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