Question

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

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

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

$\therefore$both the roots are real