+0  
 
0
54
2
avatar

Part a) In this multi-part problem, we will consider this system of simultaneous equations: 

3x+5y-6z=2,
5xy-10yz-6xz=-41
xyz=6.

Let a=3x, b=5y, and c=-6z.

Determine the monic cubic polynomial in terms of a variable t whose roots are a, b, and c. Make sure to enter your answer in terms of t and only t, in expanded form.

 

Part b) Given that (x,y,z) is a solution to the original system of equations, determine all distinct possible values of x+y.

 

I've solved part a and got the equations a+b+c=2, ab+bc+ac=-123, and abc=-540. Can someone help with part b?

Guest Aug 20, 2018
 #1
avatar
+1

b)

My CAS system gives the follwing values:
x = -4, y = 1, z = -3/2
x = -4, y = 9/5,  z = -5/6
x = 5/3, y = -12/5,  z = -3/2
x = 5/3,  y = 9/5,  z = 2
x = 3,  y = -12/5,  z = -5/6
x = 3,  y = 1,  z = 2

Guest Aug 20, 2018
edited by Guest  Aug 20, 2018
 #2
avatar+19992 
+2

Part a)
In this multi-part problem, we will consider this system of simultaneous equations:

 

\(\begin{array}{lcr} 3x + 5y - 6z &=& 2 \\ 5xy - 10yz - 6xz &=& -41 \\ xyz &=& 6 \\ \end{array}\)

 

\(\text{Let $a = 3x$, $b = 5y$, and $c = -6z.$}\)

 

Determine the monic cubic polynomial in terms of a variable \(t\) whose roots are \(a\), \(b\), and \(c\).
Make sure to enter your answer in terms of \(t\) and only \(t\), in expanded form.

 

\(\begin{array}{|rcll|} \hline (t-a)(t-b)(t-c) &=& 0 \quad | \quad a=3x, \quad b = 5y \quad c=-6z \\ \left(t^2-t(a+b)+ab\right)(t-c) &=& 0 \\ t^3-t^2c-t^2(a+b)+t(a+b)c+tab-abc &=& 0 \\ t^3-t^2(a+b+c)+t(ac+bc+ab)-abc &=& 0 \\ \\ \hline a+b+c &=& 3x + 5y -6z \\ &=& 2 \\ \\ \hline ac+bc+ab &=& -18xz-30yz+15xy \\ &=& 3\cdot(5xy-10yz-6xz) \\ &=& 3\cdot(-41) \\ &=& -123 \\ \\ \hline abc &=& 3x5y(-6z) \\ &=& -90xyz \\ &=& -90\cdot 6 \\ &=& -540 \\ \\ \hline t^3-t^2(a+b+c)+t(ac+bc+ab)-abc &=& 0 \\ t^3-t^2\cdot 2+t(123)-(-540) &=& 0 \\ \mathbf{t^3-2t^2 -123t +540} & \mathbf{=} & \mathbf{0} \\ \text{The roots are } t_1 = -12, t_2 = 5, t_3 = 9 \\ \hline \end{array}\)

 

 

Part b)
Given that \((x,y,z)\) is a solution to the original system of equations,

determine all distinct possible values of \(x+y\).

 

Roots are \(a\), \(b\), and \(c\)

 

\(\mathbf{x =\ ?}\)

\(\begin{array}{|rcll|} \hline a = 3x_1 &=& t_1 \\ 3x_1 &=& -12 \\ x_1 &=& -4 \\ \\ a = 3x_2 &=& t_2 \\ 3x_2 &=& 5 \\ x_2 &=& \frac53 \\\\ a = 3x_3 &=& t_3 \\ 3x_3 &=& 9 \\ x_3 &=& 3 \\\\ \hline x &=& \{-4,\frac53,3 \} \\ \hline \end{array} \)

 

\(\mathbf{y =\ ?}\)

\(\begin{array}{|rcll|} \hline b = 5y_1 &=& t_1 \\ 5y_1 &=& -12 \\ y_1 &=& -\frac{12}{5} \\ \\ b = 5y_2 &=& t_2 \\ 5y_2 &=& 5 \\ y_2 &=& 1 \\\\ b = 5y_3 &=& t_3 \\ 5y_3 &=& 9 \\ y_3 &=& \frac95 \\\\ \hline y &=& \{-\frac{12}{5},1,\frac95 \} \\ \hline \end{array}\)

 

\(\mathbf{z =\ ?}\)

\(\begin{array}{|rcll|} \hline c = -6z &=& t_1 \\ -6z &=& -12 \\ z_1 &=& 2 \\ \\ c = -6z &=& t_2 \\ -6z &=& 5 \\ z_2 &=& -\frac{5}{6} \\\\ c = -6z &=& t_3 \\ -6z &=& 9 \\ z_3 &=& -\frac32 \\\\ \hline z &=& \{2,-\frac{5}{6},-\frac32 \} \\ \hline \end{array}\)

 

Solutions:

\(\begin{array}{|r|r|r|r|} \hline x & y & z & \text{all distinct possible values of } x+y \\ \hline -4 & 1& -\frac32 & -3 \\ \hline -4 & \frac95& -\frac56 & -2.2 \\ \hline \frac53&-\frac{12}{5}& -\frac32 & -\frac{11}{15} \\ \hline \frac53& \frac95& 2 & \frac{52}{15} \\ \hline 3&-\frac{12}{5}& -\frac56 & \frac35 \\ \hline 3& 1& 2 & 4 \\ \hline \end{array}\)

 

laugh

heureka  Aug 21, 2018

15 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.