Correct option is (C) b + c + 1 = 0
Let \(\alpha\) be common root of both given equations.
\(\therefore\) \(\alpha^2+b\alpha+c=0\) ____________(1)
and \(\alpha^2+cx+b=0\) ____________(2)
Subtract equation (1) from (2), we get
\((\alpha^2+c\alpha+b)-(\alpha^2+b\alpha+c)=0\)
\(\Rightarrow\) \((c-b)\alpha+b-c=0\)
\(\Rightarrow\) \((c-b)(\alpha-1)=0\)
\(\Rightarrow\) \(\alpha-1=0\) \((\because b\neq c\Rightarrow b-c\neq0\Rightarrow c-b\neq0)\)
\(\Rightarrow\) \(\alpha\) = 1
Put \(\alpha\) = 1 in equation (1), we obtain
1+b+c = 0