Correct option is (C) 4
Let \(\alpha\;and\;\beta\) are roots of equation \(3x^2+(2k+1)\,x-(k+5)=0.\)
\(\therefore\) Sum of roots \(=\frac{-(2k+1)}3\)
\(\Rightarrow\) \(\alpha+\beta\) \(=\frac{-(2k+1)}3\) _____________(1)
And product of roots \(=\frac{-(k+5)}3\)
\(\Rightarrow\) \(\alpha\beta\) \(=\frac{-(k+5)}3\) _____________(2)
Given that sum of roots = Product of roots
\(\Rightarrow\) \(\frac{-(2k+1)}3\) \(=\frac{-(k+5)}3\)
\(\Rightarrow\) 2k+1 = k+5
\(\Rightarrow\) 2k - k = 5 - 1 = 4
\(\Rightarrow\) k = 4