(x^2+2x+ 1)+(y^2-6y+9)=0
(x+ 1)^2+(y-3)^2=0
The square is greater than or equal to 0.
The total is 0.
So both squares are equal to 0.
So x+ 1 = 0 and y-3 = 0.
x=- 1,y=3
xy=-3
(ax+2)(x^2+ 12x- 1 1)
Item X 2 is
ax* 12x+2*x^2=( 12a+2)x^2
So 12a+2=0.
a=- 1/6
(a-2)^2(a^2-6a-9)-a(a^2-2a- 15)
=(- 1/6-2)^2*( 1/36+ 1-9)-(- 1/6)*(a-5)(a+3)
= 169/36*(-287/36)+ 1/6*(- 1/6-5)*(- 1/6+3)
=-5 1665/ 1296
Did you copy this question wrong?