What is the sum of α + β in these given equations?

[anemone]

Hi members of the forum,

Problem:
The real numbers $\displaystyle \alpha$, $\displaystyle \beta$ satisfy the equations

$\displaystyle \alpha^3-3\alpha^2+5\alpha-17=0,$

$ \displaystyle \beta^3-3\beta^2+5\beta+11=0.$

Find $\displaystyle \alpha + \beta$.

This problem has me stumped. It's not that I didn't try, believe you me, but I kept returning back to square one no matter how hard I tried. It's really annoying and I don't believe the author wants us to use the cubic formula to find each root explicitly (or even a numeric root-finding technique) to get the sum, but at this moment, I just don't see any way to manipulate the given two cubic polynomials to find the desired sum.

Could you please shed some light on this problem for me?

Thanks in advance.

 

[Sudharaka]

Hi members of the forum,

Problem:
The real numbers $\displaystyle \alpha$, $\displaystyle \beta$ satisfy the equations

$\displaystyle \alpha^3-3\alpha^2+5\alpha-17=0,$

$ \displaystyle \beta^3-3\beta^2+5\beta+11=0.$

Find $\displaystyle \alpha + \beta$.

This problem has me stumped. It's not that I didn't try, believe you me, but I kept returning back to square one no matter how hard I tried. It's really annoying and I don't believe the author wants us to use the cubic formula to find each root explicitly (or even a numeric root-finding technique) to get the sum, but at this moment, I just don't see any way to manipulate the given two cubic polynomials to find the desired sum.

Could you please shed some light on this problem for me?

Thanks in advance.

Hi anemone, :)

Let me outline the method of solving this problem. Firstly you don't know whether each of these equations have any real roots. So use the discriminant of each equation find out how many real roots each equation has (Refer >>this<<). You will see that each equation has only one real root.

Let \(\alpha+\beta=\lambda\). Add the two equations together and rearrange it to get an cubic equation of \(\lambda\). You should get,

\[\lambda^3-3\alpha\beta(\lambda-2)-3\lambda^2+5\lambda-6=0.\]

Try to find a value for \(\lambda\) that satisfies the above equation (by trial and error). Do you think that the value you obtained for \(\lambda\) is unique? If so why?

Kind Regards,
Sudharaka.

 

[Fernando Revilla]

Another way. Define $f(x)=x^3-3x^2-5x$. We have $f(\alpha)+f(\beta)=6$.We can exoress $f(x)=(x-1)^3+2(x-1)+3$, as a consequence
$$f(\alpha)-3=(\alpha-1)^3+2(\alpha -1)\\
f(\beta)-3=(\beta-1)^3+2(\beta-1)$$
Adding the above equalities we obtain $0=(\alpha-1)^3+(\beta-1)^3+2(\alpha+\beta-2)$. Now, decompose to get $0=(\alpha +\beta-2)g(\alpha,\beta)$ where $g(\alpha,\beta)>0$, so $\alpha+\beta=2$.

 

[anemone]

Hi anemone, :)

\[\lambda^3-3\alpha\beta(\lambda-2)-3\lambda^2+5\lambda-6=0.\]

Try to find a value for \(\lambda\) that satisfies the above equation (by trial and error). Do you think that the value you obtained for \(\lambda\) is unique? If so why?

Kind Regards,
Sudharaka.

Hi Sudharaka, my first instinct is to try the value of $\displaystyle \lambda=2$ and it does work. I think the value of $\displaystyle \lambda$ that I obtained is unique since the other two roots from both of these functions are imaginary roots. Am I correct?

Thanks for your answer and I appreciate it!:D

Another way. Define $f(x)=x^3-3x^2-5x$. We have $f(\alpha)+f(\beta)=6$.We can exoress $f(x)=(x-1)^3+2(x-1)+3$, as a consequence
$$f(\alpha)-3=(\alpha-1)^3+2(\alpha -1)\\
f(\beta)-3=(\beta-1)^3+2(\beta-1)$$
Adding the above equalities we obtain $0=(\alpha-1)^3+(\beta-1)^3+2(\alpha+\beta-2)$. Now, decompose to get $0=(\alpha +\beta-2)g(\alpha,\beta)$ where $g(\alpha,\beta)>0$, so $\alpha+\beta=2$.

