Correct option is (C) x2 – 6x + 4 = 0
If \(3+\sqrt{5}\) is one root of quadratic equation.
Then \(3-\sqrt{5}\) is other root of that quadratic equation.
\(\therefore\) Required quadratic equation is
\((x-(3+\sqrt5))(x-(3-\sqrt5))=0\)
\(\Rightarrow((x-3)-\sqrt5)((x-3)+\sqrt5)=0\)
\(\Rightarrow(x-3)^2-(\sqrt5)^2=0\) \((\because(a-b)(a+b)=a^2-b^2)\)
\(\Rightarrow x^2-6x+9-5=0\) \((\because(a-b)^2=a^2-2ab+b^2)\)
\(\Rightarrow x^2-6x+4=0\)
