Answer:
The equations are :
y=x²+x-2
x+y=1⇒ y= 1-x
multiply(y=1-x) by -1 then add it to y= x²+x-2
you get : x²+x-2+x-1=0⇒ x²+2x-3 = 0
let Δ be the dicriminat of this equation :
Δ = 2²-4*1*(-3)= 16
so this equation has two solutions
x= (-2-4)/2 = -3 or x=(-2+4)/2 = 1
then :
- if x= -3 y= 4 y= 1+3
- if x = 1 y= 0
now the possible solutions are (-3,4) and (1,0)
the possible values of xy are :
- -12 : -3*4= -12
- 0 : -3*0= 0
- 4 : 4*1= 4
- 0 : 1*0=0
so xy could be : 0 or 4 or -12
