If the quadratic equation $ax^{2}+bx+a^{2}+b^{2}+c^{2}-ab-bc-ca=0$, where $a, b, c$ are distinct real numbers, has imaginary roots, then
- $2\left ( a-b \right )+\left ( a-b \right )^{2}+\left ( b-c \right )^{2}+\left ( c-a \right )^{2}> 0$
- $2\left ( a-b \right )+\left ( a-b \right )^{2}+\left ( b-c \right )^{2}+\left ( c-a \right )^{2}< 0$
- $2\left ( a-b \right )+\left ( a-b \right )^{2}+\left ( b-c \right )^{2}+\left ( c-a \right )^{2}= 0$
-
none of these
The quadratic equation has imaginary roots if the discriminant D < 0. D = b^2 - 4a(a^2+b^2+c^2-ab-bc-ca). This simplifies to a condition related to the sum of squares, which is always positive for distinct real numbers, leading to the conclusion in option A.
For the quadratic to have imaginary roots, the discriminant b squared minus 4a times the constant term must be strictly negative. Multiplying by negative 3 reverses the inequality to 4a times (ab plus bc plus ca) is greater than 3b times (a squared plus b squared plus c squared). This expression factors perfectly into 2ab plus 2bc plus 2ca minus a squared minus b squared minus c squared is greater than 0. Adding 2a minus 2b to both sides gives 2 times (a minus b) plus (a minus b) squared plus (b minus c) squared plus (c minus a) squared is greater than 0.