Explanation:
Okay, so I need to figure out how many distinct possible values there are for the digit B in the two-digit number B6, which is a perfect square. Let me start by understanding the problem.

First, the number is two digits, and the tens digit is B, while the units digit is 6. So the number can be written as 10*B + 6. And this number is a square of some positive integer. My job is to find all possible values of B (which is a single digit, so from 0 to 9) such that 10B + 6 is a perfect square. Then count how many distinct B's there are.

Let me think. So first, let me note that B is a digit, so B can be 0 through 9. However, since it's a two-digit number, B can't be 0, right? Because if B is 0, then the number would be 06, which is just 6, a single-digit number. So B must be from 1 to 9. Wait, but maybe the problem allows B to be 0? Let me check the problem statement again.

It says "the two-digit number B6", so if B is 0, then it's 06, which is technically not a two-digit number. So B must be from 1 to 9. So possible values of B are 1 through 9. But maybe I need to confirm that.

But let me not get stuck here. Let me think about the possible squares that end with 6. Because the number is B6, so the units digit is 6. So the square number ends with 6. Therefore, the square root must be a number whose square ends with 6. Let me recall that squares modulo 10 can only be certain digits. Let me recall that the last digit of a square number depends on the last digit of the original number.

So, if a square ends with 6, then the original number must end with either 4 or 6, because:

- 4 squared is 16, ends with 6
- 6 squared is 36, ends with 6
- 12 squared is 144, ends with 4
- 14 squared is 196, ends with 6
- 16 squared is 256, ends with 6
- etc.

So, the square roots of numbers ending with 6 must end with 4 or 6. Therefore, the square roots are numbers ending with 4 or 6.

So, since the number B6 is a two-digit number, the square root must be a number such that when squared, it's between 10 and 99 (since B6 is between 16 and 96, since B is from 1 to 9). Wait, actually, B is from 1 to 9, so the number is from 16 (when B=1) up to 96 (when B=9). Therefore, the square roots would be numbers whose squares are between 16 and 96.

So let me find all integers n such that n² is between 16 and 96. Let me compute the square roots of 16 and 96.

sqrt(16) is 4, sqrt(96) is approximately 9.798. So n can be integers from 4 up to 9, since 10² is 100, which is already three digits. Wait, but hold on. Wait, if n is 4, then n² is 16, which is two digits. If n is 9, n² is 81, which is two digits. If n is 10, n² is 100, which is three digits. Therefore, n can be integers from 4 to 9 inclusive. However, earlier I thought that the square roots must end with 4 or 6. So let me check which numbers between 4 and 9 end with 4 or 6.

Numbers from 4 to 9: 4,5,6,7,8,9. So numbers ending with 4 or 6 are 4, 6. Wait, 4 ends with 4, 6 ends with 6. So only 4 and 6? Wait, but 14 squared is 196, which is three digits. But 14 is outside of our range. Wait, but maybe there are numbers between 4 and 9 that end with 4 or 6. Let me check:

Numbers from 4 to 9:

- 4: ends with 4
- 5: ends with 5
- 6: ends with 6
- 7: ends with 7
- 8: ends with 8
- 9: ends with 9

So only 4 and 6. Therefore, the square roots of numbers ending with 6 in the two-digit range must be 4 or 6? Wait, but let me check. For example, 14 squared is 196, which ends with 6, but 14 is outside of our range. Similarly, 16 squared is 256, which ends with 6, but again, 16 squared is three digits. Wait, but maybe there are numbers between 4 and 9 whose squares end with 6?

Let me check:

4² = 16 → ends with 6. Yes.

5² = 25 → ends with 5.

6² = 36 → ends with 6.

7² = 49 → ends with 9.

8² = 64 → ends with 4.

9² = 81 → ends with 1.

So only 4² and 6² end with 6. Therefore, the only two-digit numbers that are squares and end with 6 are 16 (4²) and 36 (6²). Wait, but 16 is 16, which is two-digit, and 36 is two-digit. But wait, 16 is B6 where B is 1, and 36 is B6 where B is 3. Therefore, B can be 1 or 3? But that would mean only two possible values for B. But the problem says "how many distinct possible values are there for B?" So maybe the answer is 2? But that seems too low. Wait, maybe I made a mistake here.

