Question

Find value of k for which (k+4)x^2+(k+1)x+1=0has equal roots

Answer 1

Answer:

D=0

b²-4ac=0

a=(k+4)

b=(k+1)

c=1

(k+1)²-4×(k+4)×1=0

k²+1+2k-4k-16=0

k²-2k-15=0

k²-5k+3k-15=0

k(k-5)+3(k-5)=0

(k-5)(k+3)=0

k=5 or -3

Mark as brainliest plz...