Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

tomiko Group Title

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

  • 2 years ago
  • 2 years ago

  • This Question is Closed
  1. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  2. tomiko Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    yes. but i don't know how to apply this to work out the answer. please help me

    • 2 years ago
  3. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    look at the first term:\[\sqrt{12}-\sqrt{11}\]if you multiply this by \(\sqrt{12}+\sqrt{11}\) what will you get?

    • 2 years ago
  4. tomiko Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    you get 1

    • 2 years ago
  5. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  6. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    *now

    • 2 years ago
  7. tomiko Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

    • 2 years ago
  8. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  9. tomiko Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\sqrt{12} + \sqrt{13}\] is bigger

    • 2 years ago
  10. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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?

    • 2 years ago
  11. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    I meant to say: you know (2) is bigger than (1)

    • 2 years ago
  12. tomiko Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  13. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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?

    • 2 years ago
  14. tomiko Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    in that case x is bigger

    • 2 years ago
  15. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    imagine you had:\[x*4=1\]\[y*8=1\]which one would you say is bigger, x or y?

    • 2 years ago
  16. tomiko Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    x

    • 2 years ago
  17. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    yes, x is bigger, therefore you know which one is bigger in your original question

    • 2 years ago
  18. tomiko Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    yes i know \[\sqrt{13} - \sqrt{12}\] is bigger.

    • 2 years ago
  19. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    look back carefully through what was derived and re-think your answer.

    • 2 years ago
  20. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    remember that \(\sqrt{13} - \sqrt{12}\) was multiplied by the bigger number \(\sqrt{13} + \sqrt{12}\) to get 1.

    • 2 years ago
  21. tomiko Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  22. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    we showed that:\[(\sqrt{12}-\sqrt{11})(\sqrt{12}+\sqrt{11})=1\]\[(\sqrt{13}-\sqrt{12})(\sqrt{13}+\sqrt{12})=1\]and\[\sqrt{13}+\sqrt{12}\gt\sqrt{12}+\sqrt{11}\]

    • 2 years ago
  23. tomiko Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    that makes \[\sqrt{12} - \sqrt{11}\] greater!

    • 2 years ago
  24. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    perfect! well done! :)

    • 2 years ago
  25. tomiko Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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 ?

    • 2 years ago
  26. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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:\[(\sqrt{11}+\sqrt{13})^2=?\]

    • 2 years ago
  27. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    do you understand?

    • 2 years ago
  28. tomiko Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    yeap...! i'm trying to do it on paper here :D

    • 2 years ago
  29. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    ok :)

    • 2 years ago
  30. tomiko Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
    1 Attachment
  31. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    you expansion of \((\sqrt{11}+\sqrt{13})^2\) is not correct. remember that:\[(a+b)^2=a^2+2ab+b^2\]

    • 2 years ago
  32. tomiko Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  33. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    that is correct so far. now take the 24 to the left hand side and then divide both sides by 2

    • 2 years ago
  34. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    also remember that:\[\sqrt{a}\times\sqrt{b}=\sqrt{ab}\]

    • 2 years ago
  35. tomiko Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    now i get (attached)

    • 2 years ago
    1 Attachment
  36. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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

    • 2 years ago
  37. tomiko Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    now the final answer!! :d

    • 2 years ago
  38. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    great! I'm glad you finally got there :)

    • 2 years ago
  39. tomiko Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    attached

    • 2 years ago
    1 Attachment
  40. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    perfect!

    • 2 years ago
  41. tomiko Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

    • 2 years ago
  42. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    no problem at all my friend - the main thing here is you obviously love to learn! :)

    • 2 years ago
  43. tomiko Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    challenge** thank you very much once again. now any question like this that comes my way i will murder it!!

    • 2 years ago
  44. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    he he - I love your "ambition" - keep up the good work! :)

    • 2 years ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.