Question 619629
Ex: Let a = 3 and b = 4 (pick any two random numbers, just make sure that neither number is 0)



sqrt( a^2+b^2) = a+b


sqrt( 3^2+4^2) = 3+4


sqrt( 9+16) = 3+4


sqrt(25) = 3+4


5 = 3+4


5 = 7


5 ≠ 7



So sqrt( a^2+b^2)≠a+b  in general.