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

pls check the question
is it 2kx - (2k-1)x + 1=0 ?
Kindly recheck the question
Aakash if it was 2kx - (2k-1)x + 1 = 0 then it on opening the bracket the equation would be x+1=0. May be the problem is somewhere else
yeah i thought about it .... it definitely lies somewhere else ...


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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)
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Do u think there is anything like (k2x)2 in maths? :)
np i guess nt an i meant that 2(k sq. x)
well only the questioner can tell us the truth ;)
Thanks for your help, but both, K and X have power of 2
And ireally it is correct question.
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.
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