Question 239516
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Strictly speaking, from the way you worded your question, there are no real values of *[tex \Large k] for which *[tex \Large kt^2\ -\ 6t\ +\ k\ =\ 0] has purely imaginary roots.  That is because the coefficient on the 1st degree term is non-zero and rational, hence any roots of the given equation must have a rational (and therefore real) component.  There is a range of values for *[tex \Large k] for which *[tex \Large kt^2\ -\ 6t\ +\ k\ =\ 0] will have a pair of complex roots of the form *[tex \Large a\ \pm\ bi] where *[tex \Large a,\,b\,\in\,\R] and *[tex \Large i^2\ =\ -1].


Remember the discriminant:  *[tex \Large \Delta\ =\ b^2\ -\ 4ac]


If *[tex \Large \Delta < 0] then the roots are a conjugate pair of complex numbers of the form *[tex \Large a\ \pm\ bi].  Only in the case where *[tex \Large b\ =\ 0] is the real part of the complex number equal to zero and the complex number purely imaginary.


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \Delta\ =\ (-6)^2\ -\ 4(k)(k)]


And for *[tex \Large \Delta < 0], *[tex \Large 4k^2 > 36], hence *[tex \Large k\ > 3] or *[tex \Large k < -3].


Actually it is possible to have purely imaginary roots if you allow *[tex \Large k\ \in\ C], but somehow I don't think that is what you were talking about when you asked the question.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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