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Which is greater?

  1. Oct 30, 2012 #1
    how do i work this out?

    2wf24go.png
     
  2. jcsd
  3. Oct 30, 2012 #2

    arildno

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    Remember that for any two numbers, a and b, we have: (a-b)*(a+b)=a^2-b^2
     
  4. Oct 30, 2012 #3
    Think of the square root function graphed.
     
  5. Oct 30, 2012 #4

    Mark44

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    I like this approach. You can see at a glance which difference is larger.
     
  6. Oct 30, 2012 #5

    arildno

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    Very good! :smile:
     
  7. Oct 30, 2012 #6
    i knot this. but how does it help? the terms are different

    i don't want to solve this graphically. (not like i even know how. I need some help here)
     
  8. Oct 30, 2012 #7

    arildno

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    sqrt(12)-sqrt(11)=(sqrt(12)-sqrt(11))*(sqrt(12)+sqrt(11))/(sqrt(12)+sqrt(11))
     
  9. Oct 30, 2012 #8

    Mark44

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    Are you saying you don't know what the graph of y = ##\sqrt{x}## looks like?
     
  10. Oct 30, 2012 #9
    oh no! i know that! but first, i don't want to solve this graphically even if i did, how does the graph of y = √x help out?

    EDIT: if i knew everything in math, i wouldn't be here.
     
    Last edited: Oct 30, 2012
  11. Oct 30, 2012 #10

    Mark44

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    The graph of y = √x is one of the first ones you learn when you learn to graph functions. If you are asking questions about square roots, it's one you should know.

    Look at the graph of this function. Does the y value on the graph change more between 11 and 12 than it does between 12 and 13, or does it change less between 11 and 12 than it does between 12 and 13?
     
  12. Oct 30, 2012 #11
    i get it now. but the problem is. i don't want to solve it geobetrically at all. is there a non-graphical way? please?
     
  13. Oct 30, 2012 #12

    dextercioby

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    Purely algebra ?

    Place ? i/o =, > or <.

    sqrt(12)-sqrt(11) ? sqrt(13)-sqrt(12)
    2sqrt(12) ? sqrt(11)+sqrt(13)
    48 ? ...

    Can you continue ?
     
  14. Nov 11, 2012 #13
    but how is 2sqrt(12) = 48? :/
     
  15. Nov 11, 2012 #14
    i think i get it. please check this image:

    3304ojc.jpg
     
    Last edited: Nov 11, 2012
  16. Nov 11, 2012 #15
    am i correct working it this way?
     
  17. Nov 11, 2012 #16

    Mark44

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    I doubt that any instructor would accept a proof in which most of the symbols are ?.
     
  18. Nov 13, 2012 #17
    Think about it like this what is √4-√1=1 obviously what is √7-√4, √9-√4=1 so less than 1, which means that the difference between higher roots is less than lower roots-you could also think that there are more roots to share between 2 numbers higher up eg.√100 to √81, for 9-10 and √4-√1, for 2-1 which shows there is a greater difference between smaller roots.
     
  19. Nov 13, 2012 #18
    If you want a purely analytical method:

    Note that [itex] \sqrt{n} [/itex] is a twice-differentiable function, yielding a second-derivative of [itex] -.25n^{-1.5} [/itex] which is negative for all positive n. Thus the first-derivative of the function decreases monotonically for positive n.

    Apply the mean value theorem to the intervals [11,12] and [12,13]. You will get an interesting result which wil give you your answer.

    BiP
     
    Last edited: Nov 13, 2012
  20. Nov 14, 2012 #19
    You could make it an equation and use the same method... i.e. let root(13)-root(12)+x=root(12)-root(11). If x>0, then the left side is greater. If x<0, the right side is greater
     
  21. Nov 14, 2012 #20

    micromass

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    If the left side is greater, then you can't write an = between the sides.

    Am I understanding you wrong? :confused:
     
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