:Is there any reason it couldn't be an irrational root of unity?
2.
Couldn't we consider degenerate intervals such as [ \ pi, \ pi ] that would require that the polynomial have an irrational root?
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Gerard of Cremona ( c . 1150 ), Fibonacci ( 1202 ), and then Robert Recorde ( 1551 ) all used the term to refer to " unresolved irrational roots ".
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I'm wondering if the absence of further solutions can be shown from the fact that as a and n increase, their ratio will tend to the irrational root 2 ? talk ) 13 : 16, 19 March 2008 ( UTC)
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The only possible rational roots would have a numerator that divides 6 and a denominator that divides 1, limiting the possibilities to ?, ?, ?, and ? . of these, 1, 2, and 3 equate the polynomial to zero, and hence are its rational roots . ( In fact these are its only roots since a cubic has only three roots; in general, a polynomial could have some rational and some irrational roots .)
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As to whether there should be a separate article on the Casus Irreducibilis, I think that would be overkill, but it wouldn't hurt if, at the end of the section on Cardano's method, you mentioned the fact that, when applied to an equation with three real, irrational roots, it will always give a solution which includes a sum of two conjugate complex numbers, such that the imaginary parts cancel out .-- vibo56 21 : 47, 25 May 2006 ( UTC)
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