A) \[r\]
B) \[2r\]
C) \[{{r}^{2}}\]
D) \[\frac{1}{r}\]
E) \[\frac{2}{r}\]
Correct Answer: B
Solution :
Given, \[\frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r}\] \[\Rightarrow \] \[r(2x+p+q)=[{{x}^{2}}+(p+q)x+pq]\] \[\Rightarrow \] \[{{x}^{2}}+(p+q-2r)x+pq-r(p+q)=0\] As we know, if roots are equal in magnitude but opposite in sign, then coefficient of\[x\]will be zero. \[\therefore \] \[p+q-2r=0\Rightarrow p+q=2r\]You need to login to perform this action.
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