A quartic function

anemone

MHB POTW Director
Staff member
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
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 anemone

MHB POTW Director
Staff member
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 $00$. Thus, $k(k-2)(k+1)<0$ $k^3-k^2-2k<0$ $k^3 Merging these two inequalities (*) and (**) yields$k^4<2k^2+4kk^4-3k^2+k-10<-k^2+5k-10$ 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 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. anemone

MHB POTW Director
Staff member
I have a slightly different approach...
Hey agentmulder, thanks for participating!

I say your method is more straightforward and smarter than mine, well done! Btw, would you mind to hide your solution? Thanks. 