The sum of the roots of the equation \(\frac{1}{x+a}+\frac{1}{x+b}=\fr

1.	The sum of the roots of the equation \(\frac{1}{x+a}+\frac{1}{x+b}=\frac{1}{c}\) is zero. What is the product of the roots of the equation ? (a) – \(\frac{(a+b)}{2}\)(b) \(\frac{(a+b)}{2}\)(c) – \(\frac{(a^2+b^2)}{2}\)(d) \(\frac{(a^2+b^2)}{2}\)
Answer» (c) – \(\frac{(a^2+b^2)}{2}\) \(\frac{1}{x+a}+\frac{1}{x+b}=\frac{1}{c}\) ⇒ \(\frac{(x+b)+(x+a)}{(x+a)(x+b)}\) = \(\frac{1}{c}\) ⇒ \(\frac{2x+b+a}{x^2+(a+b)x+ab}\) = \(\frac{1}{c}\) ⇒ 2cx + (a + b)c = x² + (a + b)x + ab ⇒ x² + (a + b – 2c)x + ab – ac – bc = 0 Let α, β be the roots of this equation. Then, α + β = – (a + b – 2c) = 0 (Given) ⇒ a + b = 2c and αβ = ab – ac – bc = ab – (a + b)c = ab - (a + b) \(\frac{(a+b)}{2}\) = \(\frac{2ab-(a^2+b^2+2ab)}{2}\) = - \(\bigg(\frac{a^2+b^2}{2}\bigg).\)

The sum of the roots of the equation \(\frac{1}{x+a}+\frac{1}{x+b}=\frac{1}{c}\) is zero. What is the product of the roots of the equation ? (a) – \(\frac{(a+b)}{2}\)(b) \(\frac{(a+b)}{2}\)(c) – \(\frac{(a^2+b^2)}{2}\)(d) \(\frac{(a^2+b^2)}{2}\)

Answer»

\(\frac{1}{x+a}+\frac{1}{x+b}=\frac{1}{c}\)

⇒ \(\frac{(x+b)+(x+a)}{(x+a)(x+b)}\) = \(\frac{1}{c}\) ⇒ \(\frac{2x+b+a}{x^2+(a+b)x+ab}\) = \(\frac{1}{c}\)

⇒ 2cx + (a + b)c = x² + (a + b)x + ab

⇒ x² + (a + b – 2c)x + ab – ac – bc = 0

Let α, β be the roots of this equation. Then,

α + β = – (a + b – 2c) = 0 (Given)

⇒ a + b = 2c and αβ = ab – ac – bc = ab – (a + b)c

= ab - (a + b) \(\frac{(a+b)}{2}\)

= \(\frac{2ab-(a^2+b^2+2ab)}{2}\) = - \(\bigg(\frac{a^2+b^2}{2}\bigg).\)