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- Feb 14, 2012

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- Thread starter anemone
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- #1

- Feb 14, 2012

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- Feb 7, 2012

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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 $a+b=4.$

- Jan 29, 2012

- 661

Another way. We have $$f(a)=16\Leftrightarrow (a-2)^3+5a-8=0\\f(b)=20\Leftrightarrow (b-2)^3+5b-12=0$$ Denoting $\alpha=a-2,\;\beta =b-2$: $$f(a)=16\Leftrightarrow \alpha^3+5\alpha+2=0\\f(b)=20\Leftrightarrow \beta^3+5\beta-2=0$$ But if $r$ is a root of $x^3+5x+2$ iff $-r$ is a root of $x^3+5x-2$ so, $$0=\alpha+\beta=a-2+b-2\Rightarrow a+b=4$$ P.S. Of course, this is valid for $a,b\in\mathbb{R}$, otherwise $a+b$ can be $\neq 4$.

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- Feb 14, 2012

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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)...

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- Mar 5, 2012

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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)...

I'm afraid your solution has a couple of flaws in it. (bh)

The solution \(\displaystyle k=4\) is only 1 of the possible solutions.

There may be more solutions (although there aren't any).

Furthermore, it is only a solution for the sum of the 2 equations being equal to zero.

There is no guarantee that the individual equations are zero.

They may still have opposite results (although they don't).

Sorry.

- Jan 29, 2012

- 661

The problem should say:

If \(\displaystyle f(x)=x^3-6x^2+17x\) and \(\displaystyle f(a)=16\) and \(\displaystyle f(b)=20\) with $a,b\in\mathbb{R}$, find \(\displaystyle a+b\).

Oherwise, we can't guarantee $a+b=4$. I solved the question with that additional hypothesis.