
If this is to be rational, also $c/(a-b)$ is rational.
A: We know that your two roots are rational:
$$
r_1 = \frac{-2c + (a - b)}{2(a - b + c)} \in \mathbb Q \qquad\text{and}\qquad
r_2 = \frac{-2c - (a - b)}{2(a - b + c)} \in \mathbb Q
$$
Hence, their sum and difference are also rational:
$$
r_1 + r_2 = \frac{c}{a - b + c} \in \mathbb Q \qquad\text{and}\qquad
r_1 - r_2 = \frac{a - b}{a - b + c} \in \mathbb Q
$$
But then their quotient is also rational:
$$
\frac{r_1 + r_2}{r_1 - r_2} = \frac{c}{a - b} \in \mathbb Q
$$
A: The roots are
$$\frac{-2c\pm \sqrt{4c^2 - 4(c+(a-b))(c-(a-b))}}{2(a+c-b)}$$
$$=\frac{-c\pm (a-b)}{a+c-b}$$
$=\frac{a-b-c}{a-b+c}$ or $\frac{-a+b-c}{a-b+c}=-1$
For the first root,
$$=\frac{a-b+c-2c}{a-b+c}=1-\frac{2c}{a-b+c}$$
So $\frac{2c}{a-b+c}$ is rational, so its inverse is also rational
$\frac{a-b+c}{2c}=\frac{a-b}{2c}+\frac 1 2$
A: Reduce this to a monic form:
$$
x^2+\frac{2c}{a+c-b}x+\frac{b+c-a}{a+c-b}
$$
then as the roots are rational so are the sum and product of the roots, so both $\frac{2c}{a+c-b}$ and $\frac{b+c-a}{a+c-b}$ are rational and so $\frac{c}{b+c-a}$ is also rational, and you should be able to take it from there.
