9 Coniche ESERCIZI To solve this system algebraically, it is sufficient to remember the relations that exist between the solutions of a second-degree equation: if given two numbers x and y we know the sum s the product p, then they are the solutions of the equation z2 sz + p = 0. Indicating with z1 and z2 the solutions of this equation, these values can be assigned both to x and to y and the solutions of the system are therefore: x1 = z1 x2 = z2 {y 1 = z 2 ; {y 2 = z 1 2. Another type of second-order symmetric system is the following: x+y=s {x2 + y2 = r2 It identifies the intersections between the circle with center O and the radius r and a straight line perpendicular to the bisector of the first and third quadrant. y O x In such a case, we can rewrite the second equation in this equivalent way: (x + y)2 2xy = r2 By substituting s with x + y we will have: s2 r2 s2 2xy = r2 2xy = s2 r2 xy = _ 2 In this way, we will obtain the following symmetric system: x+y=s s2 r2 with k = _ 2 For example, let s solve in the set of real numbers R the second-degree symmetric system: x2 + y2 = 290 {x + y = 2 In this specific case r2 = 290 and s2 = 4. The system is equivalent to the following: xy = 143 {x + y = 2 The solutions of the system can be traced back to those of the equation: z2 + 2z 143 = 0 which are z1 = 13 e z2 = 11. So, the system solutions are: x1 = 13 x2 = 11 ; {y1 = 11 {y2 = 13 {xy = k Exercises Solve the following symmetric systems: x+y=5 x+y=2 1. 2. {xy = 8 {x2 + y2 13 = 0 497