Question 380139
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{2x\ +\ 3}\ -\ \sqrt{x\ +\ 1}\ =\ 1]


Add *[tex \Large \sqrt{x\ +\ 1}] to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{2x\ +\ 3}\ =\ \sqrt{x\ +\ 1}\ +\ 1]


Square both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ +\ 3\ =\ x\ +\ 1\ +\ 2\sqrt{x\ +\ 1}\ +\ 1]


Add *[tex \Large -x\ -\ 2] to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 1\ =\ 2\sqrt{x\ +\ 1}]


Square both sides again:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 2x\ +\ 1\ =\ 4(x\ +\ 1)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 2x\ +\ 1\ =\ 4x\ +\ 4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 2x\ -\ 3\ =\ 0]


Finally solve the factorable quadratic.  Check both answers to account for possible extraneous roots.


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