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Aryang

  • 2 years ago

prove that x^8 -x^7 + x^2 -x +15 has no real root

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  1. mukushla
    • 2 years ago
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    for \(x\le0\) its obvious that \(x^8 -x^7 + x^2 -x +15 >0\) u just prove for \(x>0\) : \(x^8 -x^7 + x^2 -x +15 >0\)

  2. Aryang
    • 2 years ago
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    and does descartes' theorem help a great deal ? according to that,,it should have max 4 +ve roots..

  3. mukushla
    • 2 years ago
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    no need to descartes...

  4. Aryang
    • 2 years ago
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    well what then ?

  5. Aryang
    • 2 years ago
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    ohh wait,,leme try..

  6. mukushla
    • 2 years ago
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    well just show that \(x^8 -x^7 + x^2 -x +15>0\) for every real number

  7. Yahoo!
    • 2 years ago
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    descartes helps a lot lol

  8. Yahoo!
    • 2 years ago
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    http://mathworld.wolfram.com/DescartesSignRule.html

  9. Aryang
    • 2 years ago
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    for x>1 ,, x^8 > x^7 and x^2 >x so f(x) > 0 for x>1 for 0<x<1 , x> x^2 > x^7 > x^8 we then have 15 - (fraction + fraction) ,,which is obviously >0 so f(x) is always >0 am i right ?

  10. mukushla
    • 2 years ago
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    exactly

  11. Aryang
    • 2 years ago
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    ohh..thank you @mukushla your little hint only helped me a lot :)

  12. mukushla
    • 2 years ago
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    np...:)

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