If the equation -x^2 + 2(k+2)x - 9k = 0 has two imaginary roots, what are the values of k?

Question

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Answer 1

Multiply through by -1: x²-2(k+2)x+9k=0,

Rewrite: x²-2(k+2)x=-9k,

x²-2(k+2)x+(k+2)²=-9k+(k+2)²,

(x-(k+2))²=-9k+k²+4k+4=k²-5k+4=(k-4)(k-1).

For two complex roots, (k-4)(k-1)<0. That means k-4<0 and k-1>0, that is, k<4 and k>1, so 1<k<4. (It's not possible for the alternative: k>4 but k<1, so we can reject this.)