Question 583145
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The discriminant is: *[tex \LARGE b^2\ -\ 4ac]


Your quadratic in standard form


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9x^2\ -\ 6x\ +\ 1\ =\ 0]


So *[tex \LARGE a\ =\ 9], *[tex \LARGE b\ =\ -6], and *[tex \LARGE c\ =\ 1]


The first thing for you to do is to write back and tell me how you managed to get *[tex \LARGE 72i] out of


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (-6)^2\ -\ 4(9)(1)]


and how you managed to get anything involving *[tex \LARGE i] when you didn't take a square root anywhere.


Use the following to determine the nature of the roots:


Find the Discriminant, *[tex \LARGE \Delta\ =\ b^2\ -\ 4ac] and evaluate the nature of the roots as follows:


No calculation quick look:  If the signs on *[tex \Large a] and *[tex \Large c] are opposite, then *[tex \LARGE \Delta > 0] guaranteed.


*[tex \LARGE \Delta > 0 \ \ \Rightarrow\ \] Two real and unequal roots. If *[tex \LARGE \Delta] is a perfect square, the quadratic factors over *[tex \LARGE \mathbb{Q}] (the rationals).


*[tex \LARGE \Delta = 0 \ \ \Rightarrow\ \] One real root with a multiplicity of two.  That is to say that the trinomial is a perfect square and has two identical factors.  Presuming rational coefficients, the root will be rational as well.


*[tex \LARGE \Delta < 0 \ \ \Rightarrow\ \] A conjugate pair of complex roots of the form *[tex \LARGE a \pm bi] where *[tex \LARGE i] is the imaginary number defined by *[tex \LARGE i^2 = -1]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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