# If the roots of the equation (a — b)x2 + (b — c) x + (c — a) = 0 are equal, prove that 2a = b + c.

Since roots are equal

d=0 (1)

(a — b)x2 + (b — c) x + (c — a) = 0

d = b2 – 4ac

d = (b–c)2 – 4 (a–b) (c–a)

d = b2 + c2 – 2bc –4 [a (c – a) – b (c – a)]

d = b2 + c2 – 2bc – 4 [ac – a2 – bc + ba]

From (1), d = 0

Equation will be:

0 = b2 + c2 – 2bc – 4ac + 4a2 + 4bc – 4ba

b2 + c2 – (2a)2 2bc + 2c (–2a) + 2(–2a)b = 0

(b + c – 2a)2 = 0

(b + c – 2a) = 0

b + c = 2a

Hence proved.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Quiz | Knowing the Nature of Roots44 mins
Take a Dip Into Quadratic graphs32 mins
Foundation | Practice Important Questions for Foundation54 mins
Nature of Roots of Quadratic Equations51 mins
Getting Familiar with Nature of Roots of Quadratic Equations51 mins
Quadratic Equation: Previous Year NTSE Questions32 mins
Champ Quiz | Quadratic Equation33 mins
Understand The Concept of Quadratic Equation45 mins
Quiz | Lets Solve Imp. Qs of Quadratic Equation43 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