SOLUTION: Please help with solution of sqrt(5x+8) - sqrt(5x-1) = 1

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Question 614248: Please help with solution of sqrt(5x+8) - sqrt(5x-1) = 1
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Here's a procedure for solving this kind of equation:
  1. Isolate a square root.
  2. Square both sides of the equation. (Be careful with this step!)
  3. If there is still a square root, repeat steps 1-3.
  4. At his point there should no longer be any square roots. Use appropriate techniques to solve the type of equation you now have.
  5. Check your solution(s). This is not optional! Whenever you square both sides of an equation (which you have done at least once to get this far), extraneous solutions may occur. Extraneous solutions are solutions that fit the squared equation but do not fit the original equation. Extraneous solutions are not a sign that you have made a mistake. Nor does it mean that you should never square both sides of an equation. If your check discovers any extraneous solutions, then you must reject them, even if it means that all your solutions get rejected. (In this case your equation has no solution.)
Let's see this in action:
sqrt%285x%2B8%29+-+sqrt%285x-1%29+=+1
1. Isolate a square root.
Adding sqrt%285x-1%29 to each side we get:
sqrt%285x%2B8%29+=+sqrt%285x-1%29++%2B+1

2. Square both sides.
%28sqrt%285x%2B8%29%29%5E2+=+%28sqrt%285x-1%29++%2B+1%29%5E2

Squaring the isolated square root is easy. Squaring the other side requires some care. Exponents do not distribute!! Use FOIL or the %28a%2Bb%29%5E2+=+a%5E2%2B2ab%2Bb%5E2 pattern to square it correctly. I prefer using the pattern:
5x%2B8+=+%28sqrt%285x-1%29%29%5E2++%2B+2%2A%28sqrt%285x-1%29%29%2A1+%2B+%281%29%5E2
5x%2B8+=+5x-1++%2B+2%2Asqrt%285x-1%29+%2B+1
5x%2B8+=+5x+%2B+2%2Asqrt%285x-1%29

3. There is still a square root. So we repeat.
3.1 Isolate a square root.
Subtracting 5x from each side we get:
8+=+2%2Asqrt%285x-1%29
The 2 in front of the square root is not a problem. The square root is sufficiently isolated. But if it bothers you, you can divide both sides by the 2 to get rid of it.

3.2 Square both sides:
%288%29%5E2+=+%282%2Asqrt%285x-1%29%29%5E2
64+=+4%2A%285x-1%29
64+=+20x-4%29
3.3 There are no longer any square roots. So we can finally proceed to step 4.

4. Solve the equation.
Adding 4 to each side:
68+=+20x
Divide by 20:
68%2F20+=+x
17%2F5+=+x

5. Check your solution(s)
Use the original equation:
sqrt%285x%2B8%29+-+sqrt%285x-1%29+=+1
Checking x+=+17%2F5:
sqrt%285%2817%2F5%29%2B8%29+-+sqrt%285%2817%2F5%29-1%29+=+1
sqrt%2817%2B8%29+-+sqrt%2817-1%29+=+1
sqrt%2825%29+-+sqrt%2816%29+=+1
5+-+4+=+1
1+=+1 Check!

So x+=+17%2F5 is the only solution to your equation.