Hi Fernando, it feels so great to receive two different methods in a day to solve my headache. Thanks enormously!

But another idea hit me when I was having my dinner tonight (I guess there is no rest for the weary.:p) and I'd really like to hear some opinions about it, if possible. (Smile)

Let $\displaystyle f(x)=x^3-3x^2+5x-17$ and $\displaystyle g(x)=x^3-3x^2+5x+11$ and if we rewrite the function of g(x) to become $\displaystyle g(x)=(x^3-3x^2+5x-17)+28$, we can see that the function of g(x) is just the function of f(x) that shifted 28 units upward, i.e. g(x) is the transformed function of f(x).

I then use a simple trick by exploiting the fact that the gradient of the function of f(x) and its transformed function, g(x) would be the same at their real root point.

That is, $\displaystyle f'(\alpha)=\displaystyle g'(\beta)$.
$\displaystyle 3\alpha^2-6\alpha+5=3\beta^2-6\beta+5$
$\displaystyle \alpha^2-2\alpha=\beta^2-2\beta$
$\displaystyle \alpha^2-\beta^2=2\alpha-2\beta$
$\displaystyle (\alpha+\beta)(\alpha-\beta)=2(\alpha-\beta)$
$\displaystyle (\alpha-\beta)(\alpha+\beta-2)=0$

Since $\displaystyle (\alpha-\beta)\ne 0$, we can conclude that $\displaystyle (\alpha+\beta-2)=0$, or $\displaystyle \alpha+\beta=2$.

Hmm...how does that sound to you?

 

[Fernando Revilla]

I then use a simple trick by exploiting the fact that the gradient of the function of f(x) and its transformed function, g(x) would be the same at their real root point.

Hmm, why? Consider for instance $f(x)=x^3-1$ and $g(x)=x^3-8$. Here $\alpha=1$ and $\beta=2$, however $f'(\alpha)=3$ and $g'(\beta)=12$ i.e. $f'(\alpha)\neq g'(\beta)$.

 

[Sudharaka]

Hi Sudharaka, my first instinct is to try the value of $\displaystyle \lambda=2$ and it does work. I think the value of $\displaystyle \lambda$ that I obtained is unique since the other two roots from both of these functions are imaginary roots. Am I correct?

Thanks for your answer and I appreciate it!:D

Yes that's correct. You are welcome! :)

 

[anemone]

Hmm, why? Consider for instance $f(x)=x^3-1$ and $g(x)=x^3-8$. Here $\alpha=1$ and $\beta=2$, however $f'(\alpha)=3$ and $g'(\beta)=12$ i.e. $f'(\alpha)\neq g'(\beta)$.

I'm so embarrassed to think of this now, as I really should have verified the result by using another pair of simple functions (such as your example) and I understand it now, these two functions (the original and its transformed function) have the same gradient at the same point.

Thanks for commenting on my idea!

Yes that's correct. You are welcome! :)

Thanks again, Sudharaka! ;)

 

[MarkFL]

I'm so embarrassed to think of this now...

Please don't feel embarrassed, we all make mistakes...I know I have made my share of mistakes, and the potential embarrassment is magnified when it is public, but when we come together in the forums to discuss a problem there will be mistakes, but we have our friends to guide us and hence we learn.

Others reading this topic can benefit from having seen your "mistake" and learn why this does not work. I feel that if one person makes an error, then there are others who will make this same error.

By being willing to take a chance and show your thinking, you have helped others.(Smile)

 

[anemone]

Hey Mark, it's so sweet and nice of you to say that!(Smile)

 

[Fernando Revilla]

I'm so embarrassed to think of this now, ...

The only way to not make mistakes is to never do anything. :)

 

[anemone]

The only way to not make mistakes is to never do anything. :)

I'm so touched by the genuine kindness shown by the community at this site, everyone here is so helpful and friendly...

Thanks to all of you, Mark, Fernando and Sudharaka because you all just made my day!:D