Question

find the value of k which equation has
real & equal roots
k² x² - 4 (k-1) x+1=0​

Answer 1

For equation to have real and equal roots, Discriminant needs to be equal to 0 i.e D=0

$f(x) =k² x² - 4 (k-1) x+1=0 \\\\ D=b^2-4ac ~needs~to~be~0\\\\ D= (4k-4)^2-4k^2=0 \\\\ 16k^2+16-32k-4k^2 = 12k^2-32k+16$

By solving by quadratic formula, k= -4/3 ± 2/3

And, Pls check the question if it is correct.