solving equations with radicals

Question

In Algebra 2, I am given the following equation:

 

√(x - 32) - √(x) = 8

 

(x - 32 is under the radical sign.)

 

My directions are to solve. I solved it myself, and got x = 36. Then, I checked it, as instructed in my book. I got -4 = 8. There is one equation in the book that doesn't check; they wrote beneath it, "Our answer does not check, so the original equation has no solution." Should I do that, or is my process incorrect?

Answer 1

Yes, x = 36 is the answer you get, but when you plug it back into the original equation, it does not not work.  Hence x = 36 is known as an "extraneous solution."  Since there is only one solution and it is extraneous, the original equation has no solution.