Question:

The equation, \(kx^2 -(k+3)x – 1 =0\), which has real coefficients having complex roots, \(a+ib\) and \(a-ib\). In order to have these roots, find the range of \(k\). If they are purely imaginary roots, find the value of \(k\).

Answer:

The discriminant of the quadratic equation, \(ax^2+bx+c=0\), is

\[  D = b^2 – 4ac  \]

For complex roots, this must be \(D<0\). Note that \(k \neq 0\). Thus,

\[  D = (k+3)^2 +4k    \]

\[    = k^2+10k+9      \]

\[    = (k+1)(k+9)  < 0  \]

The range of \(k\) is

\[  -9 < k < -1 \quad \mathrm{but} \quad k \neq 0  \]

If the roots are purely imaginary, we let the roots be \(ib\) and \(-ib\). We have the following relationship of the coefficients and solutions:

\[  ib + (-ib) = – \frac{-(k+3)}{k}  \]

This is from the relationship between the roots and coefficients. Then,

\[  0 = k +3  \]

\[  \rightarrow k = -3  \]

This satisfies \(-9 < k < -1\).