Question

for what value of k the roots are real and equal for the equation k2x2 -2(2k-1)x + 1=0

Answer 1

I guess the question is correct, i misinterpreted it
Well the logic is: for equal roots of quadratic equation $b^2-4ac=0$ 
So $(4k-2)^2-4(k^2)(1)=0 \\16k^2-16k+4-4k^2=0 \\12k^2-16k+4=0 \\3k^2-4k+1=0 \\(k-1)(3k-1)=0 \\(k=1) or (k=1/3)$

Answer 2

reframe the question correctly. well if it is k^2x^2 - 2(2k-1)x +1=0,then one root is
$x=(2(2k-1)+ \sqrt{(4k-2)^2-4*1*k^2})/2 \\ =2k-1+ \sqrt{4k^2-4k+1-k^2} \\ =2k-1+ \sqrt{3k^2-4k+1} \\ =2k-1+ \sqrt{(k-1)(3k-1)}$
and the other root is $x=2k-1- \sqrt{(k-1)(3k-1)}$.
hope that helps.