Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Callisto

  • 3 years ago

Tutor question #3 Simplify: \[\sqrt{5-\sqrt{21}}\] (conditions: pen, paper, no calculator, done in 2 minutes) *

  • This Question is Closed
  1. lgbasallote
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what if x = \(\sqrt{5-\sqrt{21}}\) then \(x^2 = 5 - \sqrt{21}\) \[x^2 - 5 = -\sqrt{21}\] \[(x^2 - 5)^2 = 21\] oh goodness quadratic o.O

  2. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    actually it is a 4th degree equation

  3. Callisto
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 4

    \[(x^2-5)^2=21\]\[x^4-10x^2+25=21\]\[x^4-10x^2+4=0\]\[x^4-10x^2+4=0\]\[x^2=\frac{10\pm \sqrt{(-10)^6-4(4)}}{2}\]\[x^2=5\pm \sqrt{21}\]Sorry... are you sure that it is 'simplified'?

  4. Callisto
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 4

    Oh.that's power 2, typo

  5. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    apparently it is also \[\sqrt{\frac{7}{2}}-\sqrt{\frac{3}{2}}\] i remember seeing something like this before but i am not sure i remember how to go from one to the other

  6. Callisto
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 4

    I need to know how to work that out.....

  7. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    cant seem to get it, i think it was a trick

  8. Callisto
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 4

    ............................ It... was ... a ... question ... asked ... when ... my friend applied for a summer job ...

  9. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    really? doing what??

  10. Callisto
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 4

    tutor - teaching high school students

  11. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hmm i wonder what answer they wanted

  12. Callisto
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 4

    Same here...

  13. Callisto
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 4

    When I was doing some exercises few days ago, I saw similar questions, but clearer, like this: express \(\sqrt{28-2\sqrt{147}}\) in the form of \(\sqrt{x}-\sqrt{y}\). I can still handle this. But that one, I failed :(

  14. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    multiplying by the conjugate give \(\frac{2}{\sqrt{5+\sqrt{21}}}\) think

  15. myininaya
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    any examples anywhere?

  16. Callisto
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 4

    example?!

  17. myininaya
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You want to show \[\sqrt{5-\sqrt{21}} \text{ equals } \frac{\sqrt{7}-\sqrt{3}}{2} ?\] was just wondering if you have an example for writing that one thing in that other form

  18. myininaya
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oops sqrt(2) on bottom

  19. myininaya
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    how did you get that identity?

  20. Callisto
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 4

    From experience...

  21. myininaya
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    lol great experience I want to prove that identity lol

  22. Callisto
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 4

    First, \((\sqrt{a} - \sqrt{b})^2\) = a + b -2\sqrt{ab} For a>b \[\sqrt{5-\sqrt{21}} = \sqrt{5-2\sqrt{\frac{21}{4}}}\] Now, a+b = 5 => a=5-b ab = 21/4 (5-b)b = 21/4 -4b^2 + 20b - 21 =0 b=1.5 or b =3.5 (rejected) a = 5 - 1.5 = 3.5 So, it is \(\sqrt{\frac{7}{2}}-\sqrt{\frac{3}{2}}\) Does that make sense?

  23. myininaya
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok i see that identity :)

  24. myininaya
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    that one is easy to prove

  25. myininaya
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    maybe because it is an actual identity, right? :p

  26. Callisto
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 4

    Perfect square is perfect :)

  27. Callisto
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 4

    Does that make sense? Apart from the latex fail...

  28. myininaya
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    very interesting i wouldn't have thought of that

  29. Callisto
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 4

    I'm going to post the link to satellite73's post then. He doesn't even come to check..

  30. myininaya
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Great work @Callisto :)

  31. Callisto
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 4

    And thank you for all your time!!!!

  32. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy