Utilizamos cookies para personalizar el contenido y la publicidad y para analizar el acceso a nuestro sitio web. Además, nuestros socios para la publicidad online reciben información pseudónima sobre su uso de nuestra página web. políticas sobre cookies y Políticas de privacidad.
 

heureka

avatar
Nombre de usuarioheureka
Puntuación22152
Stats
Preguntas 11
Respuestas 4232

 #1
avatar+22152 
+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

14-may-2019
 #2
avatar+22152 
+2
14-may-2019
 #1
avatar+22152 
+2

Let
\(f(x)=(x^2+6x+9)^{50}-4x+3\), and let \(r_1,r_2,\ldots,r_{100}\) be the roots of \(f(x)\).
Compute \((r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100}\).

 

\(\begin{array}{|rcll|} \hline \mathbf{f(x)} &=& \mathbf{(x^2+6x+9)^{50}-4x+3} \\ &=& \left((x+3)^2\right)^{50}-4x+3 \\ \mathbf{f(x)} &=& \mathbf{\left( x+3 \right)^{100}-4x+3} \\ \hline \end{array} \)

 

\(\begin{array}{|lrcll|} \hline & f(r_1) =0 &=& \left( r_1+3 \right)^{100}-4r_1+3 \\ & f(r_2) =0 &=& \left( r_2+3 \right)^{100}-4r_2+3 \\ & \ldots \\ & f(r_{100}) =0 &=& \left( r_{100}+3 \right)^{100}-4r_{100}+3 \\ & \hline \\ \text{sum} & 0 &=& (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} \\ & && -4(r_1+r_2+\ldots + r_{100}) + 3\cdot 100 \\ \hline \end{array} \)

 

\(\begin{array}{|lrcll|} \hline (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} &=& 4(r_1+r_2+\ldots + r_{100}) - 300 \\ \hline \end{array} \)

 

\(\mathbf{\text{vieta:}}\)

For any polynomial equation
\(0=x^n+a_{n-1}·x^{n-1}+...+a_2·x^2+a_1·x^1+a_0 \)
with the solutions \(r_1\dots r_n\), the relatively simple formulas for \(a_0\) and \(a_{n-1}\) are:

   \(a_0=(-1)^n \prod\limits_{k=1}^{n} r_k \\ a_{n-1}= -\sum \limits_{k=1}^{n} r_k\)

 

\(\begin{array}{|rcll|} \hline && \left( x+3 \right)^{100} \\ \\ &=& x^{100} + \binom{100}{1}x^{99}\cdot 3 + \ldots \\ \\ &=& x^{100} + \underbrace{300}_{ \underbrace{=a_{n-1}}_{= -(r_1+r_2+\ldots + r_{100})}} x^{99} + \ldots \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} &=& 4(r_1+r_2+\ldots + r_{100}) - 300 \\ && \boxed{r_1+r_2+\ldots + r_{100} = -300} \\ (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} &=& 4(-300) - 300 \\ \mathbf{(r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100}} &=& \mathbf{-1500} \\ \hline \end{array} \)

 

 

laugh

14-may-2019