Question

find the values of k for which the equation has real amd equal roots
$k {}^{2} {x}^{2} - 2(k - 1)x + 4 = 0$

Answer 1

k^2x^2 - 2(k-1)x + 4 =0

If it have equal roots than
b^2 - 4ac = 0
( k-1)^2 - 4(k^2)(4) =0
k^2+1-2k -16k^2 =0
15k^2 + 2k - 1 =0 ( Factorise it)
15k^2 +5k -3k -1 = 0
5k(3k+1) -1(3k+1) = 0
(5k-1) ( 3k+1) = 0

K = (1/5) , (-1/3)

Thanks
Mark brainlist if possible