Question

Solve the following system of n quadratic equations in n variables $x_n, \: i=1,2,3,...,n-1,n.$

Given a ≠ 0. a, b and c ∈ R.

$a x_1^2+bx_1+c=x_2\\ax_2^2+bx_2+c=x_3\\ax_3^2+bx_3+c=x_4\\ \ : \ : \ \ \ \ :\\ax^2_{n-1}+bx_{n-1}+c=x_n\\ax^2_n+bx_n+c=x_1$

Answer 1

    We are given a set of n quadratic simultaneous equations in n unknown variables $x_i$. These are all symmetric in the variables. Important is that all equations have the same coefficients.

$Given \ a \neq 0. \: and \: a, b, c \in R.\\a x_1^2+bx_1+c = x_2 \\ax_2^2+bx_2+c = x_3 \\ax_3^2+bx_3+c = x_4\\ \ : \ : \ \ \ \ :\\ ax^2_{n-1}+bx_{n-1}+c =x_n \\ax^2_n+bx_n+c = x_1$

   Since the symmetry is wrt all variables and equations are cyclic, it is clear that the solution exists for $x_1=x_2=x_3=... \: = x_i \: .. \ =x_{n-1}=x_n$

Thus the equations can be written as :

$a \: x_i^2 + b x_i + c = x_i\\a x_i^2+(b-1)x_i+c=0\\x_i=roots=\frac{1}{2}[1-b+\sqrt{(b-1)^2-4ac} \: ] or \: \frac{1}{2a}[1-b+\sqrt{(b-1)^2-4ac} \: ]$

Now there exist three cases:    (all roots have same solutions)

1) (b-1)² > 4 a c

      The roots are real. So solution exists for given equations. There are two solutions for each variable. 

2)  (b-1)² = 4 a c
 
     The roots are real. There is only one solution, as roots are equal.
       each x = (1-b)/2a

3)  (b-1)² < 4 a c 

     The roots are imaginary. There exist two imaginary complex conjugate solutions for each variable.

Answer:
     Solutions are:   For i = 1, 2, 3, .... n
              x_i = [ (1-b) + √{(b-1)² - 4 a c } / 2a ]

Answer 2

★ QUADRATIC RESOLUTION ★