Question 437355
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Remember that *[tex \Large x^n\ \cdot\ x^m\ =\ x^{n\,+\,m}]


So *[tex \Large x^{\frac{3}{2}}\ =\ x^{1\,+\,\frac{1}{2}}]


Factor out *[tex \Large x^{\frac{1}{2}]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^{\frac{1}{2}}(x\ -\ 3)\ =\ 0]


Zero Product Rule



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^{\frac{1}{2}}\ =\ 0]


Square both sides


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 0]


Or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ 3\ =\ 0]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 3]


Next time -- One question per post.



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{2}\ =\ \frac{2}{4}\ =\ 2\,\cdot\,\frac{1}{4}]


Let *[tex \Large u\ =\ x^{\frac{1}{4}}]


Then



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u^2\ -\ 2u\ +\ 1\ =\ 0]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (u\ -\ 1)^2\ =\ 0]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u\ =\ 1]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^{\frac{1}{4}}\ =\ 1]


Raise both sides to the 4th power:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 1]


Go root around in <a href="http://www.purplemath.com/">Purple Math</a> or <a href="http://www.cliffsnotes.com/math-study-guides.html">Cliff's Notes</a>


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