A) \[a+c=b+d\]
B) \[2a+b=2c+d\]
C) \[b+2c=2a+d\]
D) \[b-2c=2a-d\]
Correct Answer: C
Solution :
Let \[f\,\,(x)={{x}^{2}}+ax+b\] and \[g\,\,(x)={{x}^{2}}+cx+d\] \[\therefore \] \[f\,\,(-\,\,2)=4-2a+b=0\] \[\Rightarrow \] \[b-2a=-\,\,4\] and \[g\,\,(-\,\,2)=4-2c+d=0\] \[\Rightarrow \] \[d-2c=-\,\,4\] \[\therefore \] \[b-2a=d-2c\] \[\Rightarrow \] \[b+2c=d+2a\]You need to login to perform this action.
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