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 + bc)

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Real

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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 get $3x^2 - 2(a+b+c)x + (ab + bc + ca) = 0$.For a quadratic equation $Ax^2 + Bx + C = 0$,the discriminant is $D = B^2 - 4AC$.Here,$D = [-2(a+b+c)]^2 - 4(3)(ab + bc + ca) = 4(a+b+c)^2 - 12(ab + bc + ca)$.$D = 4[a^2 + b^2 + c^2 + 2ab + 2bc + 2ca - 3ab - 3bc - 3ca] = 4[a^2 + b^2 + c^2 - ab - bc - ca]$.This can be written as $D = 2[(a-b)^2 + (b-c)^2 + (c-a)^2]$.Since the sum of squares 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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