The discussion revolves around finding the sum \(a + b\) given the function \(f(x) = x^3 - 6x^2 + 17x\) and the conditions \(f(a) = 16\) and \(f(b) = 20\). Participants explore various methods to approach the problem, including graphical interpretations, algebraic manipulations, and assumptions about the nature of \(a\) and \(b\).
Participants generally arrive at the conclusion that \(a + b = 4\) through various methods, but there is disagreement regarding the uniqueness of this solution and the assumptions required for it to hold true.
Some participants express that the solution may depend on the assumption that \(a\) and \(b\) are real numbers, indicating that without this condition, the conclusion may not be valid.
The graph of $f$ has a rotational symmetry about the point $(2,18)$. In fact, you can check that $f(2-x) + f(2+x) = 18+18 =36$. But $36 = 16+20$, and it follows that if $f(a) = 16$ and $f(b) = 20$ then $a=2-x$ and $b=2+x$ for some $x$. Therefore $a+b=4.$anemone said:
...anemone said:As always, thanks to Opalg and Fernando for participating in this problem and to be honest, I learned a lot from both of the solutions.
I solved this problem differently and here goes my solution:
From
$$a^3-6a^2+17a-16=0$$ and $$b^3-6b^2+17b-20=0$$,
We add the equations to find that
$$a^3+b^3-6a^2-6b^2+17a+17b-16-20=0$$
$$(a^3+b^3)-6(a^2+b^2)+17(a+b)-36=0$$
$$((a+b)^3-3ab(a+b))-6((a+b)^2-2ab)+17(a+b)-36=0$$
By replacing $$k=a+b$$ yields
$$(k^3-3abk)-6(k^2-2ab)+17k-36=0$$
This simplifies to
$$k^3-6k^2-36+(17-3ab)k+12ab=0$$
$$k^3-6k^2-36-(3ab-17)k+4(3ab-17)+4(17)=0$$
$$k^3-6k^2+32+(3ab-17)(4-k)=0$$
$$(k+2)(k-4)^2+(3ab-17)(4-k)=0$$
Therefore, we can conclude that $$k=4$$ must be true.
This implies $$a+b=4$$.
But hey, it is very obvious that my method is the least impressive/worst one...(bh)