document.write( "Question 239516: for what value(s) of k does kt^2 - 6t + k = 0 have imaginary roots? \n" ); document.write( "

Algebra.Com's Answer #175794 by solver91311(24713) $\"\"$ $\"About$

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\n" ); document.write( "\n" ); document.write( "Strictly speaking, from the way you worded your question, there are no real values of

for which

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

for which

will have a pair of complex roots of the form

where

and

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\n" ); document.write( "\n" ); document.write( "Remember the discriminant:

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\n" ); document.write( "\n" ); document.write( "If

then the roots are a conjugate pair of complex numbers of the form

. Only in the case where

is the real part of the complex number equal to zero and the complex number purely imaginary.\r
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\n" ); document.write( "\n" ); document.write( "So:\r
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\n" ); document.write( "\n" ); document.write( "And for

36\">, hence

3\"> or

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\n" ); document.write( "\n" ); document.write( "Actually it is possible to have purely imaginary roots if you allow

, but somehow I don't think that is what you were talking about when you asked the question.\r
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\n" ); document.write( "\n" ); document.write( "John
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