MHB Find m+n in $P(x)=x^3-6x^2+17x$ given $P(m)=16$ and $P(n)=20$

[anemone]

$P(x)=x^3-6x^2+17x$. If $x=m$, $P(m)=16$ and when $x=n$, $P(n)=20$.

Evaluate $m+n$.

Note: If you happen to see this problem posted by me before, I apologize (I did google it and I found no such thread exists) and I would appreciate it if you give me the link to it. If that is the case, I'd delete this one.

 

[kaliprasad]

Spoiler

we have $P(x) = x^3 - 6x^2 + 17x$

so we have $P(x+2) = x^3 + 5x + 18$ ( I take x+2 to eliminate the $x^2$ term to see in case we get odd function)

now
$P(m) = (m-2)^3 + 5(m-2) + 18 = 16$

or $(m-2)^3 +5 (m-2) = - 2\cdots (1)$

$P(n) = (n-2)^3 + 5(n-2) + 18 = 20$

or $(n-2)^3 + 5(n-2) = 2\cdots (2)$

from (1) and (2) as

$f(x) = x^3 + 5x$

$f(m-2) + f(n-2) = 0$ so $ m- 2 + n -2 = 0$

or $m+n = 4 $

 

[Albert1]

$P(x)=x^3-6x^2+17x$. If $x=m$, $P(m)=16$ and when $x=n$, $P(n)=20$.

Evaluate $m+n$.

Note: If you happen to see this problem posted by me before, I apologize (I did google it and I found no such thread exists) and I would appreciate it if you give me the link to it. If that is the case, I'd delete this one.

Spoiler

$p(m)=m^3-6m^2+17m=16----(1)$
$p(n)=n^3-6n^2+17n=20----(2)$
from (1) and (2) we get:
$p(m+n)=36+3mn(m+n-4)---(3)$
since $mn\neq 0$
from (3) :$p(4)=36$
we conclude $m+n$ must =4

 

[kaliprasad]

Spoiler

$p(m)=m^3-6m^2+17m=16----(1)$
$p(n)=n^3-6n^2+17n=20----(2)$
from (1) and (2) we get:
$p(m+n)=36+3mn(m+n-4)---(3)$
since $mn\neq 0$
from (3) :$p(4)=36$
we conclude $m+n$ must =4

how is

Spoiler

$p(4) = 36$

 

[Albert1]

how is

Spoiler

$p(4) = 36$

from (3)
p(m+n)=36+3mn(m+n-4)
when m+n=4
p(4)=36+3mn(4-4)=36
you can also put x=4 to :
$P(x)=x^3-6x^2+17x$ and check it
in fact there is no need to do it ,this must be true!

 

[anemone]

Spoiler

we have $P(x) = x^3 - 6x^2 + 17x$

so we have $P(x+2) = x^3 + 5x + 18$ ( I take x+2 to eliminate the $x^2$ term to see in case we get odd function)

now
$P(m) = (m-2)^3 + 5(m-2) + 18 = 16$

or $(m-2)^3 +5 (m-2) = - 2\cdots (1)$

$P(n) = (n-2)^3 + 5(n-2) + 18 = 20$

or $(n-2)^3 + 5(n-2) = 2\cdots (2)$

from (1) and (2) as

$f(x) = x^3 + 5x$

$f(m-2) + f(n-2) = 0$ so $ m- 2 + n -2 = 0$

or $m+n = 4 $

Well done, kaliprasad! You've made full use of the concept of the odd function in this problem and I learned something so useful from you today! Thanks for participating, thanks for showing us how we could think creatively and critically while solving the intriguing problem, thanks for everything, kali!:)

Spoiler

$p(m)=m^3-6m^2+17m=16----(1)$
$p(n)=n^3-6n^2+17n=20----(2)$
from (1) and (2) we get:
$p(m+n)=36+3mn(m+n-4)---(3)$
since $mn\neq 0$
from (3) :$p(4)=36$
we conclude $m+n$ must =4

Well done, Albert! I approached it using the same method as yours and once we know $P(4)=4^3-6(4)^2+17(4)=36$ and $P(m+n)=36+3mn(m+n-4)$, we can deduce that $m+n=4$.

 

[kaliprasad]

Well done, kaliprasad! You've made full use of the concept of the odd function in this problem and I learned something so useful from you today! Thanks for participating, thanks for showing us how we could think creatively and critically while solving the intriguing problem, thanks for everything, kali!:)

You are welcome. Persons like you always boost my morale and keep me active in MHB.