<Question>
Compute the product of the number $5+\sqrt{3}$ and its radical conjugate.
</Question>

<think>
<Plan_1>
Okay, so I need to compute the product of the number 5 + √3 and its radical conjugate. Hmm, first, I should remember what a radical conjugate is. From what I recall, radical conjugates are expressions where the radical part is reversed in sign. So, if the original expression is a + √b, then its radical conjugate would be a - √b. That makes sense because conjugates are used to rationalize denominators and to find products that eliminate the radical.

Let me confirm that. For example, the conjugate of 3 + √2 is 3 - √2. Multiplying them together gives (3 + √2)(3 - √2) = 9 - (√2)^2 = 9 - 2 = 7. Right, so the product is a rational number. So, applying that to the given problem, the radical conjugate of 5 + √3 should be 5 - √3. Then, multiplying them should give a rational number as well.

So, the problem is to compute (5 + √3)(5 - √3).

Let me write that out:
(5 + √3)(5 - √3)

I can use the formula for the difference of squares here, which is (a + b)(a - b) = a² - b². Let's apply that.
Here, a is 5 and b is √3. So, expanding this, it should be 5² - (√3)².

Calculating each part:
5² = 25
(√3)² = 3

So, subtracting these gives 25 - 3 = 22.

Wait, that seems straightforward. Let me check again to make sure I didn't make a mistake.

Alternatively, I can multiply it out step by step without using the difference of squares formula, just to verify.

<prompt_1.1> Verify the product by expanding (5 + √3)(5 - √3) step by step. </prompt_1.1>

Therefore, the product is 22.

Is there any chance I messed up the signs? Let me see.

<prompt_1.2> Check the signs and terms in the product (5 + √3)(5 - √3). </prompt_1.2>

Wait, but just to ensure that I identified the correct radical conjugate. The term "radical conjugate" usually refers to changing the sign of the radical term.

<prompt_1.3> Confirm the identification of the radical conjugate and the final result. </prompt_1.3>
</Plan_1>

<Execution_1>
<execution_1.1>
Multiply (5 + √3)(5 - √3):

First, use the distributive property (FOIL method):

First terms: 5 * 5 = 25

Outer terms: 5 * (-√3) = -5√3

Inner terms: √3 * 5 = 5√3

Last terms: √3 * (-√3) = - (√3 * √3) = -3

Now, add all these together:

25 - 5√3 + 5√3 - 3

Combine like terms:

The -5√3 and +5√3 cancel each other out, leaving 25 - 3 = 22.
</execution_1.1>

<execution_1.2>
If I expand (5 + √3)(5 - √3), it's 5*5 + 5*(-√3) + √3*5 + √3*(-√3). Calculating each term:

5*5 = 25

5*(-√3) = -5√3

√3*5 = 5√3

√3*(-√3) = - (√3 * √3) = -3

Now, adding all these terms together: 25 - 5√3 + 5√3 - 3. The middle terms, -5√3 and +5√3, cancel each other out, leaving 25 - 3 = 22.
</execution_1.2>

<execution_1.3>
So if the original number is 5 + √3, then the conjugate is 5 - √3. If it were, say, a more complicated expression like 5 + 3√3, then the conjugate would be 5 - 3√3. But in this case, since it's just a single radical term, the conjugate is straightforward. So, yes, 5 - √3 is correct.

Another way to think about it: when you have a binomial with a radical, the conjugate is formed by changing the sign between the terms. So, if it's a sum, it becomes a difference and vice versa. Here, since it's 5 + √3, the conjugate is 5 - √3. So, multiplying them should eliminate the radical, resulting in a rational number, which in this case is 22.

Let me just recheck the arithmetic:

5 squared is 25. √3 squared is 3. 25 minus 3 is 22. Yep, that's correct. So, the answer should be 22.
</execution_1.3>
</Execution_1>

<Plan_2>
Based on execution_1.1, execution_1.2, execution_1.3:
I think that's thorough. I considered using the difference of squares, verified by expanding manually, checked the signs, confirmed the definition of a radical conjugate, and even thought about the purpose of conjugates in rationalizing expressions.
All steps lead to the conclusion that the product is 22.
</Plan_2>

<Execution_2>
No prompt to execute.
</Execution_2>

<Final_answer>
Therefore, the product of $5+\sqrt{3}$ and its radical conjugate is $\boxed{22}$.
</Final_answer>
</think>