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Homework Help: Algebra problem involving square roots

  1. Sep 9, 2015 #1
    • Member warned about posting with no template
    So I stumbled upon this problem:

    And I have attempted to solve it, however my solution doesn't match that of the book.
    The solution should be x+4, but I don't have a clue how to get to that. I'm sorry if this seems pretty rudimentary, but I just want some help because it's the first time I've dealt with an equation like this (1<x<3 what should I do with that?). Any help is appreciated. Thanks!
  2. jcsd
  3. Sep 9, 2015 #2
    When the last sq.root goes,what remains is the absolute value,so you are left woth (-x+3) because x<3
  4. Sep 9, 2015 #3
    So the solution is going to be |3x-2|? Sorry if I'm not getting this, but could you please do it step by step, I'm really bad at this.
  5. Sep 9, 2015 #4


    Staff: Mentor

    Technically, you aren't "solving" anything -- that term is applied to equations or inequalities. What you are doing is simplifying the given expression.
    In your work you have
    $$\sqrt{(x + 2)^2} + \sqrt{(x - 1)^2} + \sqrt{(x - 3)^2}, 1 < x < 3$$
    $$= |x + 2| + |x - 1| + |x - 3|$$
    You need to take the absolute value expressions on the given interval, 1 < x < 3, into consideration.
  6. Sep 9, 2015 #5
    What is meant is that $$\sqrt{x^2}=|x|$$

    Then you have to check if your x in your case x+2, x-1, x-3 are positve or negative in the domain 1<x<3.
    Last edited: Sep 9, 2015
  7. Sep 9, 2015 #6
    So, because we need to take the given interval into consideration, we take the absolute value of each term? So that would be the final "simplification" and not x+4?
  8. Sep 9, 2015 #7
    If you consider only the interval given it will give x+4

    But on other intervals the solution will be different.

    And sometimes we have to choose the sign of the square root depending on the definition.
  9. Sep 9, 2015 #8
    But how do I check if my x in x+2, x-1, x-3 is positive or negative? And could you give an example of how it would be if the domain was slightly different? Apologies if I may be annoying you with too many questions, but I want to truly understand.
  10. Sep 9, 2015 #9
    For example if 1<x<3 is x+2 positive or negative ?
  11. Sep 9, 2015 #10
  12. Sep 9, 2015 #11
    Yes it runs from 3 to 5, so |x+2|=x+2 for x in that domain.

    Now you do the same reasoning for x-1 and x-3 so you find ?
  13. Sep 9, 2015 #12


    Staff: Mentor

    I'm not sure what you're saying here. By definition and long usage, the square root of a positive number is positive. In symbols, if x > 0, ##\sqrt{x} > 0##.
  14. Sep 9, 2015 #13
    So what happens if it's negative, for x-1 and x-3? How would I do it now?
  15. Sep 9, 2015 #14
    You notice that x-1 is positive whereas x-3 is negative for 1<x<3

    Hence |x-1|=x-1 and |x-3|=-(x-3)=3-x

    Then you make the sum : x+2+x-1+3-x=?
  16. Sep 9, 2015 #15
    Oh, now I see. Well thank you and everyone else for clarifying and helping me understand this problem. Thanks a bunch!
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