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How many ordered triplets \((a,b,c)\) of rational numbers are there where \(a,b,c\) are the roots of \(x^3 + ax^2 + bx + c = 0\)?

 May 12, 2019
 #1
avatar+22260 
+4

How many ordered triplets \((a,b,c) \) of rational numbers are there where\( a,b,c\) are the roots of \(x^3 + ax^2 + bx + c = 0\) ?

 

I assume:

\(\begin{array}{|rcll|} \hline x^3 + ax^2 + bx + c = 0 &=& (x-a)(x-b)(x-c) \\ &=& x^3\underbrace{-(a+b+c)}_{=a}x^2\underbrace{+(ab+ac+bc)}_{=b}x\underbrace{-abc}_{=c} \\ \hline \mathbf{-abc} &=& \mathbf{c} \\ -ab &=& 1 \\ \mathbf{ab} &=& \mathbf{-1} \\ \mathbf{b} &=& \mathbf{-\dfrac{1}{a}} \\\\ \mathbf{-(a+b+c)} &=&\mathbf{ a } \\ -a-b-c &=& a \\ b+c &=& -2a \quad | \quad \cdot a \\ \mathbf{ab+ac} &=& \mathbf{-2a^2} \quad | \quad ab=-1 \\ -1+ac &=& -2a^2 \\ \mathbf{ ac }&=& \mathbf{1-2a^2} \\\\ \mathbf{ab+ac+bc} &=& \mathbf{b} \quad | \quad ab+ac= -2a^2 \\ -2a^2+bc &=& b \\ -2a^2 &=& b- bc \\ -2a^2 &=& b(1-c) \quad | \quad b=-\dfrac{1}{a} \\ -2a^2 &=&-\dfrac{1}{a} (1-c) \\ 2a^2 &=& \dfrac{1}{a} (1-c) \\ 2a^3 &=& 1-c \\ \mathbf{c} &=& \mathbf{1-2a^3} \\\\ \mathbf{ ac }&=& \mathbf{1-2a^2} \quad | \quad c=1-2a^3 \\ a(1-2a^3)&=& 1-2a^2 \\ -2a^4+a &=& 1-2a^2 \\ \mathbf{-2a^4+2a^2+a-1} &=& \mathbf{0} \\ \hline \end{array}\)

 

\(\begin{array}{|lcll|} \hline \mathbf{-2a^4+2a^2+a-1=0},\ b=-\dfrac{1}{a},\ c=1-2a^3 \\\\ a = 1,\ b = -1,\ c = -1 \\ \text{triplet}_1 ~(a,b,c) = (1,\ -1,\ -1 ) \\\\ a =0.56519771738363939644,\ b=-1.7692923542386314152,\ c=0.6388969194713526224 \\ \text{triplet}_2 ~(a,b,c) = (0.56519771738363939644,\ -1.7692923542386314152,\ 0.6388969194713526224 ) \\ \hline \end{array} \)

 

check:

\(\begin{array}{|lcll|} \hline \mathbf{x^3+ x^2-x-1=0} \\ x = -1 \\ x = 1 \\ \hline \mathbf{x^3+ 0.56519771738363939644x^2-1.7692923542386314152x+0.638896919471352622=0} \\ x=-1.76929235423863142 \\ x=0.5651977173836394 \\ x=0.6388969194713526 \\ \hline \end{array}\)

 

laugh

 May 14, 2019
 #2
avatar
-2

Heureka there is only one such triplet

Guest May 14, 2019
 #3
avatar+101709 
+2

If you are going to call Heureka out you need justify yourself.

Melody  May 14, 2019
 #4
avatar
+1

Just tidying up, (I'm not guest #2).

The second value of a is

\(\displaystyle \frac{1}{9u}+u-\frac{1}{3}, \text{ where }\quad u=\frac{(3\sqrt{57}+23)^{1/3}}{3\times4^{1/3}}\).

So, not rational.

 May 15, 2019

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