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Solving Radical Equations

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  1. Jan 9, 2017 #1
    1. The problem statement, all variables and given/known data
    √3x - 5 +2 = -3
    Underlined is under square root.

    √2x - 3 = -x + 3
    Underlined is under square root.
    2. Relevant equations


    3. The attempt at a solution
    √3x - 5 +2 = -3
    -2 -2
    (√3x - 5)2 = (-5)2
    3x - 5 = 25
    3x = 30
    x = 10
    Solution says no answer but I got one... I tried 3x - 5 ≠ 0 → x ≠5/3

    2nd Question
    √2x - 3 = -x + 3
    (√2x - 3)2 = (-x + 3)2
    2x - 3 = (-x + 3)(-x + 3)
    2x - 3 = x2 - 6x + 9
    0 = x2 - 8x + 12
    0 = (x - 2)(x - 6)
    x = 6
    x = 2
    Solution only says 2 not 6. I tried 2x - 3 ≠ 0 → x ≠ 3/2

    Wondering what I did wrong.
     
  2. jcsd
  3. Jan 9, 2017 #2

    Mark44

    Staff: Mentor

    Or equivalently, ##\sqrt{3x - 5} = -5##
    This equation has no solution. A square root evaluates to a number that is greater than or equal to zero.
    When you square both sides of an equation, you have to check to see if you have introduced extraneous roots. Check both of your solutions in the original equation. I'll bet that 6 is not a solution of the original equation.
     
  4. Jan 9, 2017 #3
    Thanks so much! I always forget to check for extraneous roots!
     
  5. Jan 9, 2017 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Instead of underlining and explaining, just use parentheses, like this: √(3x-5) +2 = -3. (However, as already explained, this equation has no (real) solutions.)
     
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