Question

find the value of k for which x2-(k+5)x+2(2k+1)​

Answer 1

(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