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- Feb 14, 2012
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If \(\displaystyle f(x)=x^3-6x^2+17x\) and \(\displaystyle f(a)=16\) and \(\displaystyle f(b)=20\), find \(\displaystyle a+b\).
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 $a+b=4.$If \(\displaystyle f(x)=x^3-6x^2+17x\) and \(\displaystyle f(a)=16\) and \(\displaystyle f(b)=20\), find \(\displaystyle a+b\).
Hey anemone!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
\(\displaystyle a^3-6a^2+17a-16=0\) and \(\displaystyle b^3-6b^2+17b-20=0\),
We add the equations to find that
\(\displaystyle a^3+b^3-6a^2-6b^2+17a+17b-16-20=0\)
\(\displaystyle (a^3+b^3)-6(a^2+b^2)+17(a+b)-36=0\)
\(\displaystyle ((a+b)^3-3ab(a+b))-6((a+b)^2-2ab)+17(a+b)-36=0\)
By replacing \(\displaystyle k=a+b\) yields
\(\displaystyle (k^3-3abk)-6(k^2-2ab)+17k-36=0\)
This simplifies to
\(\displaystyle k^3-6k^2-36+(17-3ab)k+12ab=0\)
\(\displaystyle k^3-6k^2-36-(3ab-17)k+4(3ab-17)+4(17)=0\)
\(\displaystyle k^3-6k^2+32+(3ab-17)(4-k)=0\)
\(\displaystyle (k+2)(k-4)^2+(3ab-17)(4-k)=0\)
Therefore, we can conclude that \(\displaystyle k=4\) must be true.
This implies \(\displaystyle a+b=4\).
But hey, it is very obvious that my method is the least impressive/worst one...(bh)...