Correct option is (B) 3b2 = 16ac
Let \(\alpha\;\&\;3\alpha\) are roots of the quadratic equation \(ax^2+bx+c=0.\)
\(\therefore\) Sum of roots \(=\frac{-b}a\)
\(\Rightarrow\) \(\alpha+3\alpha\) \(=\frac{-b}a\)
\(\Rightarrow\) \(\alpha\) \(=\frac{-b}{4a}\) _______________(1)
Product of roots \(=\frac ca\)
\(\Rightarrow\) \(\alpha.3\alpha\) \(=\frac ca\)
\(\Rightarrow\) \(3\alpha^2\) \(=\frac ca\)
\(\Rightarrow\) \(3(\frac{-b}{4a})^2\) \(=\frac ca\) (From (1))
\(\Rightarrow\) \(\frac{3b^2}{16a^2}\) \(=\frac ca\)
\(\Rightarrow\) \(3b^2=\frac{16a^2c}a=16ac\)