Question

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?

Answer 1

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 :

• a= 1
• b= 2
• c= -3

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