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Can you please give me some hints as to how I might approach this problem? Thanks!

Given the polynomial $f = 2X^3 - aX^2 - aX + 2, a \mathbb \in R$ and roots $x_1, x_2$ and $x_3,$ find $a$ such that $|x_1| + |x_2| + |x_3| = 3.$

Edit: We know $-1$ is one of the roots of that polynomial, regardless of the value of $a$. So, in essence, what we have to demonstrate is that $|x_2| + |x_3| = 2.$

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Without the absolute values, if we were just looking for $a$ such that $x_1+x_2+x_3=3$, I'd look to use the fact that in $a_3x^3+a_2x^2+a_1x+a_0$, the sum of the roots is $-\frac{a_2}{a_3}$ (think about expanding $a_3(x-x_1)(x-x_2)(x-x_3)$). With the absolute values, I'm less sure. –  Isaac May 22 '11 at 18:46
    
I suppose you meant to write $2X^3 - aX^2 - aX + 2$? Given this, you know that this must be equal to $2(X-x_1)(X-x_2)(X-x_3)$. Try expanding this latter expression and comparing the coefficients of $X^2, X,$ and $1$. –  Brad May 22 '11 at 18:48
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4 Answers

up vote -1 down vote accepted

The way to solve this is actually quite simple, apparently:

$$|x_2| + |x_3| =2 \implies -2 \le x_1 + x_1 \le 2$$

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I'm afraid I don't follow. Can you explain more? –  Michael Chen Jun 16 '11 at 14:11
    
@Michael I'm not sure what's to explain. :) Given any two real numbers $a$ and $b$, if $|a| + |b| = n$, then it's only logical that $-n \le a + b \le n$. (Where $n$ is obviously a positive real number.) –  Paul Manta Jun 16 '11 at 14:29
    
@Michael Or, in case you mean how this helps me solve the problem... The correct result (which I knew beforehand) is an interval. You have to arrive to this interval from $|x_2| + |x_3| = 2$. You rewrite that expression as I described above, and the result will be an interval, as expected. (You'll also have to make use of Fermat's relations to solve this, but those are also very simple.) –  Paul Manta Jun 16 '11 at 14:31
    
Why was this downvoted? –  Paul Manta Sep 15 '11 at 19:20
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We first have the following factorization,

$ 2x^3 - ax^2 - ax + 2 = (x+1)(2x^2 - (2+a)x + 2).$

Suppose $x_1 = -1$, then $|x_2| + |x_3| = 2$ and they are solutions to the quadratic equation $2x^2 - (2+a)x + 2 = 0$. Hence,

$ x_2 + x_3 = 1 + a/2 $ and

$ x_2 x_3 = 1$.

By observing the second equation, $x_2$ and $x_3$ are either both positive or both negative, so we have

$1 + a/2 = x_2 + x_3 = 2$ or $-2$.

Therefore, $a = 2$ or $-6$.

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You haven't explained why $x_2 + x_3$ is equal to either $2$ or $-2$. (You also seem to be assuming that they're real. I don't know if the OP is making that assumption or not.) –  Qiaochu Yuan May 23 '11 at 10:09
    
@Qiaochu Yuan Indeed, I don't make the assumption that they real. Hawii's demonstration, however -- using Viete's equations -- seems like a very simple approach. Why do you think it's incorrect? Remember, I have to find $a$ such that the sum is $2.$ –  Paul Manta May 23 '11 at 12:19
    
@Paul: I didn't say it was incorrect; it's incomplete. No justification is given for the claim that $x_2 + x_3 = 2$ or $-2$. Sivaram's argument shows that this is false in general, and in the case that $x_2, x_3$ are required to be real this is true but requires an argument that hasn't been given (AM-GM, for example). I am interpreting the question to mean "find all $a$ such that...." –  Qiaochu Yuan May 23 '11 at 12:20
    
@Qiaochu Yuan You are right, I did assume the roots are real. Thanks for pointing it out. –  Hawii May 24 '11 at 2:07
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HINT

$|x_1| + |x_2| + |x_3| \geq 3 \sqrt[3]{|x_1||x_2||x_3|} = 3$

Hence, we have $|x_1| + |x_2| + |x_3| \geq 3$. Equality holds implies $|x_1| = |x_2| = |x_3| = 1$

We have $x_1 + x_2 + x_3$, $x_1x_2 + x_2x_3 + x_3x_2$ and $x_1 x_2 x_3$ to be real, and further $x=-1$ satisfies the equation.

Hence $f(x) = 2 \left( x+1 \right) \left( x^2-\left(1 + \frac{a}{2} \right)x + 1 \right)$. We have $x_3 = -1$. And $|x_1| = 1$ and $|x_2| = 1$. Hence, $x_1 = e^{i \theta}$ and $x_2 = e^{i \phi}$.

$(x-x_1)(x-x_2) = \left( x^2-\left(1 + \frac{a}{2} \right)x + 1 \right)$

$x_1 x_2 = 1 \implies \phi = -\theta$

$x_1 + x_2 = 2 \cos(\theta) = 1 + \frac{a}{2}$

Hence, $\frac{a}{2} = 2 \cos(\theta) - 1 \implies a = 4 \cos(\theta) - 2$.

Hence, $a = 4 \cos(\theta) - 2$ and the roots are $-1,e^{i \theta},e^{-i \theta}$

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You can find such $a$ by inspection: Put $a=2$ and the roots are given by $$x^3 - x^2 - x + 1 = 0 \Leftrightarrow (x-1)^2(x+1) = 0.$$

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And, also by inspection, $a=-6$. –  André Nicolas May 22 '11 at 20:16
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