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Can the solution to a polynomial function be all imaginary numers or "i"
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if when u say solution u mean x values that results in a y=0 value then yes it can. But this also means there is no real number where the graph crosses the x axis
Sure, for example x^2+1=0 has solutions i and -i. Both imaginary.
If the coefficients are restricted to the rational numbers, then if any complex number is in the solution set, then so is its complex conjugate. So. there is no polynomial over the rational numbers whose only roots are i, i, i, and i.
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