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mukushlaBest ResponseYou've already chosen the best response.3
for \(x\le0\) its obvious that \(x^8 x^7 + x^2 x +15 >0\) u just prove for \(x>0\) : \(x^8 x^7 + x^2 x +15 >0\)
 one year ago

AryangBest ResponseYou've already chosen the best response.0
and does descartes' theorem help a great deal ? according to that,,it should have max 4 +ve roots..
 one year ago

mukushlaBest ResponseYou've already chosen the best response.3
no need to descartes...
 one year ago

mukushlaBest ResponseYou've already chosen the best response.3
well just show that \(x^8 x^7 + x^2 x +15>0\) for every real number
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
descartes helps a lot lol
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
http://mathworld.wolfram.com/DescartesSignRule.html
 one year ago

AryangBest ResponseYou've already chosen the best response.0
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 ?
 one year ago

AryangBest ResponseYou've already chosen the best response.0
ohh..thank you @mukushla your little hint only helped me a lot :)
 one year ago
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