Question

find the value of k for which (will be marked brainliest)

Answer 1

Refer to the attachment.


Hope this helps.

Answer 2

$kx {}^{2} - 3kx + 1 = 0$

Now for real and equal roots D = 0

$b {}^{2} - 4ac = 0$

a = k ,. b = -3k ,. c =1

$( - 3k) {}^{2} - 4(k)(1) = 0$

$9k {}^{2} - 4k = 0$

$k(9k - 4) = 0$

Hence k = 0. and k = 4/9


Thanks