Question 32335
y^2 -6y + a
=y^2 - 2*(3y) + a
=[y^2 - 2*y*3 + 3^2] + (a - 3^2)
=(y-3)^2 + (a-9)
For above to be a complete square, the remainder (a-9) has to be zero
Thus a-9 = 0
=> a=9