Question 321811
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I think you mean:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{6x\ +\ 1}\ =\ x\ -\ 1]


Square both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6x\ +\ 1\ =\ x^2\ -\ 2x\ +\ 1]


Collect like terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 8x\ =\ 0]


So:


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


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 8]


Check:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{6(0)\ +\ 1}\ =\ 1\ \neq\ 0\ -\ 1\ =\ -1]


Exclude this extraneous root.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{6(8)\ +\ 1}\ =\ \sqrt{49}\ =\ 7\ =\ 8\ -\ 1\ =\ 7]


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