Question

If a+b+c=0 and a, b, care rational, prove that the roots of the equation (b+c-a)$x^{2}$ + (c+a-b)x + (a+b-c)=0 are rational.

Answer 1

Putting x=1, we see that the equation is satisfied. So 1 is a root of the equation. This is verification. In fact, you can find the conditions when 1 will be a root of the general quadratic equation Ax2+Bx+C=0, by, x=−B±B2−4AC√2A=1, B2−4AC=(2A+B)2, B2−4AC=4A2+B2+4AB, 4A(A+B+C)=0, A+B+C=0, where the last equality comes from the fact that in a quadratic equation, A is non-zero. So 1 is a root if and only if the sum of the coefficients is 0. (Again this can be verified by putting x=1 in the general equation directly.)