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I agree with mirhaque on answer choice (E). How did you get 3 for r, thearch? The equation can be expanded to:

x^2+ax+b = x^2+2rx+r^2

I canceled out x^2 on both sides, leaving:

ax+b = 2rx+r^2

Statement (I) says b=9, but that still leaves you with 3 unknowns.
Statement (II) says a=-6, and that still leaves you with 3 unknowns.
Combinining (I) and (II) still leaves 2 unknowns.

my reasoning was:
distinguish between two sides of the equation
so you resolve the first
x^2-6x+9=0 --> only one root because 36-4*9=0
so the solution of it is (6+-0)/2=3 (X found)
then you have
(x+r)^2=0 -> r=-3

but I'm now considering that the stem doesn't tell us that x^2-6x+9=(x+r)^2=0 and without that you remain with two unknowns so it should be (E)

Soln: Lets first open up the given equation, => x^2+ax+b=(x+r)^2 => x^2+ax+b=x^2+2rx+r^2

now since coefficients of x^2 are equal, so matching the other coefficients on both sides, ax = 2rx => a = 2r => eq(1) b = r^2 => eq(2)

Considering statement 1 alone, since b = 9 and we have b = r^2 therefore r = +3 or -3 Since we have two values of r, Statement 1 alone is insufficient

Considering statement 2 alone, since a = -6 and we have a = 2r therefore r = -3 Since we have just one unique value of r, Statement 2 alone is sufficient