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This is my exercise

$x^3 - 3 x^2 + 6 x - a$, the polynom is from $\mathbb{R}[X]$.

And my job is to calculate $x_1^2+x_2^2+x_3^2$. I solved it and I get $-3$ as a result.

My question is: can 3 numbers $a^2+b^2+c^2$ be lower than $0$? Because the power is supposted to make all 3 of them positive.

AugSB
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B.Noc
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  • This is hard to follow. What are $x_1,x_2,x_3?$ what is it you solved? What are $a,b,c$? What are you asking? – lulu Apr 07 '17 at 16:02
  • those 3 are the roots. i calculated x1^2+x2^2+x3^2 and i don't understand how can their sum be lower than 0 in our current case -3 – B.Noc Apr 07 '17 at 16:02
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    But the roots need not all be real. – quasi Apr 07 '17 at 16:02
  • isn't that R[X] supposted to mean that the roots are real? – B.Noc Apr 07 '17 at 16:04
  • No, it means the coefficients are real. Example: The polynomial $x^2 +1$ is an element of $\mathbb{R}[x].$ – quasi Apr 07 '17 at 16:04
  • The sum of the roots is $3$, obviously. Just expand $(x-x_1)(x-x_2)(x-x_3)$. But you appear not to want the sum of the roots but the sum of the squares. Maybe you could edit your question? Explain your calculation? – lulu Apr 07 '17 at 16:05
  • Since the coefficient of $x$ equals $x_1x_2+x_1x_3+x_2x_3 = 6 $ we have that the sum of the squares of the roots is $3^2-2.6 = -3$ – Marc Bogaerts Apr 08 '17 at 03:45

1 Answers1

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your answer is correct

your polynomial is : $x^3 - 3 x^2 + 6 x - a $

and its coefficients are : $a_0=a$, $a_1=6$, $a_2=-3$, $a_3=1$

$\sigma_k = (-1)^k \frac{a_{3-k}}{a3} =(-1)^ka_{3-k} $ for $k=1,2,3$

$$x_1+x_2+x_3=\sigma_1 = -a_2=3$$ $$x_1x_2+x_1x_3+x_2x_3 =\sigma_2 = a_1=6$$ $$(x_1+x_2+x_3)^2= x_1^2+x_2^2+x_3^2+2(x_1x_2+x_1x_3+x_2x_3) \implies x_1^2+x_2^2+x_3^2 = \sigma_1^2 - 2\sigma_2= 9-12 =-3 $$

and to answer your question yes it can be less than $0$ if $a,b,c \in \mathbb{C}$

for example : $a = i, b = 2i, c = -1$ $a^2+b^2+c^2= i^2+4i^2+(-1)^2=-1-4+1=-4$

the_firehawk
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