A) 0
B) 1
C) \[-1\]
D) None of these
Correct Answer: C
Solution :
Given, equations are\[{{x}^{2}}+ax+b=0\]and\[{{x}^{2}}+bx+a=0\] \[\Rightarrow \] \[(a-b)x+(b-a)=0\] \[\Rightarrow \] \[(a-b)(x-1)=0\] \[\Rightarrow \] \[a=b,x=1\] but \[a\ne b\] \[\therefore \]\[x=1\] Put\[x=1\]in first equation, \[1+a+b=0\] \[a+b=-1\]You need to login to perform this action.
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