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

Correct answer: D. ( (-2 b c)/(b+c) ) ( (1)/(x+a)-(1)/(x+b)=(1)/(x+c) ) or x²+(a+b) x+a b=(b-a) x+(b-a) c or ( x^ 2+2 ax+ab+ca-bc= 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) )

BITSAT 2009

Exam: BITSAT
Year: 2009
Subject: Mathematics
Chapter: Complex Number

BITSAT 2009 · Mathematics · Complex Number

For the equation ( (1)/(x+a)-(1)/(x+b)=(1)/(x+c) ), if the product...

For the equation (

\frac{1}{x+a}-\frac{1}{x+b}=\frac{1}{x+c}

), if the product of roots is zero, then the sum of roots is

Options

A0
B( $\frac{2 \mathrm{ab}}{\mathrm{b}+\mathrm{c}}$ )
C( $\frac{2 b c}{b+c}$ )
D( $\frac{-2 b c}{b+c}$ )

Correct answer

D. ( $\frac{-2 b c}{b+c}$ )

Step-by-step solution

(

\frac{1}{x+a}-\frac{1}{x+b}=\frac{1}{x+c}

) or

x²+(a+b) x+a b=(b-a) x+(b-a) c

or (

\mathrm{x}^

2+2

\mathrm{ax}+\mathrm{ab}+\mathrm{ca}-\mathrm{bc}=

0) Since product of the roots

=0

(a b+c a-b c=0 ; a=

\frac{b c}{b+c}

) Thus sum of roots (=-2 a=

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

)

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