Correct option is (C) 1/√2, 1/√2
Given quadratic equation is
\(2x^2-2\sqrt{2}x+1=0\)
\(\Rightarrow2x^2-\sqrt{2}x-\sqrt{2}x+1=0\)
\(\Rightarrow\sqrt{2}x(\sqrt{2}x-1)-1(\sqrt{2}x-1)=0\)
\(\Rightarrow(\sqrt{2}x-1)(\sqrt{2}x-1)=0\)
\(\Rightarrow\sqrt{2}x-1=0\) \(or\;\sqrt{2}x-1=0\)
\(\Rightarrow\) \(x=\frac1{\sqrt2}\) or \(x=\frac1{\sqrt2}\)
Hence, the roots of given quadratic equation are \(\frac{1}{\sqrt{2}}\;and\;\frac{1}{\sqrt{2}}.\)
