For the equation (1/x+a)-(1/x+b)=(1/x+c), if the product of roots is zero, then sum of roots is

Question

For the equation (1/x+a)-(1/x+b)=(1/x+c), if the product of roots is zero, then sum of roots is (A) -(2 b c/b+c) (B) (2 ca / c + a ) (C) (b c/c+a) (D) (-b c/b+c)

Tardigrade Admin · Accepted Answer

(1/x+a)-(1/x+b)=(1/x+c) (b-a/x2+(b+a) x+a b)=(1/x+c) or x2+(a+b) x+a b=(b-a) x+(b-a) c or x2+2 a x+a b+c a-b c=0 Since product of the roots =0 a b+c a-b c=0 ⇒ a=(b c/b+c) Thus, sum of roots =-2 a=(-2 b c/b+c)

Q. For the equation $\frac{1}{x + a} - \frac{1}{x + b} = \frac{1}{x + c}$ , if the product of roots is zero, then sum of roots is

$- \frac{2 b c}{b + c}$

$\frac{2 c a}{c + a}$

$\frac{b c}{c + a}$

$\frac{- b c}{b + c}$

Q. For the equation x+a1​−x+b1​=x+c1​, if the product of roots is zero, then sum of roots is

−b+c2bc​

c+a2ca​

c+abc​

b+c−bc​

x+a1​−x+b1​=x+c1​ x2+(b+a)x+abb−a​=x+c1​ or x2+(a+b)x+ab=(b−a)x+(b−a)c or x2+2ax+ab+ca−bc=0 Since product of the roots =0 ab+ca−bc=0 ⇒a=b+cbc​ Thus, sum of roots =−2a=b+c−2bc​

Q. For the equation $\frac{1}{x + a} - \frac{1}{x + b} = \frac{1}{x + c}$ , if the product of roots is zero, then sum of roots is

$- \frac{2 b c}{b + c}$

$\frac{2 c a}{c + a}$

$\frac{b c}{c + a}$

$\frac{- b c}{b + c}$