Question 310370
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Let *[tex \Large u] represent *[tex \Large \sqrt{x}] then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u^2\ +\ 6u\ -\ 7\ =\ 0]


Factor:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (u\ +\ 7)(u\ -\ 1)\ =\ 0]


Hence,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u\ =\ -7]


or 


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


And then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{x}\ =\ -7]


But since *[tex \Large \sqrt{a}\ >\ 0] by definition -- the negative square root being denoted *[tex \Large -\sqrt{a}] -- discard this root.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{x}\ =\ 1]


So


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


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