document.write( "Question 384711: \n" ); document.write( "
Algebra.Com's Answer #272365 by jsmallt9(3758)\"\" \"About 
You can put this solution on YOUR website!
\"sqrt%28x%5E2%2B2x-2sqrt%28x%5E2-2x%2B10%29%29+=+x\"
\n" ); document.write( "Solving equations where the variable is inside one or more square roots can be done by:
  1. Isolate a square root.
  2. Square both sides of the equation.
  3. If there are any remaining square roots, repeat steps 1 and 2.
  4. At this point there should be no square roots remaining. Solve the equation using appropriate techniques for the type of equation.
  5. Check your answer(s)! This is not optional. Squaring both sides of an equation (which has been done at least once) can introduce what are called extraneous solutions. Extraneous solutions are solutions that fit the squared equation but do not fit the original equation! Extraneous solutions can occur even if no mistakes have been made. So you must check your answers and extraneous solutions, if any, must be rejected. (It is even possible that all \"solutions\" are extraneous meaning that all \"solutions\" are rejected leaving no solution to your equation!)

\n" ); document.write( "Let's see this in action:
\n" ); document.write( "1) Isolate a square root.
\n" ); document.write( "The \"big\" square root on the left side is already isolated.
\n" ); document.write( "2) Square both sides:
\n" ); document.write( "\"%28sqrt%28x%5E2%2B2x-2sqrt%28x%5E2-2x%2B10%29%29%29%5E2+=+%28x%29%5E2\"
\n" ); document.write( "This simplifies to:
\n" ); document.write( "\"x%5E2%2B2x-2sqrt%28x%5E2-2x%2B10%29+=+x%5E2\"
\n" ); document.write( "3) There is still a square root so we must repeat steps 1 and 2.
\n" ); document.write( "Isolate a square root.
\n" ); document.write( "Subtracting \"x%5E2\" and 2x from each side we get:
\n" ); document.write( "\"-2sqrt%28x%5E2-2x%2B10%29+=+-2x\"
\n" ); document.write( "Dividing by -2 we get:
\n" ); document.write( "\"sqrt%28x%5E2-2x%2B10%29+=+x\"
\n" ); document.write( "Square both sides:
\n" ); document.write( "\"%28sqrt%28x%5E2-2x%2B10%29%29%5E2+=+%28x%29%5E2\"
\n" ); document.write( "which simplifies to:
\n" ); document.write( "\"x%5E2-2x%2B10+=+x%5E2\"
\n" ); document.write( "The square roots are gone. We can proceed to step 4.
\n" ); document.write( "4) Solve the equation.
\n" ); document.write( "This appears to be a quadratic equation. So we want one side to be zero. Subtracting \"x%5E2\" from each side we get:
\n" ); document.write( "-2x+10 = 0
\n" ); document.write( "Since the squared terms disappeared this is no longer a quadratic equation. Now we just isolate x. Subtracting 10 from each side we get:
\n" ); document.write( "-2x = -10
\n" ); document.write( "Dividing by -2 we get:
\n" ); document.write( "x = 5
\n" ); document.write( "5) Check you answer(s).
\n" ); document.write( "Always us the original equation to check:
\n" ); document.write( "\"sqrt%28x%5E2%2B2x-2sqrt%28x%5E2-2x%2B10%29%29+=+x\"
\n" ); document.write( "Checking x = 5:
\n" ); document.write( "\"sqrt%28%285%29%5E2%2B2%285%29-2sqrt%28%285%29%5E2-2%285%29%2B10%29%29+=+%285%29\"
\n" ); document.write( "This simplifies as follows:
\n" ); document.write( "\"sqrt%2825%2B2%285%29-2sqrt%2825-2%285%29%2B10%29%29+=+5\"
\n" ); document.write( "\"sqrt%2825%2B+10+-2sqrt%2825+-+10%2B10%29%29+=+5\"
\n" ); document.write( "\"sqrt%2835+-2sqrt%2825%29%29+=+5\"
\n" ); document.write( "\"sqrt%2835+-2%2A5%29+=+5\"
\n" ); document.write( "\"sqrt%2825%29+=+5\"
\n" ); document.write( "5 = 5 Check! \n" ); document.write( "
\n" );