Question 1159695
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{x\,+\,4}\ >\ 2\ +\ \sqrt{x}]


Square both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 4\ >\ 4\ +\ 4\sqrt{x}\ +\ x]


Add *[tex \Large -x\ -\ 4] to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ 0\ >\ 4\sqrt{x}]


Multiply both sides by *[tex \Large \frac{1}{4}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0\ >\ \sqrt{x}]


Square both sides


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


Which is to say:


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


However, this result must be discarded because the most restrictive domain is the domain of the right-hand side expression, namely


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \text{dom }2\ +\ \sqrt{x}\ =\ {x\,\in\,\mathbb{R}\,|\,x\,>\,0]


Therefore the solution set is the null set over the reals.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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