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Why doesn't this radical equation have a solution?

  1. Jun 3, 2013 #1
    I dont understand why this is a no solution



    original problem
    √(x-5) - √(x-8) = 3




    My solution attempt

    add +√(x-8) to both sides:
    √(x-5) - √(x-8) = 3

    √(x-5) = 3 + √(x-8)

    squaring both sides:
    (√(x-5))^2 = (3 + √(x-8))^2

    FOIL out the right side:
    x - 5 = 3^2 + 2(3√(x-8)) + √(x-8)^2

    simplify:
    x - 5 = 9 + 6√(x-8) + x - 8

    combine like terms:
    x - 5 = x + 1 + 6√(x-8)

    Subtract x and 1 from both sides:
    x - 5 = x + 1 + 6√(x-8)

    -6 = 6√(x-8)

    divide both sides by 6:
    -1 = √(x-8)

    At this point, is it the -1 that means no solution?

    I continued with squaring both sides again:
    (-1)^2 = √(x-8)^2

    1 = x - 8
    +8

    x = 9
     
  2. jcsd
  3. Jun 3, 2013 #2

    Mark44

    Staff: Mentor

    Correct.
    A simpler problem is this:
    √w - √(w - 3) = 3
    If you sketch the graphs of y = √w and y = √(w - 3), you will see that there are no values of w for which the first graph is 3 units above the second graph. That's why there is no solution to this equation or to the one you posted.
     
  4. Jun 3, 2013 #3

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    welcome to pf!

    hi ryn droma! welcome to pf! :smile:
    it's because of the definition of √

    √ always means the positive square root

    if you put x = 9 into the original equation, you get

    √(9 - 5) - √(9 - 8) = 3, which is false

    if you allow negative square roots, then

    √(9 - 5) - √(9 - 8) = 2 - (-1) = 3, which is true​

    (every time you square an equation, you get extra solutions …

    x = 9 is your extra solution :wink:)
     
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