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