SOLUTION: sqrt(2x-5)-sqrt(x-3)=1

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Question 33877: sqrt(2x-5)-sqrt(x-3)=1
Found 2 solutions by fz_shurtugal, MathTherapy:
Answer by fz_shurtugal(3) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt(2x-5)-sqrt(x-3)=1
Simplify this equation first, so basically you have sqrt(2x-5-x+3)=1
now simplify the radicand which would equall to sqrt(x-2),making the equation sqrt(x-2)=1.Now you want to get rid of the radical sign to solve for x, therefore you would square both sides of the equation, sqrt(x-2)^2=1^2,and then simplify, which would equall to, x-2=1. You now solve this equality which would be x=3.

Answer by MathTherapy(10692) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt(2x-5)-sqrt(x-3)=1
======================
Whatever the other person who responded did, doesn't make sense, at all, to this author. Yet,
it's quite surprising that he/she got one of the solutions. But, there's another one!

  sqrt%282x+-+5%29+-+sqrt%28x+-+3%29+=+1, with x+%3E=+3  
          sqrt%282x+-+5%29+=+1+%2B+sqrt%28x+-+3%29 ----- Adding sqrt%28x+-+3%29 to both sides
       %28sqrt%282x+-+5%29%29%5E2+=+1%5E2+%2B+2sqrt%28x+-+3%29+%2B+%28sqrt%28x+-+3%29%29%5E2
            2x+-+5+=+1+%2B+2sqrt%28x+-+3%29+%2B+x+-+3
            2x+-+5+=+2sqrt%28x+-+3%29+%2B+x+-+2
      2x+-+5+-+x+%2B+2+=+2sqrt%28x+-+3%29
             x+-+3+=+2sqrt%28x+-+3%29
           %28x+-+3%29%5E2+=+%282sqrt%28x+-+3%29%29%5E2 ---- Squaring each side
        x%5E2+-+6x+%2B+9+=+4%28x+-+3%29
        x%5E2+-+6x+%2B+9+=+4x+-+12
x%5E2+-+6x+%2B+9+-+4x+%2B+12+=+0
     x%5E2+-+10x+%2B+21+=+0
  (x - 7)(x - 3) = 0
   x - 7 = 0     OR    x - 3 = 0
      x = 7    OR      x = 3

As 7 > 3 and 3 = 3, the above constraint, x+%3E=+3 is satisfied. Therefore, both solutions are ACCEPTABLE!!