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tomiko Group Title

is my working correct? i need to find which number is bigger. i concluded that 2*sqrt38 is bigger of the two since 38^2 = 1444

  • 2 years ago
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  1. tomiko Group Title
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    attched is my working

    • 2 years ago
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  2. jasonxx Group Title
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    seems correct

    • 2 years ago
  3. tomiko Group Title
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    thanks a lot!

    • 2 years ago
  4. jasonxx Group Title
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    (:

    • 2 years ago
  5. asnaseer Group Title
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    yes - correct again. BTW: you can make use of this fact:\[a^2\gt a^2-1\]\[\therefore a^2\gt (a+1)(a-1)\]In this case a=38 so you know that:\[38^2\gt (38+1)(38-1)\]

    • 2 years ago
  6. tomiko Group Title
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    how did you just decide a = 38? @asnaseer

    • 2 years ago
  7. asnaseer Group Title
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    in your last step you had:\[38^2?39\times37\]

    • 2 years ago
  8. asnaseer Group Title
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    so can conclude ? is ">" without actually evaluating \(38^2\)

    • 2 years ago
  9. tomiko Group Title
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    but 38^2 is "a" but 39 x 37 is not = (a+1)(a-1) so how can you compare that way?

    • 2 years ago
  10. asnaseer Group Title
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    no, a=38, remember the identity I showed was:\[a^2\gt (a+1)(a-1)\]so, if a=38 we get:\[38^2\gt(38+1)(38-1)\]\[\therefore 38^2\gt39\times37\]

    • 2 years ago
  11. tomiko Group Title
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    wow!! i see it now. where did you know this. i never came across this before?

    • 2 years ago
  12. asnaseer Group Title
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    nor did I - I just suddenly "spotted" the pattern - that is one of the beauty of maths, you never know what subtleties you will discover. :)

    • 2 years ago
  13. tomiko Group Title
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    thanks...

    • 2 years ago
  14. asnaseer Group Title
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    yw :)

    • 2 years ago
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