Answer :
I guess the question is correct, i misinterpreted it
Well the logic is: for equal roots of quadratic equation [tex]b^2-4ac=0[/tex]
So [tex](4k-2)^2-4(k^2)(1)=0 \\16k^2-16k+4-4k^2=0 \\12k^2-16k+4=0 \\3k^2-4k+1=0 \\(k-1)(3k-1)=0 \\(k=1) or (k=1/3)[/tex]
Well the logic is: for equal roots of quadratic equation [tex]b^2-4ac=0[/tex]
So [tex](4k-2)^2-4(k^2)(1)=0 \\16k^2-16k+4-4k^2=0 \\12k^2-16k+4=0 \\3k^2-4k+1=0 \\(k-1)(3k-1)=0 \\(k=1) or (k=1/3)[/tex]
reframe the question correctly. well if it is k^2x^2 - 2(2k-1)x +1=0,then one root is
[tex]x=(2(2k-1)+ \sqrt{(4k-2)^2-4*1*k^2})/2 \\ =2k-1+ \sqrt{4k^2-4k+1-k^2} \\ =2k-1+ \sqrt{3k^2-4k+1} \\ =2k-1+ \sqrt{(k-1)(3k-1)} [/tex]
and the other root is [tex]x=2k-1- \sqrt{(k-1)(3k-1)} [/tex].
hope that helps.
[tex]x=(2(2k-1)+ \sqrt{(4k-2)^2-4*1*k^2})/2 \\ =2k-1+ \sqrt{4k^2-4k+1-k^2} \\ =2k-1+ \sqrt{3k^2-4k+1} \\ =2k-1+ \sqrt{(k-1)(3k-1)} [/tex]
and the other root is [tex]x=2k-1- \sqrt{(k-1)(3k-1)} [/tex].
hope that helps.