Algebra More... XD

ab(a+b) = 2

ab = 2/(a+b)

let c = a+b

a^2 + b^2 = 2

(a+b)^2 - 2ab = 2

c^2 - 2(2/c) = 2

c^3 - 4 = 2c

c^3 - 2c - 4 = 0

(c-2)(c^2 + 2c + 2) = 0

c = 2

Reject c^2 + 2c + 2 = 0 because it will give a not real solution:

Therefore:

a+b = 2

ab = 1

Going to:

x^2 - 2x + 1 = 0

(x-1)^2 = 0

x = 1

That means:

So, 1 is the 1st root:

a+b = 2 ----> 1+b = 2 -----. b = 1

Therefore:

a = b = 1

Easiest way is to simply plug in the values and eliminate some options by observation .A and B can not be zero because $0^2+0^2\neq2$ In the second option,A cannot be 2 because $2^2+0^2\neq2$ If we plug in the value of 1 for both A and B in both the equation both equations are satisfied so the answer is $\boxed{1\;and\;1}$