anonymous
  • anonymous
prove that x^8 -x^7 + x^2 -x +15 has no real root
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
for \(x\le0\) its obvious that \(x^8 -x^7 + x^2 -x +15 >0\) u just prove for \(x>0\) : \(x^8 -x^7 + x^2 -x +15 >0\)
anonymous
  • anonymous
and does descartes' theorem help a great deal ? according to that,,it should have max 4 +ve roots..
anonymous
  • anonymous
no need to descartes...

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anonymous
  • anonymous
well what then ?
anonymous
  • anonymous
ohh wait,,leme try..
anonymous
  • anonymous
well just show that \(x^8 -x^7 + x^2 -x +15>0\) for every real number
anonymous
  • anonymous
descartes helps a lot lol
anonymous
  • anonymous
http://mathworld.wolfram.com/DescartesSignRule.html
anonymous
  • anonymous
for x>1 ,, x^8 > x^7 and x^2 >x so f(x) > 0 for x>1 for 0 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 ?
anonymous
  • anonymous
exactly
anonymous
  • anonymous
ohh..thank you @mukushla your little hint only helped me a lot :)
anonymous
  • anonymous
np...:)

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