Q. 6

# Show that the roots of the equation(x – a) (x – b) + (x – b) (x – c) + (x – c) (x – a) = 0 are always real and they cannot be unless a = b = c.

Answer :

(x a) (x b) + (x b) (x c) + (x c) (x a) = 0

x(x b) a(x b) + x(x c) b(x c) + x(x a) c(x a) = 0

x2 – bx ax + ab + x2 – cx bx + bc + x2 – ax cx + ac = 0

3x2 – 2x(a + b + c) + ab + bc + ac = 0

D = b2 – 4ac

D = (a + b + c)2 – 4(3)(ab + bc + ac) = 0

D = 4(a2 + b2 + c2 + 2ab + 2bc + 2ca – 3ab – 3bc – 3ca)

D = 4(a2 + b2 + c2 – ab – bc – ca)

D = 2[(a –b)2 + (b – c)2 + (c – a)2]

Which is always greater than zero so the roots are real.

Roots are equal if D = 0

i.e. (a – b)2 + (b + c)2 + (c – a)2 = 0

since sum of three perfect square is equal to zero so each of them separately equal to zero.

So, a – b = 0, b – c = 0, c – a = 0

a = b , b = c, c = a

so, a = b = c.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Relationship Between Coefficients and Zeros44 mins
Quiz on Number of Solutions of an Equation24 mins
Important Algebra Questions46 mins
Quiz on Important Algebra Questions26 mins
Interactive Quiz: Algebra Important Questions38 mins
Genius Quiz | NTSE Questions Algebra27 mins
Interactive Quiz: Nature of Roots51 mins
Know Some Interesting Proofs in Algebra42 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses