Question

Both roots of the equation $(x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)=0$ are always :

• A. Positive
• B. None
• C. Real
• D. Negative

Answer 1

The correct option is C

Real

The equation can be written as

$3x^2-2(a+b+c)x+(bc+ca+ab)=0$

$\therefore B^2-4AC=4(a+b+c{)}^2-12(bc+ca+ab)$

$=2\left[\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]\ge 0$

$\therefore$ Both the roots are real.