Both the roots of the given equation (x - a)( | vedclass

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(C) The given equation is $(x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0$. Expanding the terms,we get: $(x^2 - (a+b)x + ab) + (x^2 - (b+c)x + b

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(C) The given equation is $(x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0$. Expanding the terms,we get: $(x^2 - (a+b)x + ab) + (x^2 - (b+c)x + bc) + (x^2 - (c+a)x + ca) = 0$. Combining like terms,we obtain the quadratic equation: $3x^2 - 2(a + b + c)x + (ab + bc + ca) = 0$. For a quadratic equation $Ax^2 + Bx + C = 0$,the discriminant is $D = B^2 - 4AC$. Here,$A = 3$,$B = -2(a + b + c)$,and $C = (ab + bc + ca)$. $D = [-2(a + b + c)]^2 - 4(3)(ab + bc + ca)$ $D = 4(a^2 + b^2 + c^2 + 2ab + 2bc + 2ca) - 12(ab + bc + ca)$ $D = 4(a^2 + b^2 + c^2 - ab - bc - ca)$ $D = 2[2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca]$ $D = 2[(a-b)^2 + (b-c)^2 + (c-a)^2]$. Since the sum of squares of real numbers is always non-negative,$D \ge 0$. Therefore,the roots are always real.

Both the roots of the given equation $(x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0$ are always

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