Welcome to our community

Be a part of something great, join today!

A quartic function

  • Thread starter
  • Admin
  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,756
Prove that there is no root exists in the interval $(0,2)$ for a quartic function $k^4-10+k-3k^2$.
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
Sturm's theorem gives 0. But I have no intention of calculating here that with whole details as I have done in my notebook, so I simply leave it to someone else :D
 
  • Thread starter
  • Admin
  • #3

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,756
Prove that there is no root exists in the interval $(0,2)$ for a quartic function $k^4-10+k-3k^2$.
My solution:

If we let $0<k<2$,

We have $k(k-2)<0$.Besides, we also have $k^3(k-2)<0$
Adding 1 to the inequality $0<k<2$ we get $1<k+1<3$.
Or simply $k+1>0$.

Thus,

$k(k-2)(k+1)<0$

$k^3-k^2-2k<0$

$k^3<k^2+2k$

$2k^3<2k^2+4k$ (*)
Expanding the inequality we get

$k^4-2k^3<0$

$k^4<2k^3$(**)

Merging these two inequalities (*) and (**) yields

$k^4<2k^2+4k$

$k^4-3k^2+k-10<-k^2+5k-10$

Graph of a quadratic function.JPG

From the graph, we can tell $-k^2+5k-10<0$ for $0<k<2$, hence, $k^4-3k^2+k-10<0$ for $0<k<2$ and we can conclude there is no root exists in the interval $(0,2)$ for a quartic function $k^4-10+k-3k^2$.
 

agentmulder

Active member
Feb 9, 2012
33
I have a slightly different approach.

$k^4 -10 + k - 3k^2$

Re-arrange ,

$f(k) = k^2(k^2 - 4) + (k^2 + k - 10)$

If we can show the function is negative for $0<k<2$ then we are done. This should be obvious from f(k) since both expressions within parenthesis are negative in the interval $0<k<2$ , therefore it cannot be 0 in that interval.

:D
 
  • Thread starter
  • Admin
  • #5

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,756
I have a slightly different approach...
Hey agentmulder, thanks for participating!

I say your method is more straightforward and smarter than mine, well done!:cool:

Btw, would you mind to hide your solution? Thanks.:eek: