$$^{\left(x+y\right)^2=\left(x+y\right)\left(x+y\right)\ =\ x^2+2xy+y^2}$$
$$^{\left(x-y\right)^2=\left(x-y\right)\left(x-y\right)\ =\ x^2-2xy+y^2}$$
$$\left(x+y\right)\left(x-y\right)\ =\ x^2-y^2$$
Statement 1 follows the top pattern and can be factored to: $$\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)=0$$
Then you can set each factor equal to zero to find: $$\sqrt{x}+\sqrt{y}=0$$
$$\sqrt{x}=-\sqrt{y}$$ When we are working with real numbers, the square root of a number is never negative. So once we know $$\sqrt{x}=-\sqrt{y}$$ we know x is negative and y is positive ( assuming neither x or y can be 0 which would likely be a stipulation on this problem). Since x is negative it will always be smaller than y , so you know the answer to the question is no. Statement 1 is enough information. Narrow the answer choices to A and D.
Statement 2 follows the bottom pattern and can be factored to: $$\left(x+y\right)\left(x-y\right)\ = 0$$
Set each factor equal to zero to find x + y = 0 and x - y = 0. So x = -y and x= y. So x and y could be equal and then the answer to the question is no. But x could be positive and y could be negative and then the answer to the question is yes. Since we don't have a definitive answer, this statement is not sufficient. So the answer is A.
Take care,
Sionainn Marcoux
BA - Stanford University, MPP - Harvard University
Instructor, tutor for Princeton Review and Airbnb host
In other words a blend of Jamie Escalante from Stand and Deliver, Julie from The Love Boat, and Schneider the Super from One Day at a Time.
