<s>[INST] 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 ?
 [/INST] 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
$\frac{a-b+c}{2c}=\frac{a-b}{2c}+\frac 1 2$
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