If $a$, $b$, $c$ are rational and $a+b+c=0$, show that the roots of $(b+c-a)x^2 + (c+a-b)x + (a+b-c)=0$ are rational.

Question

If $a$, $b$, $c$ are rational and $a+b+c=0$, show that the roots of $(b+c-a)x^2 + (c+a-b)x + (a+b-c)=0$ are rational.

My Attempt:

Given, $$(b+c-a)x^2 + (c+a-b)x + (a+b-c)=0$$ Comparing above equation with $Ax^2+Bx+C=0$, we get: \begin{align} A&=b+c-a=-2a \\ B&=c+a-b=-2b \\ C&=a+b-c=-2c \end{align} Now, \begin{align} B^2-4AC &= (-2b)^2-4(-2a)(-2c) \\ &= 4b^2-16ac. \end{align}

Answer 1

\begin{align}4b^2-16ac &=4(b^2-4ac) \\&=4(b^2+4a(a+b))\\ &=4(b^2 +4ab+4a^2)\\ &=(2(b+2a))^2 \end{align}

Hence the roots are $$\frac{-B\pm2(b+2a)}{2A}$$

which is rational.

Note that if $A=0$, the equation is linear.

Remark: something to think about, what happens if $a=b=c=0$, are all the roots still rational?

Answer 2

Multiplying by $-1/2$, the equation can be rewritten as $$ ax^2+bx+c=ax^2+bx-(a+b)=(x-1)(ax+a+b). $$

Answer 3

Observe that $x_1=1$ is a root. If $a=0$, it is the only root. Else, the product of the roots is

$$1.x_2=\frac {a+b-c}{b+c-a} $$ $$=\frac {c}{a}\in \Bbb Q $$