Question 237808
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Perhaps it will be easier to see if you use a substitution.


Let *[tex \Large a^2\ = (x\ -\ 3)^2].  Then we can say *[tex \Large a\ = (x\ -\ 3)]


Let *[tex \Large b^2\ = 4z^2].  Then we can say *[tex \Large b\ = 2z]


Now, we know that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^2\ -\ b^2\ =\ (a\ +\ b)(a\ -\ b)]


So just do the substitutions:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 3)^2\ -\ 4z^2\ =\ ((x\ -\ 3)\ +\ 2z)((x\ -\ 3)\ -\ 2z)]


Why didn't I worry about the negative square root when I was defining *[tex \Large a] and *[tex \Large b]?  Think about it -- if we had defined *[tex \Large b], for example, as *[tex \Large b\ = -2z], then the above would just have come out:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 3)^2\ -\ 4z^2\ =\ ((x\ -\ 3)\ +\ (-2z))((x\ -\ 3)\ -\ (-2z))\ =\ ((x\ -\ 3)\ -\ 2z)((x\ -\ 3)\ +\ 2z)]
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ =\ ((x\ -\ 3)\ +\ 2z)((x\ -\ 3)\ -\ 2z)]


That is to say, same answer.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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