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tomiko

  • 2 years ago

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

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

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

  3. asnaseer
    • 2 years ago
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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?

  4. tomiko
    • 2 years ago
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    you get 1

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

  6. asnaseer
    • 2 years ago
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    *now

  7. tomiko
    • 2 years ago
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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.

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

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

  10. asnaseer
    • 2 years ago
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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?

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

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

  13. asnaseer
    • 2 years ago
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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?

  14. tomiko
    • 2 years ago
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    in that case x is bigger

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

  16. tomiko
    • 2 years ago
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    x

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

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

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

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

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

  22. asnaseer
    • 2 years ago
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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\[\sqrt{13}+\sqrt{12}\gt\sqrt{12}+\sqrt{11}\]

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

  24. asnaseer
    • 2 years ago
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    perfect! well done! :)

  25. tomiko
    • 2 years ago
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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 ?

  26. asnaseer
    • 2 years ago
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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:\[(\sqrt{11}+\sqrt{13})^2=?\]

  27. asnaseer
    • 2 years ago
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    do you understand?

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

  29. asnaseer
    • 2 years ago
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    ok :)

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

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

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

  33. asnaseer
    • 2 years ago
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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

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

  35. tomiko
    • 2 years ago
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    now i get (attached)

    1 Attachment
  36. asnaseer
    • 2 years ago
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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

  37. tomiko
    • 2 years ago
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    now the final answer!! :d

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

  39. tomiko
    • 2 years ago
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    attached

    1 Attachment
  40. asnaseer
    • 2 years ago
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    perfect!

  41. tomiko
    • 2 years ago
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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.

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

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

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

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