Find the value of k for which the given equation has real and equal roots: x² + k(4x+ k-1) + 2 = 0.
Attachments:
Answers
Answered by
1
Answer:
X²+K(4X+K-1)+2=0
X²+4KX+K²-K+2=0
Comparing with ax²+bx+-c=0
a=1 b=4k c= k²-k+2
b²-4ac=0
4k²-4(1)(k²-k+2)=0
4k²-4k²-4k-8=0
-4k=0+8
k=8/-4
k=-2
pls mark me as brainliest
Similar questions