Question

A10/171 Sa se rezolve sistemele de ecuatii folosind notatia de tipul x+y=s, xy=p.
d)2x^2+3xy+2y^2=4
x+y-xy=1

Answer 1

Metoda simplă:

2x²+3xy+2y² = 4

x+y-xy = 1

x+y-xy = 1 => x+y(1-x) = 1 =>

=> y(1-x) = 1-x => y = 1 când x ≠ 1.

=> 2x²+3xy+2y² = 4

=> y = 1

=> 2x²+3x+2 = 4

=> 2x²+3x-2 = 0

Delta = 9 + 16 = 25

=> x1,2 = (-3±5)/4

=> x1 = -2, x2 = 1/2

=> (x,y) = { ( -2 ; 1) , (1/2, 1) }

Dar sistemul e simetric =>

=> (x,y) = { (-2 ; 1), (1; -2), (1/2; 1), (1, 1/2) }

Metoda cu s si p:

2x²+3xy+2y² = 4

x+y-xy = 1

x²+2xy+y²+x²+2xy+y² - xy = 4 =>

=>(x+y)²+(x+y)² - xy = 4

2(x+y)² - xy = 4

x+y - xy = 1

x+y = s

xy = p

=> 2s²-p = 4

=> s - p = 1

(scadem)

2s² - s = 3 => 2s² - s - 3 = 0

Delta = 1+24 = 25 => s1,2 = (1±5)/4

=> s1 = -1, s2 = 3/2

=> -1 - p1 = 1 => p1 = -2

=> 3/2 - p2 = 1 => p2 = 1/2

(1) s1 = -1, p1 = -2

=> t²-(-1)t-2 = 0 => t²+t-2 = 0

Delta = 1+8 = 9 => t1,2 = (-1±3)/2

=> t1 = -2, t2 = 1

(2) s2 = 3/2, p2 = 1/2

=> t²-(3/2)t+1/2 = 0 =>

= 2t²-3t+1 = 0

Delta = 9-8 = 1 => t1,2 = (3±1)/4 =>

=> t1 = 1/2, t2 = 1

=> (x,y) = { (-2 ; 1), (1; -2), (1/2; 1), (1, 1/2) }