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$(x - a) (x - b) + (x - c) (x - a) + (x - c) (x - b) = 0$ then its roots are

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Real

Complex

Imaginary

Rational

answer is A.

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Detailed Solution

It is given that

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

.
Simplifying the given equation,

\Rightarrow 3 x^{2} - 2 (a + b + c) x + ab + bc + ca = 0

Since the equation is in a standard quadratic form we can compare it with

a x^{2} + bx + c = 0

and substitute the discriminant formula with the values of a, b and c.
Hence,

D = 4 (a + b + c)^{2} - 12 (ab + bc + ca)

= 4 a^{2} + 4 b^{2} + 4 c^{2} + 8 ab + 8 bc + 8 ac - 12 (ab + bc + ca)

= 4 (a^{2} + b^{2} + c^{2} - ab - bc - ca)

= 2 [(a - b)^{2} + (b - c)^{2} + (c - a)^{2}]

Since, the discriminant >0 the roots are real.
Hence, the correct option is 1.

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