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Homework Help: Decimals and surds

  1. Jun 1, 2007 #1
    Hi, I was wondering what is the best method (without using a calculator) to workout somthing like 0.5 x [tex]\sqrt{61}[/tex]
    Is there some logical steps/order you can take

  2. jcsd
  3. Jun 1, 2007 #2


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    As i've mentioned in another thread. On non-calculator papers, its always best to use fractions, so the first step will to turn the 0.5 into a fraction which is easy enough.

    With surds one must remember the following.

    [tex] \sqrt{a} \times \sqrt{b} = \sqrt{ab} [/tex]

    So what you would do is take the number under the square root and try and split it into two different numbers, one of which will be a square number. For example:

    [tex] \sqrt{60} = \sqrt{4\times 15} = 2\sqrt{15} [/tex]

    In the example you gave there is not much you could do with the 61.

    If the question is asking you to estimate an answer as GCSE questions often do then use the square root of 60 as in my example along with the half as a fraction.

    What exactly is the question you're working on?
  4. Jun 1, 2007 #3
    I just came out with an answer [tex]\frac{\sqrt{61}}{2}[/tex] for a question and I was wondering wether I could simplify it and/or pick up any new techniques. Thanks for the reply
  5. Jun 1, 2007 #4


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    Well theres not much you can do with a root 61 but if you can simplify then do so.
  6. Jun 9, 2007 #5
    u can use linearization to estimate square root of 61 but at some place u will need square root of 3 which we don't actually know without using a calculator
  7. Jun 9, 2007 #6

    Gib Z

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    Well to find the square root of 3 i would iterate newtons method, or just happen to know roughly as it as i do: 1.73205
  8. Jun 10, 2007 #7
    ok then if u can do it without a calculator use linearization and state the function as sqrt(64-3)
  9. Jun 11, 2007 #8
    Why? -- the whole point of linearization is to approximate with a linear function.

    For example, expand about a = 64

    [tex]f(x) = f(a) +f'(a) \cdot (x-a) + \dots [/tex]


    [tex]f(a) = a^{1/2} = 8 [/tex]

    [tex]f'(a) = (1/2) a^{-1/2} = 1/2 \cdot 1/8[/tex]

    [tex]f(61) \approx 8 + \frac{1}{16} \cdot (61-64) = 8 - 3/16 =7.8125[/tex]

    and that's within about .03% relative error.
    Last edited: Jun 11, 2007
  10. Jun 11, 2007 #9


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    Just to calm the OP poster a bit, linerization isn't part of the GCSE syllabus. I believe the last exam for GCSE maths was today though.
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