Wait, hold on. Wait, maybe there are more numbers. Wait, let me think again. Because maybe there are numbers outside of 4 to 9 whose squares are two-digit numbers ending with 6. Wait, but if n is 14, n² is 196, which is three-digit. Similarly, n=16 is 256, three-digit. So numbers larger than 9 will have squares that are three-digit or more. Therefore, the only two-digit squares ending with 6 are 16 and 36. Therefore, B can be 1 or 3. Therefore, only two possible values. But that seems too few. Wait, but let me check again.

Wait, let me check all two-digit squares and see which ones end with 6.

Two-digit squares are:

4² = 16

5² = 25

6² = 36

7² = 49

8² = 64

9² = 81

10² = 100 (three-digit)

So the two-digit squares are 16, 25, 36, 49, 64, 81. So among these, which end with 6? 16 and 36. So yes, only 16 and 36. Therefore, B is 1 and 3. Therefore, two possible values. But the problem says "how many distinct possible values are there for B?" So answer is 2? But maybe I'm missing something here.

Wait, but maybe there are other squares that are two-digit numbers ending with 6? Let me check numbers like 14² is 196, which is three-digit. 12² is 144, ends with 4. 11² is 121, ends with 1. 13² is 169, ends with 9. 14² is 196, ends with 6. But 14² is 196, which is three-digit. So no. So only 16 and 36. Therefore, B is 1 and 3. Therefore, two possible values. But maybe I need to check if there are other numbers?

Wait, but maybe negative numbers? But the problem says "positive integer", so square roots are positive integers. So no negative numbers. So only 4 and 6. Therefore, only 16 and 36. Therefore, B is 1 and 3. Therefore, answer is 2? But maybe I need to check again.

Wait, but let me think again. Wait, the problem says "the two-digit number B6", so B is the tens digit. So for example, if the number is 16, then B is 1. If the number is 36, then B is 3. Are there any other two-digit squares ending with 6?

Wait, let me check 24² is 576, which ends with 6, but that's three-digit. 26² is 676, also three-digit. 14² is 196, three-digit. 16² is 256, three-digit. So no, there are no other two-digit numbers that are squares ending with 6. Therefore, only 16 and 36. Therefore, B can be 1 or 3. Therefore, two possible values. Therefore, the answer is 2? But I feel like maybe I'm missing something here. Let me check again.

Wait, maybe there are numbers where the square is two-digit, ends with 6, but the square root is not between 4 and 9? Wait, but if the square is two-digit, then the square root must be between 4 and 9, since 4²=16 and 9²=81. So numbers from 4 to 9. So square roots are 4,5,6,7,8,9. As we saw, only 4 and 6 have squares ending with 6. Therefore, only 16 and 36. Therefore, B is 1 and 3. Therefore, two values. Therefore, answer is 2? But maybe the problem is in another base? No, the problem says "two-digit number", so base 10. So I think that's correct.

Wait, but let me check once more. Let me list all two-digit numbers ending with 6 and check if they are perfect squares.

Numbers ending with 6 between 10 and 99:

16, 26, 36, 46, 56, 66, 76, 86, 96.

Now check which of these are perfect squares.

16: 4², yes.

26: Not a square.

36: 6², yes.

46: Not a square.

56: Not a square.

66: Not a square.

76: Not a square.

86: Not a square.

96: Not a square.

So only 16 and 36. Therefore, B is 1 and 3. Therefore, two possible values. Therefore, the answer is 2. But the problem says "how many distinct possible values are there for B?" So 2? But maybe I made a mistake here?

Wait, but let me check if there are any other squares. Wait, 0² is 0, but that's not two-digit. 1² is 1, same. 2² is 4, same. 3² is 9, same. 4² is 16, which is two-digit. So yes, only 16 and 36. Therefore, answer is 2. But maybe the problem is trickier?

Wait, but let me check if there are numbers where B is a digit, but the square is two-digit. For example, if B is 0, then the number is 06, which is 6, not two-digit. So B can't be 0. So only 1 and 3. Therefore, answer is 2. But maybe the problem is in another interpretation?

Wait, the problem says "the two-digit number B6, where B is the tens digit, is the square of a positive integer." So maybe there are other squares where the units digit is 6, but the number is two-digit. But as we saw, only 16 and 36. Therefore, answer is 2. But maybe I need to check if there are other squares?

