A) \[{{b}^{2}}=8ac\]
B) \[3{{b}^{2}}+16ac=0\]
C) \[3{{b}^{2}}=16ac\]
D) \[{{b}^{2}}+3ac=0\]
Correct Answer: C
Solution :
Let the roots are a and 3a \ \[\alpha +3\alpha =\frac{-b}{a}\Rightarrow \alpha =-\frac{b}{4a}\] and \[\alpha .3\alpha =\frac{c}{a}\] Þ \[\frac{c}{a}=3.\frac{{{b}^{2}}}{16{{a}^{2}}}\] Þ \[16ac=3{{b}^{2}}\].
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