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

positive

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negative

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real

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none of these

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Solution

The correct option is C real
$(x - a) (x - b) + (x - b) (x - c) + (x - c) (x - a) = 0$
$3 x^{2} - 2 (a + b + c) x + (a b + b c + c a) = 0$
$D = 4 (a + b + c)^{2} - 12 (a b + b c + c a)$
$= 4 [a^{2} + b^{2} + c^{2} - a b - b c - c a]$
$D = 2 [(a - b)^{2} + (b - c)^{2} + (c - a)^{2}]$
$D \geq 0$
Root must be real.

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Similar questions

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

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then show that the roots of the equation

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Q. For

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both the roots of the equation

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

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

The correct option is C real (x−a)(x−b)+(x−b)(x−c)+(x−c)(x−a)=0 3x2−2(a+b+c)x+(ab+bc+ca)=0 D=4(a+b+c)2−12(ab+bc+ca) =4[a2+b2+c2−ab−bc−ca] D=2[(a−b)2+(b−c)2+(c−a)2] D≥0 Root must be real.