Question 531331
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Let *[tex \Large p] and *[tex \Large q] represent arbitrary constants.


Consider the quadratic function *[tex \Large \varphi(x)\ =\ ax^2\ +\ bx\ +\ c]


If *[tex \Large (-2,0)] is a point on the parabola, then *[tex \Large \varphi(-2)\ =\ 0] which means that *[tex \Large p(x\ +\ 2)\ =\ 0] for any arbitrary constant *[tex \Large p].  Further *[tex \Large q(x\ -\ 7)\ =\ 0] for any arbitrary constant *[tex \Large q].  And finally,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ p(x\ +\ 2)\,\cdot\,q(x\ -\ 7)\ =\ ax^2\ +\ bx\ +\ c]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \Phi(x)\ =\ pqx^2\ -\ 5pqx\ -\ 14pq]


For simplicity's sake name a new arbitrary constant *[tex \Large k\ =\ pq], then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \Phi(x)\ =\ k\left(x^2\ -\ 5x\ -\ 14\right)]


Then 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \forall\,k\,\in\,\mathbb{R},\,\Phi(-2)\ =\ 0\text{ and }\Phi(7)\ =\ 0]


Having said all of that, since you asked for <i><b>an</b></i> equation with the given zeros, pick any value for *[tex \Large k] that you like (*[tex \Large k\ =\ 1] works fine), and write the specific equation that suits your mood.


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