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What's $\alpha+\beta$ if we have $\alpha^3-6\alpha^2+13\alpha=1$ and $\beta^3-6\beta^2+13\beta=19$? Here $\alpha$ and $\beta$ are real.

Firstly, I subtracted the two equations and got the following: $$\alpha^3-\beta^3-6(\alpha^2-\beta^2)+13(\alpha-\beta)=-18$$ Then I tried to factorize the left hand side as: $$(\alpha-\beta)(\alpha^2+\alpha\beta+\beta^2-6\alpha-6\beta+13)=-18$$ At this point it seems we can't go on! Then I tried adding the two equations as: $$(\alpha+\beta)(\alpha^2-\alpha\beta+\beta^2-6\alpha-6\beta+13)+12\alpha\beta=20$$ unfortunately, again I can't continue!
Is there any special creativity needed for solving this question?

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  • $\begingroup$ By numerical computation, the answer seems to be 4. $\endgroup$ Feb 18, 2016 at 10:35
  • $\begingroup$ Thank you , but this is a pure algebraic question $\endgroup$ Feb 18, 2016 at 10:36
  • $\begingroup$ If you're really desperate to solve this, you can always simply find $\alpha$ and $\beta$ using the cubic formula. $\endgroup$ Feb 18, 2016 at 10:52
  • $\begingroup$ Have you tried simply solving each equation seperately, thus obtaining $\alpha$ and $\beta$? $\endgroup$
    – Eric S.
    Feb 18, 2016 at 11:02
  • $\begingroup$ Yes,but again it seems difficult! $\endgroup$ Feb 18, 2016 at 11:03

2 Answers 2

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In order to get rid of the quadratic terms we write $\alpha:=a+2$, $\beta:=b+2$. Then $$1=\alpha^3-6\alpha^2+13\alpha=(a+2)^3-6(a+2)^2+13(a+2)=a^3+a+10\ ,$$ and similarly $$19=\beta^3-6\beta^2+13\beta=b^3+b+10\ .$$ It follows that $a$ and $b$ satisfy the equations $$a^3+a=-9,\quad b^3+b=9\ .$$ Now (an unexpected) symmetry comes to our rescue: As $t\mapsto t^3+t$ is monotonically increasing there is exactly one $b=:b_*>0$ satisfying the second of these equations, and it is then obvious that $a:=-b_*$ is the unique solution of the first of these equations. It follows that necessarily $$\alpha+\beta=-b_*+2 +b_*+2=4\ .$$

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    $\begingroup$ How could you guess that substitution? $\endgroup$ Feb 18, 2016 at 11:29
  • $\begingroup$ A revision of last part of Christian's method: add the two equations as: $$a^3+b^3+a+b=0$$ , then factorize left hand as: $$(a+b)(a^2-ab+b^2+1)=0$$ , so $a+b=0$ , then: $\alpha+\beta=4$ $\endgroup$ Feb 18, 2016 at 11:42
  • $\begingroup$ @HamidRezaEbrahimi, I think the idea is to use : en.wikipedia.org/wiki/… $\endgroup$ Feb 18, 2016 at 12:03
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Let $f(x)=x^3-6x^2+13x,$ Then $f'(x)=3x^2-12x+13 = 3[(x-2)^2]+1>0$

So $f(x)$ is strictly increasing function.

Now above we have given $f(\alpha)=1$ and $f(\beta) = 19\;,$ Where $\alpha<\beta$

Then $f(\alpha)<f(\beta)$, because the function is increasing.

Now \begin{align*}f(4-x)&=(4-x)^3-6(4-x)^2+13(4-x)\\&=64-x^3-48x+12x^2-96-6x^2+48x+52-13x\\&=-x^3+6x^2-13x+20=-f(x)+20\end{align*} So if $x=\alpha$, Then $f(4-\alpha)=-f(\alpha)+20=19=f(\beta)$

The function is continuous and monotonically increasing, thus $f(x)=f(y)\implies x=y$, hence $$4-\alpha=\beta\Rightarrow \boxed{\alpha+\beta=4}$$

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  • $\begingroup$ Is there any simpler technique for this question?? Because the audience might not know derivative!! $\endgroup$ Feb 18, 2016 at 11:01
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    $\begingroup$ To Hamid Reza Ebrahimi , Iam also trying for that, Thanks $\endgroup$
    – juantheron
    Feb 18, 2016 at 11:04
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    $\begingroup$ Why did you choose to consider f(x-4) ? $\endgroup$
    – Saikat
    Feb 18, 2016 at 11:16
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    $\begingroup$ @user230452, $f(x)=x^3-6x^2+13x=x\left[(x-3)^2+4\right]$, thus $f(q-x)=\dots=-x^3+(3q-6)x^2-(3q^2-12q+13)x+f(q)$. Let's assume we want to get $f(q-x)=f(q)-f(x)$, thus we need $-x^3+(3q-6)x^2-(3q^2-12q+13)x=-x^3+6x^2-13x$. If we equating coefficients we get $q=4$, thus $f(4-x)=20-f(x) \Rightarrow f(4-\alpha)=20-f(\alpha)=f(\beta)\Rightarrow 4-\alpha=\beta$. $\endgroup$
    – Galc127
    Feb 18, 2016 at 12:55

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