- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,840

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

- Thread starter anemone
- Start date

- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,840

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

- Mar 22, 2013

- 573

- Thread starter
- Admin
- #3

- Feb 14, 2012

- 3,840

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

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

- Feb 9, 2012

- 33

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.

- Thread starter
- Admin
- #5

- Feb 14, 2012

- 3,840

HeyI have a slightly different approach...

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

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