Find the value of $\sqrt{2p^2+4}$ such that a 6-digit number $52p304$ is divisible by 11.

A number is divisible by 11 if the difference between the sum of digits at odd places and the sum of digits at even places is either 0 or a multiple of 11.For the number $52p304$:Sum of digits at odd positions (from right or left):
$$\text{Odd positions (1st, 3rd, 5th from right)}=4+3+2=9$$
Sum of digits at even positions (2nd, 4th, 6th from right):
$$\text{Even positions}=0+p+5=5+p$$
The difference:
$$\text{Difference}=(5+p)-9=p-4$$
For divisibility by 11, $p-4$ must be 0 (since $p$ is a single digit between 0 and 9):
$$p-4=0⟹p=4$$
Now, substitute $p=4$ into $\sqrt{2p^2+4}$:
$$\sqrt{2(4{)}^2+4}=\sqrt{2(16)+4}=\sqrt{32+4}=\sqrt{36}=6$$