Question 611018
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x}{2}\ -\ \frac{18}{x}\ +\ \frac{5}{2}\ =\ 0]


Multiply by *[tex \LARGE 2x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 5x\ -\ 36\ =\ 0]


This factors, so let's solve it that way as a check on our work with the quadratic formula.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 9)(x\ -\ 4)\ =\ 0]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ -9] or *[tex \LARGE x\ =\ 4]


Using the quadratic formula:


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


Plug in the numbers:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-(5)\ \pm\ \sqrt{(5)^2\ -\ 4(1)(-36)}}{2(1)}]


All that is left is for you to do the arithmetic.  If you get something different than we got when we solved it by factoring, then you have an arithmetic error.


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