Question 394082
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The first step is to stop disrespecting me by typing "u" when you mean "you."


Since there is no pair of integers *[tex \Large p] and *[tex \Large q] such that *[tex \Large p\ +\ q\ =\ 3] and *[tex \Large pq\ =\ -15], use the quadratic formula.


For


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 3x\ -\ 15\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ 1\ \ ], *[tex \LARGE b\ =\ 3\ \ ], and *[tex \LARGE c\ =\ -15]


Plug the values into:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-b\ \pm\ \sqrt{b^2\ -\ 4ac}}{2a}]


And do the necessary arithmetic to simplify it.  Hint:  *[tex \Large \sqrt{69}] cannot be simplified further.  Leave your answer in radical form if you need the exact answers, or use your calculator to derive the appropriate numerical approximations.


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