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.

(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.

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