Question

find
has
the value of k for which the
equation x² + k (2x+k-1) +2
real and equal noots​

Answer 1

Step-by-step explanation:

+2/-2

Hope it will be helpful

Answer 2

$\huge\color{red}\boxed{AnSwEr}$

if roots are real and equal then D = 0

by simplification

x² + 2kx +k² -k +2

$where \: \\ \\ \large\color{brown}\boxed{D={b}^{2}-4ac}$

where b = 2k c = k² -k +2 and a = 1

by putting value in formula

$D = {2k}^{2} - 4(1)( {k}^{2} - k + 2) = 0 \\ \\ D = {4k}^{2} - {4k}^{2} - 4k + 8 = 0 \\ \\ - 4k - 8 = 0 \\ \\ - 4k = 8 \\ \\ k = \frac{ + 8}{ - 4} \\ \\ k = - 2$

hope it helps you

be brainly