A) 8
B) \[-8\]
C) 16
D) \[-16\]
Correct Answer: C
Solution :
Since \[x=2\]is a root of the equation \[{{x}^{2}}+bx+12=0\] \[\Rightarrow \] \[{{(2)}^{2}}+b(2)+12=0\] \[\Rightarrow \] \[2b=-16\] \[\Rightarrow \] \[b=-8\] Then, the equation \[{{x}^{2}}+bx+q\] becomes \[{{x}^{2}}-8x+q=0\] (ii) Since (1) has equal roots \[\Rightarrow \] \[{{b}^{2}}-4ac=0\] \[\Rightarrow \] \[{{(-8)}^{2}}-4(1)q=0\] \[\Rightarrow \] \[q=16\]You need to login to perform this action.
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