Question 1118951
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Let's begin with a quick review of terminology.  The quadratic trinomials that you have listed are NOT quadratic equations.  Equations have an equals sign somewhere in them; hence the name.  Had you written:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x^2\ -\ 2x\ +\ 5\ =\ 0]


(or with any other expression in the right-hand side), you would have been correct to call it a quadratic equation.  For the time being, I'll simply assume that is what you meant (and probably what was written in your textbook, assignment sheet, test paper, or whatever).


Given that assumption, note the following:


*[tex \Large \sqrt{-19}\ =\ \sqrt{(-1)(19)}\ =\ \sqrt{-1}\,\cdot\,\sqrt{19}\ =\ i\sqrt{19}].


Where *[tex \Large i] is the imaginary number defined by *[tex \Large i^2\ =\ -1]


Then take the common factor of 2 out of the numerator and denominator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_1\ =\ \frac{1\ -\ i\sqrt{19}}{4}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_2\ =\ \frac{1\ +\ i\sqrt{19}}{4}]


As for the second problem, correcting the terminology error:


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


Your application of the quadratic formula, given that you actually had an equation to start with, is correct and reduced to lowest terms.  Since the lead coefficient and the constant term have opposite signs, you are guaranteed to have two distinct real roots. (*[tex \Large b^2] must be non-negative and *[tex \Large 4ac] must be negative given opposite signs on *[tex \Large a] and *[tex \Large c].  Then, since you are subtracting the negative value *[tex \Large 4ac], the entire result under the radical must be positive and therefore real.  The consequence to this problem is that there is no imaginary part to the complex number representations of the two roots to the given equation.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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