A) \[-2<a<8\]
B) \[-2<a<8\]
C) \[-2<a<8\]
D) \[2<a<8\]
Correct Answer: A
Solution :
\[\because \]Roots of the equation \[{{x}^{2}}-8x+{{a}^{2}}-6a=0\]are real. \[\therefore \] \[{{B}^{2}}-4AC\ge 0\] \[\Rightarrow \] \[{{(8)}^{2}}-4(1)({{a}^{2}}-6a)\ge 0\] \[\Rightarrow \] \[64-4{{a}^{2}}+24a\ge 0\] \[\Rightarrow \] \[16-{{a}^{2}}+6a\ge 0\] \[\Rightarrow \] \[{{a}^{2}}-6a-16\le 0\] \[\Rightarrow \] \[(a+2)(a-8)\le 0\] \[\Rightarrow \] \[-2\le a\le 8\]
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