perplexus.info :: Just Math : Real Root Rejection

Real Root Rejection (Posted on 2024-11-28)

Show that the polynomial x⁸-x⁷+x²-x+15 has no real roots.

No Solution Yet Submitted by Danish Ahmed Khan

No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)

Possible Solution

| Comment 2 of 8 |

x⁸-x⁷+x²-x+15 can be rewritten as (x⁸-x⁷)+(x²-x)+15.

so we are looking for the roots when (x⁸-x⁷)+(x²-x)=-15, with the even powers always positive.

If x>1, then we have two positive terms in parentheses equalling a negative number, so no real roots. The same if x<-1.

if 0<x<1, then the parts in parentheses will again be positive, so again no real roots. The same if -1<x<0.

if x=1, the equation resolves to 0=-15, which is not a solution.

Posted by broll on 2024-11-29 03:08:39

Please log in:

Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (5)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:


perplexus dot info