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tomiko

how do i work out which number is greater? (sqrt12 - sqrt11), (sqrt13 - sqrt12)

  • one year ago
  • one year ago

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  1. asnaseer
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    @tomiko are you aware of this factorisation?\[a^2-b^2=(a+b)(a-b)\]

    • one year ago
  2. tomiko
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    yes. but i don't know how to apply this to work out the answer. please help me

    • one year ago
  3. asnaseer
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    look at the first term:\[\sqrt{12}-\sqrt{11}\]if you multiply this by \(\sqrt{12}+\sqrt{11}\) what will you get?

    • one year ago
  4. tomiko
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    you get 1

    • one year ago
  5. asnaseer
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    correct. nor similarly multiply the second expression with \(\sqrt{13}+\sqrt{12}\) - what do you get?

    • one year ago
  6. asnaseer
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    *now

    • one year ago
  7. tomiko
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    for that too i get 1. but one is definately greater. cause when i use the calculate i get 2 different values, making one of the pairs greater than the other.

    • one year ago
  8. asnaseer
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    good. now, out these two terms, which one is bigger:\[1. \sqrt{12}+\sqrt{11}\]\[2.\sqrt{12}+\sqrt{13}\]

    • one year ago
  9. tomiko
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    \[\sqrt{12} + \sqrt{13}\] is bigger

    • one year ago
  10. asnaseer
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    good. so you now know that: a) x times (1) = 1 b) y times (2) = 1 and you know (2) is bigger than one. therefore what can you conclude about x and y?

    • one year ago
  11. asnaseer
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    I meant to say: you know (2) is bigger than (1)

    • one year ago
  12. tomiko
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    yes. 2 is bigger than one so that makes \[\sqrt{13} - \sqrt{12}\] bigger?? :D

    • one year ago
  13. asnaseer
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    no - in terms of x and y that I listed, x is being multiplied by a smaller number than y is being multiplied by, but both end up as 1. therefore is x or y bigger?

    • one year ago
  14. tomiko
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    in that case x is bigger

    • one year ago
  15. asnaseer
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    imagine you had:\[x*4=1\]\[y*8=1\]which one would you say is bigger, x or y?

    • one year ago
  16. tomiko
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    x

    • one year ago
  17. asnaseer
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    yes, x is bigger, therefore you know which one is bigger in your original question

    • one year ago
  18. tomiko
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    yes i know \[\sqrt{13} - \sqrt{12}\] is bigger.

    • one year ago
  19. asnaseer
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    look back carefully through what was derived and re-think your answer.

    • one year ago
  20. asnaseer
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    remember that \(\sqrt{13} - \sqrt{12}\) was multiplied by the bigger number \(\sqrt{13} + \sqrt{12}\) to get 1.

    • one year ago
  21. tomiko
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    yes!! i get it now. had to look at my solutions again using the different of 2 squares.

    • one year ago
  22. asnaseer
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    we showed that:\[(\sqrt{12}-\sqrt{11})(\sqrt{12}+\sqrt{11})=1\]\[(\sqrt{13}-\sqrt{12})(\sqrt{13}+\sqrt{12})=1\]and\[\sqrt{13}+\sqrt{12}\gt\sqrt{12}+\sqrt{11}\]

    • one year ago
  23. tomiko
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    that makes \[\sqrt{12} - \sqrt{11}\] greater!

    • one year ago
  24. asnaseer
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    perfect! well done! :)

    • one year ago
  25. tomiko
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    i undertand now. thank you very much! but you have any idea about this approach? Purely algebra ? Place ? i/o =, > or <. sqrt(12)-sqrt(11) ? sqrt(13)-sqrt(12) 2sqrt(12) ? sqrt(11)+sqrt(13) 48 ? ... Can you continue ?

    • one year ago
  26. asnaseer
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    ok what this approach is doing is bringing together the two square roots of 12 and and then squaring both sides. so:\[(2\sqrt{12})^2=4*12=48\]now you just need to square the right hand side:\[(\sqrt{11}+\sqrt{13})^2=?\]

    • one year ago
  27. asnaseer
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    do you understand?

    • one year ago
  28. tomiko
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    yeap...! i'm trying to do it on paper here :D

    • one year ago
  29. asnaseer
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    ok :)

    • one year ago
  30. tomiko
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    and this shows \[\sqrt{12} - \sqrt{11}\] is bigger. you can see my working attached as an image

    • one year ago
    1 Attachment
  31. asnaseer
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    you expansion of \((\sqrt{11}+\sqrt{13})^2\) is not correct. remember that:\[(a+b)^2=a^2+2ab+b^2\]

    • one year ago
  32. tomiko
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    so if i correct it. i get \[48 ? 24 + 2\times \sqrt{13}\times \sqrt{11}\]

    • one year ago
  33. asnaseer
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    that is correct so far. now take the 24 to the left hand side and then divide both sides by 2

    • one year ago
  34. asnaseer
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    also remember that:\[\sqrt{a}\times\sqrt{b}=\sqrt{ab}\]

    • one year ago
  35. tomiko
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    now i get (attached)

    • one year ago
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  36. asnaseer
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    that is correct - just one step left now - square both sides BTW: It is better to save your images as .png format if you can - they are smaller than .jpg files

    • one year ago
  37. tomiko
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    now the final answer!! :d

    • one year ago
  38. asnaseer
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    great! I'm glad you finally got there :)

    • one year ago
  39. tomiko
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    attached

    • one year ago
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  40. asnaseer
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    perfect!

    • one year ago
  41. tomiko
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    sorry for all the bother. thanks for the help. i finished high school 6 years ago and didn't do any math at all and now just started uni. it's going to be a little change for me. i'm doing computer science now. lots of math in my class!! but i love subject.

    • one year ago
  42. asnaseer
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    no problem at all my friend - the main thing here is you obviously love to learn! :)

    • one year ago
  43. tomiko
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    challenge** thank you very much once again. now any question like this that comes my way i will murder it!!

    • one year ago
  44. asnaseer
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    he he - I love your "ambition" - keep up the good work! :)

    • one year ago
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