Question 1034962
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Not even close.  What you did was say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{a^2\ -\ b^2}\ =\ a\ -\ b]


However, in general, that is a false statement


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (a\ -\ b)(a\ -\ b)\ =\ a^2\ -\ 2ab\ +\ b^2\ \not =\ a^2\ -\ b^2]


So, take your answer and go back one step to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{4\ -\ \frac{x^2}{4}}]


Which could be expressed as


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\sqrt{1\ -\ \frac{x^2}{16}]


which could be more convenient depending on context.


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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