A quartic function

[anemone]

Prove that there is no root exists in the interval $(0,2)$ for a quartic function $k^4-10+k-3k^2$.

 

[mathbalarka]

Spoiler

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]

Prove that there is no root exists in the interval $(0,2)$ for a quartic function $k^4-10+k-3k^2$.

My solution:

Spoiler

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

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

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



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]

I have a slightly different approach.

Spoiler

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

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]

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.