Question

If the roots of the equation \( (x-a)(x-b) + (x-b)(x-c) + (x-c)(x-a) = 0 \) are equal, then \( a^2 + b^2 + c^2 = \)

• A. \( a+b+c \)
• B. \( 2a+b+c \)
• C. \( 3abc \)
• D. \( ab+bc+ca \)
• E. \( abc \)

Hint:

Equal roots ⇒ discriminant zero. Always expand fully before applying condition.

Answer 1

The Correct Option is D

Concept: Equal roots imply discriminant = 0.

Step 1: Expand expression
\[ (x-a)(x-b) = x^2 -(a+b)x + ab \] Similarly sum: \[ = 3x^2 -2(a+b+c)x + (ab+bc+ca) \]

Step 2: Equation
\[ 3x^2 -2(a+b+c)x + (ab+bc+ca)=0 \]

Step 3: Equal roots condition
\[ D = b^2 -4ac = 0 \] \[ [-2(a+b+c)]^2 - 4(3)(ab+bc+ca)=0 \]

Step 4: Simplify
\[ 4(a+b+c)^2 = 12(ab+bc+ca) \] \[ (a+b+c)^2 = 3(ab+bc+ca) \]

Step 5: Expand
\[ a^2+b^2+c^2 +2(ab+bc+ca) = 3(ab+bc+ca) \] \[ a^2+b^2+c^2 = ab+bc+ca \] \[ \boxed{ab+bc+ca} \]