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mukushla Group Title

dont need weird mathematics :) \[f(x)=\sqrt{x^2-2x+10}-\sqrt{x^2-2x+2}\] find all \(x \in \mathbb{R}\) such that \(f(x) \in \mathbb{N}\)

  • one year ago
  • one year ago

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  1. mukushla Group Title
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    @sauravshakya @ganeshie8 @eliassaab @TuringTest

    • one year ago
  2. nipunmalhotra93 Group Title
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    x=1,\[\frac{ -1-\sqrt{5} }{2 }\], \[\frac{-1+\sqrt{5} }{2 }\]

    • one year ago
  3. nipunmalhotra93 Group Title
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    is that it?

    • one year ago
  4. mukushla Group Title
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    let me check it :)

    • one year ago
  5. mukushla Group Title
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    u are close :) but dont tell the answer plz...so others can think on that :)

    • one year ago
  6. nipunmalhotra93 Group Title
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    Ok fine. Thanks :)

    • one year ago
  7. sauravshakya Group Title
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    So only when x=1 or are there any other value..... I think yes because irrational-irrational can be integer

    • one year ago
  8. mukushla Group Title
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    some other values...

    • one year ago
  9. sauravshakya Group Title
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    I mean are there any other values of x

    • one year ago
  10. mukushla Group Title
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    yeah there are.

    • one year ago
  11. mukushla Group Title
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    dont post ur solution :) just values of x

    • one year ago
  12. sauravshakya Group Title
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    Ok

    • one year ago
  13. sauravshakya Group Title
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    So I delete it and send u msg for the vaues of x

    • one year ago
  14. mukushla Group Title
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    no give the values of x here :)

    • one year ago
  15. sauravshakya Group Title
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    |dw:1347712962196:dw|

    • one year ago
  16. nipunmalhotra93 Group Title
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    oh crap! I misread x^2 as x^3 :\

    • one year ago
  17. sauravshakya Group Title
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    So was my answer correct @nipunmalhotra93

    • one year ago
  18. sauravshakya Group Title
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    @mukushla can u PLZ CHECK MY ANSWER

    • one year ago
  19. mukushla Group Title
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    completely right :)

    • one year ago
  20. mukushla Group Title
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    so i think nipun got it also :)

    • one year ago
  21. sauravshakya Group Title
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    Actually f(x)=2 and 1

    • one year ago
  22. mukushla Group Title
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    yup

    • one year ago
  23. sauravshakya Group Title
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    GREAT QUESTION

    • one year ago
  24. sauravshakya Group Title
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    Took me a lot to do this......

    • one year ago
  25. nipunmalhotra93 Group Title
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    :D the font is too small I thought it was x^3 lol :D

    • one year ago
  26. mukushla Group Title
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    everyone is one here...lol

    • one year ago
  27. sauravshakya Group Title
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    @mukushla CAN U PLZ GIVE SOME QUESTION LIKE THIS PLZ.......

    • one year ago
  28. mukushla Group Title
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    of course

    • one year ago
  29. sauravshakya Group Title
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    GREAT

    • one year ago
  30. nipunmalhotra93 Group Title
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    @sauravshakya I'd like to know your approach. Would you mind telling it? This is how I got it. \[\sqrt{p+10}-\sqrt{p+2}=A\] A is natural so, \[p+10=A ^{2}+p+2+2A \sqrt{p+2}\] \[10-A^2-2=2A \sqrt{p+2}\] so, \[\frac{ 4}{A }-\frac{ A }{ 2 }=\sqrt{p+2}\] So, A can only be 1 or 2 (as A is natural) for \[\sqrt{p+2}\] to be real. Put p=x^2-2x and solving \[\sqrt{p+2}=1 or \sqrt{p+2}=7/2\] , we get the answer. This is a good question that's why I wanna know your method. :)

    • one year ago
  31. sauravshakya Group Title
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    Rationalize and u will get|dw:1347715001113:dw|

    • one year ago
  32. sauravshakya Group Title
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    Right?

    • one year ago
  33. mukushla Group Title
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    \[f(x)=\sqrt{x^2-2x+10}-\sqrt{x^2-2x+2}=\frac{8}{\sqrt{x^2-2x+10}+\sqrt{x^2-2x+2}}\]\[f(x)=\frac{8}{\sqrt{(x-1)^2+9}+\sqrt{(x-1)^2+1}}\le\frac{8}{\sqrt{9}+\sqrt{1}}=2\]so f(x)=1 or 2 same as saura's method

    • one year ago
  34. sauravshakya Group Title
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    That will make it easy to solve

    • one year ago
  35. sauravshakya Group Title
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    opps. Now, when f(x)=1 (x^2-2x+10)^1/2 - (x^2-2x+2)^1/2=1........i (x^2-2x+10)^1/2 + (x^2-2x+2)^1/2=8........ii

    • one year ago
  36. sauravshakya Group Title
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    So, 2*(x^2-2x+10)^1/2=9

    • one year ago
  37. sauravshakya Group Title
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    now, solve for x

    • one year ago
  38. mukushla Group Title
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    thats it

    • one year ago
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