Wait, let me check 16, 36. What about 100? No, that's three-digit. 121 is three-digit. So no. Therefore, only 16 and 36. Therefore, answer is 2. But maybe the problem is expecting more? Maybe I need to check numbers where the square is two-digit, but the square root is not between 4 and 9? Wait, but if the square is two-digit, then square root is between 4 and 9. Because sqrt(10) is approx 3.16, so 4 is the first integer. So numbers from 4 to 9. Therefore, only 4 and 6. Therefore, answer is 2. Therefore, the number of distinct possible values for B is 2. Therefore, answer is \boxed{2}?

But wait, let me check once again. Maybe I made a mistake in thinking that only numbers ending with 4 or 6 can have squares ending with 6. Let me check.

Let me recall that the last digit of a square number depends on the last digit of the original number. Let me check all digits 0-9:

0² = 0 → ends with 0

1² = 1 → ends with 1

2² = 4 → ends with 4

3² = 9 → ends with 9

4² = 16 → ends with 6

5² = 25 → ends with 5

6² = 36 → ends with 6

7² = 49 → ends with 9

8² = 64 → ends with 4

9² = 81 → ends with 1

Therefore, only numbers ending with 4 or 6 have squares ending with 6. Therefore, square roots must end with 4 or 6. Therefore, in the two-digit numbers, the square roots are numbers from 4 to 9, so only 4 and 6. Therefore, their squares are 16 and 36. Therefore, B is 1 and 3. Therefore, answer is 2. Therefore, I think that is correct.

But maybe there's something else? Wait, maybe numbers like 14, 24, etc., but their squares are three-digit numbers. For example, 14² is 196, which is three-digit. So even though 14 ends with 4, its square is three-digit. Therefore, not applicable here. Therefore, only 4 and 6. Therefore, answer is 2. Therefore, the number of distinct possible values for B is 2. Therefore, answer is \boxed{2}.

But wait, hold on. Wait, let me check if there are any other numbers. For example, if the square is 16, which is 4², so B is 1. If the square is 36, which is 6², so B is 3. Are there any other squares? Let me check 25, which is 5², ends with 5. 49 is 7², ends with 9. 64 is 8², ends with 4. 81 is 9², ends with 1. So none of these end with 6. Therefore, only 16 and 36. Therefore, answer is 2. Therefore, I think that is correct.

But maybe the problem is in another base? But the problem didn't specify, so it's base 10. Therefore, answer is 2. Therefore, I think the answer is 2. But maybe I need to check once again.

Alternatively, maybe there are numbers where B is a digit, but the square is two-digit, but the square root is not an integer? No, the problem says it's the square of a positive integer. Therefore, the square must be a perfect square. Therefore, only 16 and 36. Therefore, answer is 2. Therefore, I think that is correct.

But maybe I need to check if there are other squares. Let me think. For example, 16, 36. What about 100? No, three-digit. 121? Three-digit. 144? Three-digit. So no. Therefore, only 16 and 36. Therefore, answer is 2. Therefore, the number of distinct possible values for B is 2. Therefore, answer is \boxed{2}.

Wait, but I just thought of something. What about numbers like 25? Wait, no, ends with 5. 49 ends with 9. 64 ends with 4. 81 ends with 1. So no. Therefore, only 16 and 36. Therefore, answer is 2. Therefore, I think that is correct.

But maybe the problem is in the way the number is written? For example, if B is a digit, but maybe there are numbers where the square is two-digit, but the square root is a two-digit number? Wait, but if the square is two-digit, then the square root is between 4 and 9, as we saw. Therefore, single-digit numbers. Therefore, no. Therefore, answer is 2. Therefore, I think that is correct.

But maybe the problem is from a different source, and the answer is different? Wait, but according to my reasoning, it's 2. Let me check once again.

List of two-digit numbers ending with 6: 16, 26, 36, 46, 56, 66, 76, 86, 96.

Check which are perfect squares:

- 16: 4² ✔
- 26: Not a square
- 36: 6² ✔
- 46: Not a square
- 56: Not a square
- 66: Not a square
- 76: Not a square
- 86: Not a square
- 96: Not a square

Yes, only 16 and 36. Therefore, B is 1 and 3. Therefore, two values. Therefore, answer is 2. Therefore, I think that is correct.

