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mukushla

  • 2 years ago

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

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  1. mukushla
    • 2 years ago
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    @sauravshakya @ganeshie8 @eliassaab @TuringTest

  2. nipunmalhotra93
    • 2 years ago
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    x=1,\[\frac{ -1-\sqrt{5} }{2 }\], \[\frac{-1+\sqrt{5} }{2 }\]

  3. nipunmalhotra93
    • 2 years ago
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    is that it?

  4. mukushla
    • 2 years ago
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    let me check it :)

  5. mukushla
    • 2 years ago
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    u are close :) but dont tell the answer plz...so others can think on that :)

  6. nipunmalhotra93
    • 2 years ago
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    Ok fine. Thanks :)

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

  8. mukushla
    • 2 years ago
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    some other values...

  9. sauravshakya
    • 2 years ago
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    I mean are there any other values of x

  10. mukushla
    • 2 years ago
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    yeah there are.

  11. mukushla
    • 2 years ago
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    dont post ur solution :) just values of x

  12. sauravshakya
    • 2 years ago
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    Ok

  13. sauravshakya
    • 2 years ago
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    So I delete it and send u msg for the vaues of x

  14. mukushla
    • 2 years ago
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    no give the values of x here :)

  15. sauravshakya
    • 2 years ago
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    |dw:1347712962196:dw|

  16. nipunmalhotra93
    • 2 years ago
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    oh crap! I misread x^2 as x^3 :\

  17. sauravshakya
    • 2 years ago
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    So was my answer correct @nipunmalhotra93

  18. sauravshakya
    • 2 years ago
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    @mukushla can u PLZ CHECK MY ANSWER

  19. mukushla
    • 2 years ago
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    completely right :)

  20. mukushla
    • 2 years ago
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    so i think nipun got it also :)

  21. sauravshakya
    • 2 years ago
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    Actually f(x)=2 and 1

  22. mukushla
    • 2 years ago
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    yup

  23. sauravshakya
    • 2 years ago
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    GREAT QUESTION

  24. sauravshakya
    • 2 years ago
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    Took me a lot to do this......

  25. nipunmalhotra93
    • 2 years ago
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    :D the font is too small I thought it was x^3 lol :D

  26. mukushla
    • 2 years ago
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    everyone is one here...lol

  27. sauravshakya
    • 2 years ago
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    @mukushla CAN U PLZ GIVE SOME QUESTION LIKE THIS PLZ.......

  28. mukushla
    • 2 years ago
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    of course

  29. sauravshakya
    • 2 years ago
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    GREAT

  30. nipunmalhotra93
    • 2 years ago
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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. :)

  31. sauravshakya
    • 2 years ago
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    Rationalize and u will get|dw:1347715001113:dw|

  32. sauravshakya
    • 2 years ago
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    Right?

  33. mukushla
    • 2 years ago
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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

  34. sauravshakya
    • 2 years ago
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    That will make it easy to solve

  35. sauravshakya
    • 2 years ago
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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

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

  37. sauravshakya
    • 2 years ago
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    now, solve for x

  38. mukushla
    • 2 years ago
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    thats it

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