Multiple choice

If the sum of the two roots of the equation $\displaystyle \frac {1}{x+a}+\frac {1}{x+b}=\frac {1}{c}$ is zero, then the product of the two roots is

  1. $0$
  2. $\displaystyle \frac {a^{2}+b^{2}}{2}$
  3. $\displaystyle \frac {a+b}{2}$
  4. $\displaystyle -\frac {(a^{2}+b^{2})}{2}$
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D Correct answer
Explanation

1/(x+a) + 1/(x+b) = 1/c => (2x + a + b) / (x^2 + (a+b)x + ab) = 1/c. 2cx + c(a+b) = x^2 + (a+b)x + ab. x^2 + (a+b-2c)x + (ab - c(a+b)) = 0. Sum of roots = -(a+b-2c) = 0 => 2c = a+b. Product of roots = ab - c(a+b) = ab - ((a+b)/2)(a+b) = ab - (a^2 + 2ab + b^2)/2 = (2ab - a^2 - 2ab - b^2)/2 = -(a^2 + b^2)/2.

AI explanation

Taking the common denominator simplifies the equation to (2c + a + b)x^2 + (ab + 2ac + 2bc)x + (abc + c^2(a + b)) = 0. Since the sum of the roots is zero, the coefficient of x must be zero, giving ab + 2c(a + b) = 0, or c = -ab / 2(a + b). The product of the roots is the constant term divided by the leading coefficient, written as (abc + c^2(a + b)) / (2c + a + b). Substituting c = -ab / 2(a + b) into this product expression and simplifying using the identity (a + b)^2 = a^2 + 2ab + b^2 gives the final product as -(a^2 + b^2) / 2.

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