Y=x2+x-2 x+y=1
If(x, y)is a solution of the system of equtions abpve which of the following is possible value of xy?

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 :

now the possible solutions are (-3,4) and (1,0)

the possible values of xy are :

so xy could be : 0 or 4 or -12

answer: 1

explanation: 1