A) \[{{a}^{2}}-9a+18=0\]
B) \[{{a}^{2}}-6a-18=0\]
C) \[{{a}^{2}}-9a+12=0\]
D) \[{{a}^{2}}-6a-12=0\]
Correct Answer: A
Solution :
\[\left( \frac{-a}{2} \right)\times \frac{1}{3}=1\,\,\Rightarrow \,a=6\] and \[\frac{1}{3}\times (-a)=-1\,\Rightarrow \,a=3\] \[\therefore \,\,{{a}^{2}}-9a+18=0\]
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