# 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 ?

• You did not write an equation, because I don't see the $=$. Anyway, if your equation is $(a+c-b)x^2 +2cx+b+c-a =0$, then your solutions are both wrong. Because the roots are $-1$ (this is easy to check) and $(a-b-c)/(a-b+c)$. Are you sure you copied the text of the problem correctly ? – Giovanni Resta Mar 24 '16 at 10:12

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 $$\frac{c}{a-b} = -\frac{\alpha + \beta}{\alpha + \beta +2} \in \Bbb{Q}$$

• Notice that you have information only about $a-b$, without any information about $a,b$. I mean, if you call $a= \pi$, $b= 1+ \pi$, their difference is rational, but these are not rational. – Crostul Mar 24 '16 at 10:14

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.

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: $$\frac{r_1 + r_2}{r_1 - r_2} = \frac{c}{a - b} \in \mathbb Q$$

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

$\frac{a-b+c}{2c}=\frac{a-b}{2c}+\frac 1 2$

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.