SOLUTION: square root of x+1 + square root of x-1 = square root of 2x+1 Please help....if you can break it down to the simplest my book does not explain how to work with one that involves t

Algebra ->  Radicals -> SOLUTION: square root of x+1 + square root of x-1 = square root of 2x+1 Please help....if you can break it down to the simplest my book does not explain how to work with one that involves t      Log On


   

Question 53082: square root of x+1 + square root of x-1 = square root of 2x+1
Please help....if you can break it down to the simplest my book does not explain how to work with one that involves three radicals. Thank you so much
Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%28x%2B1%29%2Bsqrt%28x-1%29=sqrt%282x%2B1%29
%28%28sqrt%28x%2B1%29%2Bsqrt%28x-1%29%29%29%5E2=%28sqrt%282x%2B1%29%29%5E2


x%2B1%2B2sqrt%28%28x-1%29%28x%2B1%29%29%2Bx-1=2x%2B1
2x%2B0%2B2sqrt%28x%5E2-1%29=2x%2B1
-2x%2B2x%2B2sqrt%28x%5E2-1%29=-2x%2B2x%2B1
2sqrt%28x%5E2-1%29=1
%28%282sqrt%28x%5E2-1%29%29%29%5E2=1%5E2
4%28x%5E2-1%29=1
4x%5E2-4=1
4x%5E2-4%2B4=1%2B4
4x%5E2=5
4x%5E2%2F4=5%2F4
x%5E2=5%2F4
sqrt%28x%5E2%29=%2B-sqrt%285%2F4%29
x=%2B-sqrt%285%29%2F2
x=-sqrt%285%29%2F2 is an extraneous solution because when you substitute it back into the original equation you cannot get a real number solution.
Therefore, the only answer you can use is:
x=sqrt%285%29%2F2