Question

Prove that both the roots of the equation
(x - a)(x - b) + (x - 1)(x - c) + (x - c)(x - a) = 0
are real but they are equal only when a = b = c.​

Answer 1

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

By solving the equation, we get it as

3x²-2x(a+b+c 1+b+c)+(ab+bc+ca) = 0 The discriminant D=b²-4ac

This equation is in the form of ax² +bx+c=0 Here a=3,b=-2(a+b+c) and c = ab+bc+ca

D=(-2a+b+c)²-4(3)(ab+bc+ca)

4(a+b+c)²-12 (ab+bc+ca)

⇒4[(a+b+c)²³ −3(ab+bc+ca)]

-> 4[a²+ b³ + c²-ab-bc-ca]

⇒ 2[2a² +26² +2c²-2ab-2bc-2ca]

2[(a−b)²+(b-c)²+(c-a)²] Here clearly 0≥0, if 0=0 then (a-b)²+(b-c)²+(c-a)² = 0 a-b=0,b-c=0,c-α=0 Hence a = b = c

Hence, it is proved.

Answer 2

Answer:

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