Okay, so I need to find the square root of 53 and express it in its simplest radical form. Hmm, let's start by recalling what a radical form is. If I remember correctly, a radical form is when you have a square root (or any other root) that can be simplified into a product of a rational number and another radical. The simplest radical form means that there's no perfect square factor left inside the square root. 

First, I should check if 53 is a prime number. Let me think. 53 divided by 2 is 26.5, so not divisible by 2. Divided by 3 is around 17.666..., not an integer. Divided by 5 is 10.6, nope. 7? 7 times 7 is 49, 7 times 8 is 56, which is over 53. So 53 is a prime number. That means it doesn't have any square factors other than 1. 

Wait, but if 53 is prime, then its square root can't be simplified further. So is the answer just √53? But the question says "simplest radical form." Let me verify. If the number inside the radical has no perfect square factors other than 1, then the radical is already in its simplest form. Since 53 is prime, it doesn't have any perfect square factors. Therefore, √53 is already simplified. 

But let me double-check in case I made a mistake. Let's list the perfect squares less than 53: 1, 4, 9, 16, 25, 36, 49. None of these divide 53 evenly. For example, 53 divided by 49 is about 1.08, which isn't an integer. 53 divided by 36 is roughly 1.47, not an integer. So none of the perfect squares can be factored out. 

Therefore, the simplest radical form of √53 is indeed √53 itself. There's no simplification needed. 

But wait, let me make sure I didn't confuse this with something else. For example, if the number had a factor that's a perfect square, like 72, which factors into 36*2, then √72 would be √36*√2 = 6√2. But since 53 is prime, there's nothing to factor out. 

Yes, so I think I'm confident here. The answer is √53.
Therefore, the square root of 53 in simplest radical form is $\boxed{\sqrt{53}}$.
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