Question 329177
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The way to attack this one is to use the idea that if *[tex \Large \alpha] is a zero of of a polynomial function, then *[tex \Large x\ -\ \alpha] must be a factor of the polynomial.  Further we know that irrational roots of polynomial equations always come in conjugate pairs.


So, let's say that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \rho\ +\ \sqrt{\sigma}]


is a root of our quadratic equation.  From that we can determine two things.  First:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \rho\ -\ \sqrt{\sigma}]


is the other zero of the function and the factors of the quadratic trinomial are:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\ -\ (\rho\ +\ \sqrt{\sigma})\right)\left(x\ -\ (\rho\ -\ \sqrt{\sigma})\right)]


Which, in order to be a quadratic equation in standard form, must be multiplied using FOIL and set equal to zero.


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


Now you have a quadratic equation in standard form, *[tex \Large ax^2\ +\ bx\ +\ c\ =\ 0] where


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ 1\ \ ], *[tex \LARGE b\ =\ -2\rho\ \ ], and *[tex \LARGE c\ =\ \rho^2\ -\ \sigma]


You can leave your answer like that, or pick a couple of numbers to substitute for *[tex \Large \rho] and *[tex \Large \sigma].  Pick any numbers you like as long as *[tex \Large \sigma] is NOT a perfect square.  (If you pick a perfect square for *[tex \Large \sigma], then your quadratic will turn out to be factorable)


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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