Question 1081674
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Solve both equations for *[tex \Large y] in terms of *[tex \Large x] 


Equation 1:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -xy\ =\ -x^2\ -\ 8]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  xy\ =\ x^2\ +\ 8]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  y\ =\ x\ +\ \frac{8}{x}]


Equation 2:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  y\ =\ -x^2\ +\ 8x]


Now that you have two expressions in *[tex \Large x] both equal to *[tex \Large y], set the two expressions equal to each other:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  -x^2\ +\ 8x\ =\ x\ +\ \frac{8}{x}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  -x^3\ +\ 8x^2\ =\ x^2\ +\ 8]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x^3\ -\ 7x^2\ +\ 8\ =\ 0]


The rational roots theorem tells us that if a rational root exists, it must be one of the following values: *[tex \Large \{\pm1,\,\pm2,\,\pm4,\,\pm8\}]


Use Synthetic Division: (that the value 1 fails is left as an exercise for the student)

<pre>
-1   |   1    -7    0    8
              -1    8   -8
     ---------------------
         1    -8    8    0
</pre>

Hence, *[tex \Large (x\ +\ 1)] and *[tex \Large (x^2\ -8x\ +8)] are factors of the cubic.


Therefore one of the roots is -1 and the other two are the roots of the quadratic factor (calculation of these roots is left as an exercise for the student).


Using  *[tex \LARGE \ \ \ \ \ \  y\ =\ -x^2\ +\ 8x] with *[tex \LARGE x\ =\ -1]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \  y\ =\ -(-1)^2\ +\ 8(-1)\ =\ -9]


One of the three points of intersection is *[tex \Large \left(-1,\,-9\right)].


The other two are left for you to calculate.



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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