MathLabs

Problem 3

Consider the following system of equations in the unknowns x1,x2,…,xnx_1, x_2, \ldots, x_n, where a,b,ca, b, c are real numbers with a≠0a \neq 0: ax12+bx1+c=x2,ax22+bx2+c=x3,…,axn−12+bxn−1+c=xn,axn2+bxn+c=x1.ax_1^2+bx_1+c=x_2,\quad ax_2^2+bx_2+c=x_3,\quad \ldots,\quad ax_{n-1}^2+bx_{n-1}+c=x_n,\quad ax_n^2+bx_n+c=x_1. Let Δ=(b−1)2−4ac\Delta=(b-1)^2-4ac. Prove that: (a) if Δ<0\Delta<0, the system has no solution; (b) if Δ=0\Delta=0, the system has exactly one solution; (c) if Δ>0\Delta>0, the system has more than one solution.
Step 2 of 5: Δ is exactly the discriminant of s
Previous stepNext step
s(t):=at2+(b−1)t+c,Δ=(b−1)2−4acs(t):=at^2+(b-1)t+c,\qquad \Delta=(b-1)^2-4ac
The auxiliary parabola s(t) = t² − t − 2, an illustrative case with a = 1, b = 0, c = −2 and Δ = 9 > 0, crossing the axis at t = −1 and t = 2
An upward-opening parabola s(t) = t squared minus t minus 2, crossing the horizontal axis at t = -1 and t = 2, staying negative between these two roots and positive outside them.
Detailed analysis

The quadratic s(t)=at2+(b−1)t+cs(t)=at^2+(b-1)t+c has discriminant (b−1)2−4ac(b-1)^2-4ac, which is exactly the Δ\Delta given in the problem. Since a≠0a\neq0, the sign of Δ\Delta completely controls how many real roots ss has: none if Δ<0\Delta<0, one (a double root) if Δ=0\Delta=0, two if Δ>0\Delta>0. Between any two real roots (or everywhere, if there are none) ss keeps a constant sign.