Question

(b+c-2a)x²+(c+a-2b)x+(a+b-2c)=0 are

• A. Rational
• B. Non-real
• C. Irrational
• D. Equal

Answer 1

Answer: (A)

Rational

Answer 2

Let the equation be $Ax^2+Bx+C=0$. The coefficients are $A = b+c-2a$, $B = c+a-2b$, $C = a+b-2c$. Given $a,b,c \in \mathbb{Q}$, so $A,B,C \in \mathbb{Q}$. Sum of coefficients: $A+B+C = (b+c-2a)+(c+a-2b)+(a+b-2c) = 0$. Since $A+B+C=0$, $x=1$ is a root, which is rational. The discriminant is $\Delta = B^2-4AC$. Since $A+B+C=0$, $B = -(A+C)$. $\Delta = (-(A+C))^2 - 4AC = (A+C)^2 - 4AC = A^2+2AC+C^2-4AC = A^2-2AC+C^2 = (A-C)^2$. $A-C = (b+c-2a) - (a+b-2c) = 3c-3a = 3(c-a)$. So, $\Delta = (3(c-a))^2 = 9(c-a)^2$. Since $a,c \in \mathbb{Q}$, $c-a \in \mathbb{Q}$, so $\Delta$ is a square of a rational number, and $\sqrt{\Delta} = 3|c-a|$ is rational. The roots are $x = \frac{-B \pm \sqrt{\Delta}}{2A}$. Since $A, B, \sqrt{\Delta}$ are rational, the roots are rational (provided $A \neq 0$). If $A=0$, the equation is linear $Bx+C=0$. Since $A+B+C=0$, if $A=0$, then $B+C=0$, so $C=-B$. The equation becomes $Bx-B=0$, or $B(x-1)=0$. If $B \neq 0$, then $x=1$, which is rational. If $B=0$, then $C=0$, meaning $A=B=C=0$, which is a degenerate case. Thus, the roots are always rational.