If x and y are numbers such that (x+4)(y-4)=0, what is the smallest possible value of x^2 + y^2 Pick: A) 0 B) 9 C) 12 D) 16 E) 18
From (x+4)(y-4)=0 it follows that either x=-4 or y=4. Thus either x^2=16 or y^2=16.   Now, if x^2=16, then the least value of y^2 is 0, so the least value of x^2 + y^2 = 16 + 0= 
<gadget id="calculator">16 + 0</gadget>
<output>16</output>
16.   Similarly if y^2=16, then the least value of x^2 is 0, so the least value of x^2 + y^2 = 0 + 16= 
<gadget id="calculator">0 + 16</gadget>
<output>16</output>
16.   Answer: 16.
<result>D</result>
If x and y are numbers such that (x+4)(y-4)=0, what is the smallest possible value of x^2 + y^2