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Extra Cedit Algebra II Problems (2 of them)

  1. Dec 7, 2004 #1
    Little help please, I'm failing and need the extra credit and have no idea how to work these problems.

    1. [tex]
    x = \sqrt{30 + \sqrt{30 + \sqrt{30 + \ldots } } }
    [/tex]

    2. Find the 4th root of -1.

    I know that this is the same thing as the square root of i. I also know that the answer is:
    sqrt(i) = sqrt2/2 + sqrt2/29(i)
    However, my teacher wants us to prove it obviously.
     
    Last edited: Dec 7, 2004
  2. jcsd
  3. Dec 7, 2004 #2
    This needs clarification. Do you mean [tex]x = \sqrt{30\sqrt{30\sqrt{30\sqrt{...}}}}[/tex] ?
    For the second, you do know that there are 4 distinct fourth roots of any number ?
     
  4. Dec 7, 2004 #3
    My bad for the first one it's x = sqrt30 + sqrt 30... starting under one sqare root.

    For the 2nd I don't know what your talking about but my teacher basically said just prove that the sqrt of i is equal to sqrt2/2 + (sqrt2/2)i.
     
    Last edited: Dec 7, 2004
  5. Dec 7, 2004 #4

    Hurkyl

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    There's a fairly direct way to show that one number is the square root of another... do you recall the definition of a square root?

    (And similarly for n-th roots)
     
  6. Dec 7, 2004 #5

    Hurkyl

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    For part 1, do you mean:

    [tex]
    x := \sqrt{30 + \sqrt{30 + \sqrt{ \ldots } } }
    [/tex]

    Have you thought of anything at all for this? Any algebraic manipulations? Substitutions? Maybe you've seen other types of problems where there was some sort of operation applied infinitely many times in a simple pattern...
     
  7. Dec 7, 2004 #6

    dextercioby

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    1;The purpose of this forum (the homework part) is to help students solve the problems on their own.So that's why i'll give only hints to what u should be doing:
    2.Square both terms of the equation and make use that u have a "neverending "sum.
    3."-1" has 4 possible solutions.To find them u need to know the 2 solutions of the eq. a^2=-1 and the formula of Euler.

    That should be more than enough to make U SOLVE THE PROBLEM,NOT ME.

    Daniel.
     
    Last edited: Dec 7, 2004
  8. Dec 7, 2004 #7
    Yes, that's what I mean. I'll edit the first post in a minute.

    Well I guess you could square both sides and that would give you:
    [tex]
    x^2 = 30 + \sqrt{30 + \sqrt{30 + \sqrt{30 \ldots } } }
    [/tex]

    For the 2nd one I still don't understandd how there are 4 answers. If you simplify the cube root of -1 it is the same thing as [tex]\sqrt{i}[/tex] right? And then I use the equation [tex]\sqrt{i} = a + bi[/tex] . But where do I get numbers for a and b?
     
    Last edited: Dec 7, 2004
  9. Dec 7, 2004 #8
    If you're going to use complex numbers, there is no longer a reason to use only principal roots. For example, i2 and (-i)2 are the two square roots of -1.
    As a further example, the fourth roots of 1 are 1, -1, i and -i. (You can check that the fourth power of each is 1.
    Given that the 4th roots of -1 are not even on the real axis, I see no reason to choose preference for one over the others.
    When working with complex numbers, you will find polar form [tex]Re^{i\theta} = R(\cos\theta + i\sin\theta)[/tex] to be much easier to work with. R is modulus of the complex number and [tex]\theta[/tex] is the argument of the complex number. The motivation for this is the view of complex numbers as vectors in a complex plane with imaginary and real axes.
     
    Last edited: Dec 7, 2004
  10. Dec 8, 2004 #9
    I still don't get all this 4 numbers business and I certainly don't know where you got that polar form equation because I've never learned anything about it. Looks more like pre-cal to me.

    But forget that problem right now. What do I do with the other one after I square both sides?
     
  11. Dec 8, 2004 #10

    Galileo

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    Since it the pattern continuous indefinately, try solving the problem by identifying x as the limit of a sequence.

    [tex]x_1=\sqrt{30}, x_2=\sqrt{30+\sqrt{30}}, x_3=\sqrt{30+\sqrt{30+\sqrt{30}}}[/tex]
     
  12. Dec 8, 2004 #11

    Integral

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    You are on the right track now do the next obvious step, subtract 30 from each side.

    Stare at it! Have you ever seen the RHS before?
     
  13. Dec 8, 2004 #12

    AKG

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    So, you get:

    [tex]x^2 = 30 + \left (\sqrt{30 + \sqrt{30 + \sqrt{30 \ldots } } }\right )[/tex]

    Does that stuff in brackets look familiar?
     
  14. Dec 8, 2004 #13
    Ya it does :p

    But when I end up with:
    [tex]x^2 - 30 = \sqrt{30 + \sqrt{30 + \sqrt{30 \ldots } } }[/tex]

    then what?

    Do I take the square root of the left side?
     
  15. Dec 8, 2004 #14

    shmoe

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    You agree it looks familiar-so what is it?
     
  16. Dec 8, 2004 #15

    dextercioby

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    My guess is that he forgot how the equation looked in the first place... :rofl:
    Anyway,the solution is unique,though u'll be gettin a second order algebraic eq.with 2 real distinct solutions.One of them won't work...Which is the one that won't work???
     
  17. Dec 8, 2004 #16
    How could I forget the original equation???

    If I knew what to do when it looks the same I would have done it already. How do I get a solvabale equation out of that mess?
     
  18. Dec 8, 2004 #17

    dextercioby

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    U didn't get the joke,u didn't get that u can put "x" instead of the radical expression in the equation gotten after sqaring both sides,what chances do you think u have to solving the equation u'd be gettin'??

    Anyway,it's fair from me to say "Better luck this time!!".

    Daniel.
     
  19. Dec 8, 2004 #18

    Integral

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    It would have been sweet if we could have let HIM get that Ah-Ha moment of seeing that substitution on his own. Maybe next time.
     
  20. Dec 8, 2004 #19
    Actually I didn't figure it out the first time I read you two's posts, so I just now figured it out. So I did get the Ah-Ha. :p

    [tex]x^2 - 30 = \sqrt{30 + \sqrt{30 + \sqrt{30 \ldots } } }[/tex]
    [tex]x^2 - 30 = x[/tex]
    [tex]x^2-x-30=0[/tex]
    [tex](x-6)(x+5)=0[/tex]
    So:
    x=6 and x=-5
    And the answer is x=6 since the square root of a positive number can't be negative. Yay, 1 down 1 to go. Thanks guys :)

    Edit: I got the other one too :)
    [tex]\sqrt{i}[/tex]
    [tex]\sqrt{i} = a + bi[/tex]
    [tex]i = (a + bi)^2[/tex]
    [tex](a^2 - b^2) + (2ab)i = 0 + 1i[/tex]
    [tex]a^2 - b^2 = 0[/tex] and [tex]2ab = 1[/tex]
    a = +b
    [tex]2(-b)b = 1[/tex]
    [tex]-2b^2 = 1[/tex]
    So: a does not = b

    [tex]2a^2 = 1[/tex]
    [tex]a = b = 1/\sqrt{2}[/tex] or [tex]a = b = -1/\sqrt{2}[/tex]
    [tex]a = b = \sqrt{2}/2[/tex] or [tex]a = b = -\sqrt{2}/2[/tex]
    So:
    [tex]\sqrt{i} = (\sqrt{2}/2) + (\sqrt{2}/2)i[/tex]
     
    Last edited: Dec 8, 2004
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