Question 434319
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We know that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ y\ <\ 0]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ xy\ =\ -135]


from which we can derive


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-135}{y}]


Substitute


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -\frac{135}{y}\ +\ y\ <\ 0]


Multiply by *[tex \Large y]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y^2\ <\ 135]


The square root of 135 is approximately 11 and 2/3.  A perfect square in the vicinity of 11 is 9.  135 divided by 9 is 15.  9 is odd, a perfect square, not a factor of 15, and when multiplied by -15 (which is odd, not a perfect square, not a factor of 9) equals -135.  Furthermore 9 + (-15) is negative. 



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