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Help with basic problem?

  1. Oct 9, 2014 #1
    1. The problem statement, all variables and given/known data
    I'm working out of Serge Lang's Basic Mathematics and in chapter 3 section 3 this problem comes up.

    [itex](3x+5) ^{-4} =64[/itex]


    2. Relevant equations


    3. The attempt at a solution
    I keep getting [itex](1-(5±\sqrt{8}))/(3±\sqrt{8})[/itex]

    The book says it is [itex](±1-5\sqrt{8})/(3\sqrt{8})[/itex]

    It is not exact. I know [itex]\sqrt{8}=2\sqrt{2}[/itex]

    Thank you for your help.

    I'm obviously forgetting something basic and it is bothering me. Thanks.
     
    Last edited: Oct 9, 2014
  2. jcsd
  3. Oct 9, 2014 #2

    LCKurtz

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    The book is correct. How do you expect us to help you find your mistake when you don't show us your steps?
     
  4. Oct 10, 2014 #3

    NascentOxygen

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    You're taking the square root of each side, then the same again?
     
  5. Oct 10, 2014 #4
    This is my attempt. I hope it is legible
    . Office Lens_20141010_120317.jpg
     
  6. Oct 10, 2014 #5

    Ray Vickson

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    Don't attach photos. PF wants you to type out your work.
     
  7. Oct 10, 2014 #6
    Ok, but why? That makes no sense. That is much more difficult and time consuming and it produces the same outcome. I would get ride of it, but it is not letting me now. Feel free to remove it if you like.

    If you don't want us to attach photos then why included the ability to add an attachment?
     
    Last edited: Oct 10, 2014
  8. Oct 10, 2014 #7

    RUber

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    It looks like you are doing alright until about 2/3 down the page.
    You get to ## 3x-5 = \frac{1}{\pm \sqrt{8}}##. Then you do something I don't understand, and that is where you get off track.
    If you are multiplying by a ## \pm \sqrt{8}## please use parentheses so you don't confuse it for adding or subtracting later on.
    As far as I can tell, your solution may be off because of what I just mentioned.

    You could also take one at a time (either plus or minus) then express the solution with a ##\pm ## at the end. For more complicated algebra, this will help you maintain some sanity, even though it might take more paper.
     
  9. Oct 10, 2014 #8
    When I get to [itex]3x+5 = \frac{1}{\pm \sqrt{8}} [/itex]

    I subtract [itex]-5[/itex] from both sides

    [itex]3x= \frac{1}{\pm \sqrt{8}} -5[/itex]

    [itex]3x=1-(5±\sqrt{8}))/(±\sqrt{8})[/itex]

    Then I divided by 3.
     
  10. Oct 10, 2014 #9

    RUber

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    I think you are trying to write ## 3x = \pm \frac{1}{\sqrt{8}} - 5 ##
    Multiply through by ##\sqrt{8}## on both sides to get:
    ##3\sqrt{8}x = \pm 1- 5\sqrt{8}##.
    This is why I mentioned it might be easier to do plus first, then minus. Your expression ##(5 \pm \sqrt{8}) ## appears like addition (or subtraction), but the only way you get there is by multiplying ##(\pm \sqrt{8}) ## onto the 5--so either your notation is wrong or your understanding of algebra is wrong. I assume the former. In any case, that is the only place your answer differs from the book answer you have above.
     
  11. Oct 10, 2014 #10
    No my algebra is definitely wrong here. I need help understanding what I did wrong. I divided by ##\pm\sqrt{8}## and I either don't know something or forgot something and my algebra is wrong. What used latex to show is what I did.

    It doesn't bother me to be wrong as long as I figure it out.
     
    Last edited: Oct 10, 2014
  12. Oct 10, 2014 #11

    RUber

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    I apologize for the tone in my earlier post. I meant to say that it looks like you are doing the right steps, but poor notation is killing you.
    The signs in the denominator are often easier to keep track of when you pull them to the front.
    ## \frac{1}{\pm \sqrt{8}} =\pm \frac{1}{ \sqrt{8}}##
    However, what you have is not wrong.
    ## 3x = \frac{1}{\pm \sqrt{8}}-5##
    There is no problem multiplying by the ## \pm## but it complicates things.
    You could from what you have there just write
    ##x = \frac13( \frac{1}{\pm \sqrt{8}}-5)##
    But I suppose you are trying you get something that looks like the book answer.
    So distribute the 1/3 in and pull the ##\pm ## out of the bottom.
    ##x = \pm \frac{1}{ 3\sqrt{8}}-\frac53##
    The book answer is just this put over a common denominator.
    You went wrong when you put :
    ##3x=(1-(5 \pm \sqrt{8}))/(\pm \sqrt{8})## when you should have written ##3x=(1-5 (\pm \sqrt{8}))/(\pm \sqrt{8})## to represent the multiplication.
    Note that this is equal to ##3x=1/(\pm \sqrt{8})-5 (\pm \sqrt{8})/(\pm \sqrt{8})## and the ##\pm ## cancel in the second fraction.
     
  13. Oct 10, 2014 #12

    jtbell

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    It's easier for someone to help when they can use the "Quote" feature to pick out specific steps in your calculation in order to comment on them, instead of having to read them off a photo and type them in, and then proofread for errors in transcription. It's enough work for them to fix mistakes in their own equations!
     
  14. Oct 10, 2014 #13
    Ok I see what you were saying now. I appreciate the help.

    Ok, fair enough. I'll keep that in mind.
     
  15. Oct 10, 2014 #14

    NascentOxygen

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    Ok, all sorted now? ;)
     
  16. Oct 10, 2014 #15

    Ray Vickson

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    Sometimes you need to attach pictures of diagrams from books or notes, for example, or maybe drawings of geometric figures, etc. Other than that, attachments are discouraged.

    Anyway, back to your problem: it can be written as ##X^4 = r^4##, where ##X = 3x+5## and ##r = 1/\sqrt{8}##. For real ##r > 0## the equation ##X^4 = r^4## has two real solutions, ##X = \pm r##, and two complex (pure imaginary) solutions ##X = \pm i \, r##, where ##i = \sqrt{-1}##. Therefore, there should be only two real solutions for ##x##. These are ##x = (r-5)/3## and ##x = (-r-5)/3##.
     
  17. Oct 11, 2014 #16

    vela

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    There's a small learning curve for LaTeX, but it's actually not all that difficult nor time-consuming. I know some of the helpers don't seem to care, but I usually pass over threads where students can't be bothered to type up their work.
     
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