Question

Prove that both the roots of the equation(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0 are real but they are equal only when a=b=c

Answer 1

3x² − 2x(a + b + c) + (ab + bc + ca) = 0

Let D be the discriminant

D = 4(a + b + c)² − 12(ab + bc + ca)

= 4[(a + b + c)² − 3(ab + bc + ca)]

= 2[2a² + 2b² + 2c² − 2ab − 2bc − 2ca]

= 2[(a − b)² + (b − c)² + (c − a)²]

If D = 0, then

(a − b)² + (b − c)² + (c − a)² = 0

⇒ a − b = 0; b = c; c = a

or a = b = c

Hence proved.

Answer 2

The given equation is (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0

Hence, it is proved.​