1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Thx
     
  2. jcsd
  3. Jun 1, 2007 #2

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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
    :smile:
     
  5. Jun 1, 2007 #4

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Homework Helper

    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]

    and

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

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

    Then
    [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

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Decimals and surds
  1. Surds Question (Replies: 2)

  2. Surd as power (Replies: 3)

  3. Solving Surds (Replies: 4)

Loading...