Multiple choice

If the equation $(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$ has real roots that are equal in magnitude and opposite in sign, then

  1. $a+b+c=0$
  2. $a+b+c < 0$
  3. $a+b+c > 0$

  4. None of these
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A Correct answer
Explanation

Expanding the equation: 3x^2 - 2(a+b+c)x + (ab+bc+ca) = 0. For roots to be equal in magnitude and opposite in sign, the sum of roots must be 0. Sum = 2(a+b+c)/3 = 0, so a+b+c=0.

AI explanation

Expand the terms of the given equation and simplify to get 3 times x squared minus 2 times the quantity (a plus b plus c) times x plus (ab plus bc plus ca) equals zero. The condition for a quadratic equation to have real roots that are equal in magnitude and opposite in sign is that the sum of its roots must be zero, which requires the coefficient of x to be zero. Therefore, 2 times (a plus b plus c) equals zero, which implies a plus b plus c equals zero.

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