Question 862057
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Presuming you meant:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 6x\ +\ 9]


rather than what you actually wrote, yes, it is a perfect square.  Any polynomial of the form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ \pm\ bx\ +\ c]


is a perfect square if and only if *[tex \LARGE c\ =\ \left(\frac{b}{2}\right)^2]


So your problem is a perfect square because *[tex \LARGE 9\ =\ \left(\frac{6}{2}\right)^2]


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