Question 895124
(p-q)^2+(q-p)^2


since q-p = -(p-q), this becomes equal to:


(p-q)^2 + (-(p-q))^2 which becomes equal to (p-q)^2 + (p-q)^2 = 2*(p-q)^2


if you expand both of these, you will see that the solution for each is exactly the same.


(p-q)^2 = p^2 - 2pq + q^2


(q-p)^2 = q^2 - 2pq + p^2


rearrange the terms and the results are the same for both.


p^2 - 2pq + q^2 is the same as q^2 - 2pq + p^2.


q-p is the negative of p-q.


if you square the positive you get a positive.
if you square the negative you get a positive.


this might be easier to see with numbers.


let p = 5 and q = 3
p-q = 2
q-p = -2
2^2 = 4 and (-2)^2 = 4


since the power of the exponent is even, the result will be positive regardless if the argument is negative or positive.