A) \[0\]
B) \[\frac{2ab}{b+c}\]
C) \[\frac{2ac}{b+c}\]
D) \[-\frac{2ac}{b+c}\]
Correct Answer: D
Solution :
Given equation can be rewritten as \[\frac{b-a}{{{x}^{2}}+(a+b)x+ab}=\frac{1}{x+c}\] \[\Rightarrow \] \[{{x}^{2}}+(a+b)x+ab=(b-a)x+(b-a)c\] \[\Rightarrow \] \[{{x}^{2}}+2ax+ab+ca-bc=0\] Since, the product of roots is zero. \[\therefore \] \[ab+ca-bc=0\] \[\Rightarrow \] \[a(b+c)=bc\] \[\Rightarrow \] \[a=\frac{bc}{b+c}\] \[\therefore \] Sum of roots \[=-2a=\frac{-2bc}{b+c}\]
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