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√x-3 + √x = 3 more simple algebra

  1. Nov 19, 2007 #1
    [SOLVED] √x-3 + √x = 3 more simple algebra

    1. The problem statement, all variables and given/known data

    √x-3 + √x = 3

    ( √x-3 + √x ) ( √x-3 + √x ) = 9


    3. The attempt at a solution

    I have the answer in my book as x=4 , but I don't understand how they got there.

    these are the next two steps which I don't understand, everything after I get so I didn't type it here.

    (x-3) + 2√x-3 √x + x = 9

    2x - 3 + 2√x²-3x = 9
    * the -3x is in the square root
     
  2. jcsd
  3. Nov 19, 2007 #2

    rock.freak667

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    Homework Helper

    (x-3) + 2√x-3 √x + x = 9

    you got this by expanding out ( √x-3 + √x ) ( √x-3 + √x )

    if you take everything else except the sq.root on the other side(with the 9) you'll get
    [itex]\sqrt{x^2-3x}=(6-x)[/itex] then just square both sides and expand again
     
  4. Nov 19, 2007 #3
    man I don't know what you mean. how do you expand out?
     
  5. Nov 19, 2007 #4
    Just as in any (a + b)(a + b), you distribute appropriately.

    (a + b)(a + b) = (a + b)(a) + (a + b)(b)

    = (a)(a) + (b)(a) + (a)(b) + (b)(b).

    That's all "expand" means.
     
  6. Nov 19, 2007 #5


    so everytime I have (a + b)(a + b) this is all I have to do is put it in the form you wrote up there? is there anything else I need to know about this?

    what is this called anyway? alot of questions i know, sorry....it's just I want to understand it, rather than just memorizing how to do it.
     
  7. Nov 19, 2007 #6
    It's just simple distribution. No special name (that I can think of off the top of my head).

    As long as you know what it means to distribute, i.e.:

    a(b + c) = ab + ac.
     
  8. Nov 19, 2007 #7
    it still doesn't add up.

    how does (√x-3 + √x) (√x-3 + √x) = 9

    distributed into form (a)(a) + (b)(a) + (a)(b) + (b)(b)

    = (x-3) + 2√x-3 √x + x = 9 ?


    wouldn't it be.... (√x-3)(√x-3) + (√x)(√x-3) + (√x-3)(√x) + (√x)(√x) ?
     
  9. Nov 19, 2007 #8
    Your last expression isn't completely simplified.

    [tex] (\sqrt{x-3})(\sqrt{x-3}) = x-3 [/tex]

    [tex] (\sqrt{x})(\sqrt{x}) = x [/tex]

    And, finally, there are two of the term:

    [tex] (\sqrt{x-3})(\sqrt{x}) [/tex]

    (which is the same as:)

    [tex] (\sqrt{x})(\sqrt{x-3}) [/tex]

    which need combining.

    That should resolve that particular issue.
     
  10. Nov 20, 2007 #9
    dude...I'm even more lost now. I hate to admit that. I use to be one of the top math students, now I'm struggling with grade 11 crap.
     
  11. Nov 20, 2007 #10
    i hate to keep bothering ya, but I read your last post no less than 20x, and I still don't understand how it works.
     
  12. Nov 20, 2007 #11
    wait...nm.........i got it now.....

    thnkx alot brother...owe u 1
     
  13. Nov 20, 2007 #12
    Oh, great! Not a problem.

    I had started writing a more detailed and fancy response for you, but now I don't need to. Thanks!

    Post again if you have any more questions!

    P.S.: Instead of triple-posting, try editing your first post next time. :)
     
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