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- #1
- 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 \(\displaystyle a+b\).
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Find a + b.
- Thread starter anemone
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- #2
- Feb 7, 2012
- 2,818
Re: 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 $a+b=4.$
- Jan 29, 2012
- 661
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 $\neq 4$.
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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- #4
- Feb 14, 2012
- 3,963
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
\(\displaystyle a^3-6a^2+17a-16=0\) 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)
...
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)
...- Admin
- #5
- Mar 5, 2012
- 9,805
Re: Find a+b.
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.
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
Re: Find a+b.
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.
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.