But maybe the problem is in another interpretation? For example, maybe B is allowed to be 0? If B is 0, then the number is 06, which is 6, which is a single-digit number. But the problem says "two-digit number", so 06 is not considered a two-digit number. Therefore, B cannot be 0. Therefore, only 1 and 3. Therefore, answer is 2. Therefore, I think that is correct.

Therefore, I think the answer is 2. Therefore, the number of distinct possible values for B is 2. Therefore, the answer is \boxed{2}.

**Final Answer**
\boxed{2}

To solve the problem, we are given a two-digit number of the form **B6**, where **B** is the tens digit and **6** is the units digit. We are told that this number is a perfect square of a positive integer. Our goal is to determine how many **distinct possible values** there are for the digit **B**.

---

### Step 1: Understand the Structure of the Number

The number **B6** can be written as:

$$
10B + 6
$$

We are told that this number is a **perfect square**. So, we are looking for all integers $ n $ such that:

$$
n^2 = 10B + 6
$$

This implies that $ n^2 $ must be a two-digit number ending in **6**.

---

### Step 2: Determine the Range of Possible Squares

Since $ n^2 $ is a two-digit number, it must be between 10 and 99. Therefore, $ n $ must be an integer such that:

$$
\sqrt{10} \leq n \leq \sqrt{99}
$$

$$
3.16 \leq n \leq 9.95
$$

So, the possible integer values of $ n $ are:

$$
n = 4, 5, 6, 7, 8, 9
$$

Now, we compute the squares of these values and check which ones end in **6**:

- $ 4^2 = 16 $ → ends in 6 ✅
- $ 5^2 = 25 $ → ends in 5 ❌
- $ 6^2 = 36 $ → ends in 6 ✅
- $ 7^2 = 49 $ → ends in 9 ❌
- $ 8^2 = 64 $ → ends in 4 ❌
- $ 9^2 = 81 $ → ends in 1 ❌

So, the only two-digit perfect squares that end in **6** are:

- $ 16 = 4^2 $ → B = 1
- $ 36 = 6^2 $ → B = 3

---

### Step 3: Count the Distinct Values of B

From the above, the possible values of **B** are:

- 1 (from 16)
- 3 (from 36)

These are the only two distinct values of **B** for which the number **B6** is a perfect square.

---

### Final Answer

$$
\boxed{2}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many distinct possible values there are for the digit B in the two-digit number B6, which is a perfect square. Let me start by understanding the problem.

First, the number is two digits, and the tens digit is B, while the units digit is 6. So the number can be written as 10*B + 6. And this number is a square of some positive integer. My job is to find all possible values of B (which is a single digit, so from 0 to 9) such that 10B + 6 is a perfect square. Then count how many distinct B's there are.

Let me think. So first, let me note that B is a digit, so B can be 0 through 9. However, since it's a two-digit number, B can't be 0, right? Because if B is 0, then the number would be 06, which is just 6, a single-digit number. So B must be from 1 to 9. Wait, but maybe the problem allows B to be 0? Let me check the problem statement again.

It says "the two-digit number B6", so if B is 0, then it's 06, which is technically not a two-digit number. So B must be from 1 to 9. So possible values of B are 1 through 9. But maybe I need to confirm that.

But let me not get stuck here. Let me think about the possible squares that end with 6. Because the number is B6, so the units digit is 6. So the square number ends with 6. Therefore, the square root must be a number whose square ends with 6. Let me recall that squares modulo 10 can only be certain digits. Let me recall that the last digit of a square number depends on the last digit of the original number.

So, if a square ends with 6, then the original number must end with either 4 or 6, because:

- 4 squared is 16, ends with 6
- 6 squared is 36, ends with 6
- 12 squared is 144, ends with 4
- 14 squared is 196, ends with 6
- 16 squared is 256, ends with 6
- etc.

So, the square roots of numbers ending with 6 must end with 4 or 6. Therefore, the square roots are numbers ending with 4 or 6.

So, since the number B6 is a two-digit number, the square root must be a number such that when squared, it's between 10 and 99 (since B6 is between 16 and 96, since B is from 1 to 9). Wait, actually, B is from 1 to 9, so the number is from 16 (when B=1) up to 96 (when B=9). Therefore, the square roots would be numbers whose squares are between 16 and 96.

