A) 1 : 2 : 3
B) 3 : 2 : 1
C) 1 : 3 : 2
D) 3 : 1 : 2
Correct Answer: A
Solution :
\[{{x}^{2}}+2x+3=0\] \[D={{2}^{2}}-4\cdot 1\cdot 3<0\] \[\Rightarrow \]Both roots complex \[\Rightarrow \]Both roots are common of \[{{x}^{2}}+2x+3=0\] \[\And \text{ }a{{x}^{2}}+bx+c=0\] \[\frac{a}{1}=\frac{b}{2}=\frac{c}{3}\]
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