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tomiko

  • 2 years ago

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

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  1. tomiko
    • 2 years ago
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    attched is my working

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  2. jasonxx
    • 2 years ago
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    seems correct

  3. tomiko
    • 2 years ago
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    thanks a lot!

  4. jasonxx
    • 2 years ago
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    (:

  5. asnaseer
    • 2 years ago
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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:\[38^2\gt (38+1)(38-1)\]

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

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

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

  9. tomiko
    • 2 years ago
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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?

  10. asnaseer
    • 2 years ago
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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:\[38^2\gt(38+1)(38-1)\]\[\therefore 38^2\gt39\times37\]

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

  12. asnaseer
    • 2 years ago
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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. :)

  13. tomiko
    • 2 years ago
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    thanks...

  14. asnaseer
    • 2 years ago
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    yw :)

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