Question

(k+4)x²+x(k+1)x+1=0. find the value of k for which the following quadratic equation has equal roots​

Answer 1

Answer:

(k+4)x2 +(k+1)x+1=0

D=b2 -4ac

=(k+1)2-4(k+4)(1)

=k2 +2k+1-4k-16

=k2 -2k-15

For equal roots,

D=0

D=0

K2 -2k-15=0

k2 -5k+3k-15=0

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

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

k+3=0 OR k-5=0

k = -3, k = 5