Finding a+b Given f(x)=x^3-6x^2+17x

[anemone]

If $$f(x)=x^3-6x^2+17x$$ and $$f(a)=16$$ and $$f(b)=20$$, find $$a+b$$.

 

[Opalg]

Re: Find a+b.

If $$f(x)=x^3-6x^2+17x$$ and $$f(a)=16$$ and $$f(b)=20$$, find $$a+b$$.

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 $a+b=4.$

 

[Fernando Revilla]

Re: Find a+b.

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 $\neq 4$.

 

[anemone]

Re: Find a+b.

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)...

 

[I like Serena]

Re: Find a+b.

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)...

Hey anemone! ;)

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

The solution $$k=4$$ 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.

 

[Fernando Revilla]

Re: Find a+b.

The problem should say:

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

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