Question 1039581
<font face="Times New Roman" size="+2">


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{x\ +\ 1}\ +\ 2\ =\ \sqrt{x\ -\ 15}]


Square both sides


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 1)\ +\ 4\sqrt{x\ +\ 1}\ +\ 4\ =\ x\ -\ 15]


Put everything but the radical term in the RHS.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4\sqrt{x\ +\ 1}\ =\ x\ -\ 15\ -\ (x\ +\ 1)\ -\ 4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4\sqrt{x\ +\ 1}\ =\ -20] *Remember this step.


Square both sides again:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 16(x\ +\ 1)\ =\ 400]


Solve for x


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 1\ =\ 25]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 24]


Since you squared both sides (twice in this case), you have the potential of having introduced an extraneous root.  Therefore, check the answer:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{24\ +\ 1}\ +\ 2\ =^?\ \sqrt{24\ -\ 15}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{25}\ +\ 2\ =^?\ \sqrt{9}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5\ +\ 2\ =\ 7\ \not =\ 3]


The root is extraneous and there was only one root, so the solution set is the empty set.  Now, look back at the step I pointed out to remember.  4 is a positive number and *[tex \Large \sqrt{x\ +\ 1}] is a positive number because an unsigned even-indexed radical is defined as the positive root.  Since a positive number multiplied by a positive number is positive, such a product cannot be equal to a negative number, specifically -20.  Therefore, we could have quit at this step because it is clear that the equation has no solution.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

</font>