Question 63816
{{{x^2+9 <= 0 }}}.
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This problem has no solutions in the real numbers because {{{x^2}}} would have to be less than or equal to -9 and in the real numbers squares are always greater than or equal to 0.
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In the complex, or imaginary, numbers x can be less than or equal to {{{3sqrt(-1)}}}, or {{{ x <= 3i}}}.