Question 1161533
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If *[tex \Large (a\ +\ b)^2\ =\ 121] then *[tex \Large a\ +\ b\ =\ \pm{11}].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ +\ b\ =\ 11\ \text{and}\ ab\ =\ 24\ \text{and}\ a\ >\ b\ \Right\ a\ =\ 8\ \text{and}\ b\ =\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ +\ b\ =\ -11\ \text{and}\ ab\ =\ 24\ \text{and}\ a\ >\ b\ \Right\ a\ =\ -3\ \text{and}\ b\ =\ -8]


Case 1:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (8\ -\ 3)(64\ +\ 24\ +\ 9)\ =\ 485]


Case 2:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (-3\ -\ (-8))(64\ +\ 24\ +\ 9)\ =\ 485]
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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