6
$\begingroup$

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 $|x_2| + |x_3| = 2.$

  • 0
    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 $1$.2011-05-22

4 Answers 4

-1

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

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

  • 0
    Why was this downvoted?2011-09-15
8

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 $-1,e^{i \theta},e^{-i \theta}$

2

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 $-6$.

  • 0
    @Qiaochu Yuan You are right, I did assume the roots are real. Thanks for pointing it out.2011-05-24
0

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

  • 0
    And, also by inspection, $a=-6$.2011-05-22