Question 444589
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Let's assume that *[tex \Large x\ =\ \frac{\alpha}{\rho}] and *[tex \Large x\ =\ \frac{\beta}{\sigma}] are the roots of some quadratic equation *[tex \Large ax^2\ +\ bx\ +\ c\ =\ 0]


Then with a little algebra we can see that *[tex \Large \rho{x}\ -\ \alpha\ =\ 0] and *[tex \Large \sigma{x}\ -\ \beta\ =\ 0].


Reversing the Zero Product Rule, we can write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\rho{x}\ -\ \alpha\right)\left(\sigma{x}\ -\ \beta\right)\ =\ 0]


Apply FOIL


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \rho\sigma{x^2}\ +\ \left(\beta\rho\ +\ \alpha\sigma\right)x\ +\ \alpha\beta\ =\ 0]


Then let *[tex \Large a\ =\ \rho\sigma\ \ ], *[tex \Large b\ =\ \beta\rho\ +\ \alpha\sigma\ \ ], and *[tex \Large c\ =\ \alpha\beta] and substitute:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ax^2\ +\ bx\ +\ c\ =\ 0]


The only fly in the ointment is that you can take any real number, say *[tex \Large k] and apply it as a factor:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ k\left(\rho{x}\ -\ \alpha\right)\left(\sigma{x}\ -\ \beta\right)\ =\ 0]


You still have the same roots to the equation, but when you multiply it out you get:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ kax^2\ +\ kbx\ +\ kc\ =\ 0]


Which means that there are an infinite number of quadratic equations with a given pair of roots (or given root with a muliplicity of 2) but they only differ by a factor common to all three terms in the trinomial representation of the quadratic.


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