find the value of k for which the given equation has real roots x2+k(4x+k-1)+2=0

Question

hgfgfcccccccccccccccccccccccccccccccccccccccc

Answer 1

x^2 +4kx +(k^2 -k+2)=0 . . . hav reel roots if b^2-4ac >=0 . . . a=1 . . . b=4k . . . . c=k^2-k+2 . . . b^2-4ac=16k^2 -4*(k^2-k+2) . . . . =16k^2 -4k^2 +4k -8 . . . =12k^2 +4k -8 . . . =4*(3k^2 +k-2) . . . . this=0 at k=-1, or k=0.66666666666... its > 0 for k >0.666666666 . . . . & for k < -1

Answer 2

hiii