SOLUTION: solve radical equation and check solution....???... sqrt(x-5)-sqrt(x-8)=3

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Question 345182: solve radical equation and check solution....???...
sqrt(x-5)-sqrt(x-8)=3
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%28x-5%29-sqrt%28x-8%29=3
Here's a procedure for solving equations like this:
  1. Isolate a square root (i.e. get it by itself on one side of the equation).
  2. Square both sides of the equation.
  3. If there are still any square roots, repeat steps #1 and #2 until they are all gone.
  4. At this point you have an equation without square roots. Use an appropriate procedure for solving the type of equation you now have.
  5. Check your answer(s)!. This is important, not just a good idea, whenever you square both sides of an equation (like you have done at least once at step #2). Squaring both sides of an equation can introduce what are called extraneous solutions. These are solutions which work in the squared equation but not in the original equation! You must check your answers, using the original equation, and reject any that do not work.

Let's use this on yhour equation.
1) Isolate a square root. Add the second square root to both sides:
sqrt%28x-5%29+=+sqrt%28x-8%29+%2B+3
2) Square both sides:
%28sqrt%28x-5%29%29%5E2+=+%28sqrt%28x-8%29+%2B+3%29%5E2
Squaring the left side is simple. But squaring the right side can easily be done incorrectly. Exponent do not distribute! We must use FOIL or the %28a%2Bb%29%5E2+=+a%5E2+%2B+2ab+%2B+b%5E2 pattern to square this correctly:
x-5+=+%28sqrt%28x-8%29%29%5E2+%2B+2sqrt%28x-8%29%283%29+%2B+3%5E2
x-5+=+x-8+%2B+6sqrt%28x-8%29+%2B+9
x-5+=+x+%2B+6sqrt%28x-8%29+%2B+1
3) We still have a square root. So we need to repeat steps #1 and #2. Isolate a suqare root. There's only one left so we'll subtract x and 1 from each side to isolate it:
-6+=+6sqrt%28x-6%29
Square both sides:
36+=+36%28x-6%29
The square roots are gone so we can move on to step 4.
4) Solve the new equation. This is a simple equation to solve. We can start by dividing by 36:
1+=+x-6
Add 6 to each side:
7+=+x
5)> Check the answer(s) in the original equation.
sqrt%28x-5%29-sqrt%28x-8%29=3
Checking x = 7:
sqrt%28%287%29-5%29+-+sqrt%28%287%29-8%29+=+3
sqrt%282%29+-+sqrt%28-1%29+=+3
At this point we should notice a problem. We cannot have a negative radicand in a square root! So x = 7 does NOT work in the original equation. It is an extraneous solution which we must reject.

Since x = 7 was the only answer we found and since we had to reject it, there are no solutions to the original equation.