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Q3 - Radicals within radicals

  1. Aug 24, 2010 #1
    1. The problem statement, all variables and given/known data

    sqrt(7 + 2sqrt(6)) - sqrt(7 - 2sqrt(6))

    2. Relevant equations

    3. The attempt at a solution

    In order to work out the radicals, I define the answer as "x". Now I can square both sides to get rid of some of the first radicals:

    sqrt(7 + 2sqrt(6)) - sqrt(7 - 2sqrt(6)) = x
    (sqrt(7 + 2sqrt(6)) - sqrt(7 - 2sqrt(6)))2 = (x)2

    Now, I hope this is the right way of starting on this problem, continuing we get:

    (7 + 2sqrt(6)) - 2 * sqrt(7+ 2 sqrt(6)) * sqrt(7 - 2 sqrt(6)) + (7 - 2 sqrt(6))

    this implies:

    14 - 2 * sqrt(7+ 2 sqrt(6)) * sqrt(7 - 2 sqrt(6)) = x2

    on moving the 14 to the RHS and squaring once more to get rid of more radicals:

    (-2 * sqrt(7+ 2 sqrt(6)) * sqrt(7 - 2 sqrt(6)) )2 = (x2-14)2

    this implies

    4 * (7 + 2 sqrt(6)) * (7 - 2 sqrt(6)) = (x2-14)2

    because the two factors with sqrts are conjugate pairs, the LHS will look like this:

    4 * (72 - (4*6) = 4 * 25 = 100


    100 = (x2-14)2

    by taking the square root of both sides:

    ±10 = ± (x2-14)
    ±10±14 = ± x2

    Now, this does not look like a proper solution. Working further I can get a couple of answers for x, but no way to check them or anything:

    sqrt(24) = 2 sqrt(6) = ± x
    sqrt(4) = 2 = ± x

    Where do I go wrong?
    Last edited: Aug 24, 2010
  2. jcsd
  3. Aug 24, 2010 #2


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    Staff: Mentor

    You can check the answer by using a calculator on the original problem line. In fact, you can use a calculator to check each line of your solution...
  4. Aug 24, 2010 #3
    Yes, however, the problem states I should be able to calculate without the use of a calculator...
  5. Aug 24, 2010 #4


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    Staff: Mentor

    Sure, I'm just saying that you can use a calculator to see if what you are doing algebraically is working...
  6. Aug 24, 2010 #5


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    Homework Helper

    Why do you make it so complicated? Simplify the left side, using that

    [tex]\sqrt{7+ 2 \sqrt6} \sqrt{7 - 2 \sqrt6} =\sqrt{(7+ 2 \sqrt6) (7 - 2 \sqrt6)}=?[/tex]

    As for the sign of x, 7+2sqrt(6)>7-2sqrt(6), and so are the square roots. The difference is positive.

  7. Aug 24, 2010 #6


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    Staff Emeritus
    Science Advisor
    Gold Member

    Why do you think there is no way to check? Isn't it obvious the original number is positive? And it's pretty easy to show that it's less than 4 too.

    Sometimes separating the solutions via inequalities requires a lot of precision, but this one only requires very coarse estimates.

    P.S. it's probably better to split into cases, rather than using ±; it's easier to avoid making mistakes that way.
  8. Aug 24, 2010 #7
    Thanks to all of you. When I read your answers everything becomes clear to me. The only thing I do not understand sometimes is how I could be so stupid not to see it...I paid special attention to seeing the rules for radicals in here but missed the simple fact at one point that sqrt(a) * sqrt(b) = sqrt(ab)

    I would practice more on my own not to ask so many seemingly stupid questions. Unfortunately I have very little time left and I need to evolve my basic understandings into a (somewhat) more advanced understanding of the basic rules.

    Just want to say I really appreciate the input and patience. Feels like I found a great place to get help with math...sometime the internet is too big to find something like that, not this time luckily :)

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