Question

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

Answer 1

for \(x\le0\) its obvious that \(x^8 -x^7 + x^2 -x +15 >0\)

u just prove for \(x>0\) : \(x^8 -x^7 + x^2 -x +15 >0\)

Answer 2

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

Answer 3

no need to descartes...

Answer 4

well what then ?

Answer 5

ohh wait,,leme try..

Answer 6

well just show that \(x^8 -x^7 + x^2 -x +15>0\) for every real number

Answer 7

descartes helps a lot lol

Answer 8

http://mathworld.wolfram.com/DescartesSignRule.html

Answer 9

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 ?

Answer 10

exactly

Answer 11

ohh..thank you @mukushla
your little hint only helped me a lot :)

Answer 12

np...:)