MHB Prove No Root in $(0,2)$ for $k^4-10k+k-3k^2$

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The discussion focuses on proving that the quartic function k^4 - 10k + k - 3k^2 has no roots in the interval (0,2). Participants share their approaches, with one user acknowledging another's method as more straightforward and effective. There is a request to hide a solution for clarity. The conversation emphasizes the importance of rigorous proof in mathematical discussions. The thread highlights collaborative problem-solving in the context of polynomial root analysis.
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Prove that there is no root exists in the interval $(0,2)$ for a quartic function $k^4-10+k-3k^2$.
 
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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
 
anemone said:
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 $0<k<2$,

[TABLE="class: grid, width: 700"]
[TR]
[TD]We have $k(k-2)<0$.[/TD]
[TD]Besides, we also have $k^3(k-2)<0$[/TD]
[/TR]
[TR]
[TD]Adding 1 to the inequality $0<k<2$ we get $1<k+1<3$.
Or simply $k+1>0$.

Thus,

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

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

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

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

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

$k^4<2k^3$(**)[/TD]
[/TR]
[/TABLE]

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

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

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

View attachment 1753

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 $k^4-10+k-3k^2$.
 

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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
 
agentmulder said:
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.:o
 
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