So let me find all integers n such that n² is between 16 and 96. Let me compute the square roots of 16 and 96.

sqrt(16) is 4, sqrt(96) is approximately 9.798. So n can be integers from 4 up to 9, since 10² is 100, which is already three digits. Wait, but hold on. Wait, if n is 4, then n² is 16, which is two digits. If n is 9, n² is 81, which is two digits. If n is 10, n² is 100, which is three digits. Therefore, n can be integers from 4 to 9 inclusive. However, earlier I thought that the square roots must end with 4 or 6. So let me check which numbers between 4 and 9 end with 4 or 6.

Numbers from 4 to 9: 4,5,6,7,8,9. So numbers ending with 4 or 6 are 4, 6. Wait, 4 ends with 4, 6 ends with 6. So only 4 and 6? Wait, but 14 squared is 196, which is three digits. But 14 is outside of our range. Wait, but maybe there are numbers between 4 and 9 that end with 4 or 6. Let me check:

Numbers from 4 to 9:

- 4: ends with 4
- 5: ends with 5
- 6: ends with 6
- 7: ends with 7
- 8: ends with 8
- 9: ends with 9

So only 4 and 6. Therefore, the square roots of numbers ending with 6 in the two-digit range must be 4 or 6? Wait, but let me check. For example, 14 squared is 196, which ends with 6, but 14 is outside of our range. Similarly, 16 squared is 256, which ends with 6, but again, 16 squared is three digits. Wait, but maybe there are numbers between 4 and 9 whose squares end with 6?

Let me check:

4² = 16 → ends with 6. Yes.

5² = 25 → ends with 5.

6² = 36 → ends with 6.

7² = 49 → ends with 9.

8² = 64 → ends with 4.

9² = 81 → ends with 1.

So only 4² and 6² end with 6. Therefore, the only two-digit numbers that are squares and end with 6 are 16 (4²) and 36 (6²). Wait, but 16 is 16, which is two-digit, and 36 is two-digit. But wait, 16 is B6 where B is 1, and 36 is B6 where B is 3. Therefore, B can be 1 or 3? But that would mean only two possible values for B. But the problem says "how many distinct possible values are there for B?" So maybe the answer is 2? But that seems too low. Wait, maybe I made a mistake here.

Wait, hold on. Wait, maybe there are more numbers. Wait, let me think again. Because maybe there are numbers outside of 4 to 9 whose squares are two-digit numbers ending with 6. Wait, but if n is 14, n² is 196, which is three-digit. Similarly, n=16 is 256, three-digit. So numbers larger than 9 will have squares that are three-digit or more. Therefore, the only two-digit squares ending with 6 are 16 and 36. Therefore, B can be 1 or 3. Therefore, only two possible values. But that seems too few. Wait, but let me check again.

Wait, let me check all two-digit squares and see which ones end with 6.

Two-digit squares are:

4² = 16

5² = 25

6² = 36

7² = 49

8² = 64

9² = 81

10² = 100 (three-digit)

So the two-digit squares are 16, 25, 36, 49, 64, 81. So among these, which end with 6? 16 and 36. So yes, only 16 and 36. Therefore, B is 1 and 3. Therefore, two possible values. But the problem says "how many distinct possible values are there for B?" So answer is 2? But maybe I'm missing something here.

Wait, but maybe there are other squares that are two-digit numbers ending with 6? Let me check numbers like 14² is 196, which is three-digit. 12² is 144, ends with 4. 11² is 121, ends with 1. 13² is 169, ends with 9. 14² is 196, ends with 6. But 14² is 196, which is three-digit. So no. So only 16 and 36. Therefore, B is 1 and 3. Therefore, two possible values. But maybe I need to check if there are other numbers?

Wait, but maybe negative numbers? But the problem says "positive integer", so square roots are positive integers. So no negative numbers. So only 4 and 6. Therefore, only 16 and 36. Therefore, B is 1 and 3. Therefore, answer is 2? But maybe I need to check again.

Wait, but let me think again. Wait, the problem says "the two-digit number B6", so B is the tens digit. So for example, if the number is 16, then B is 1. If the number is 36, then B is 3. Are there any other two-digit squares ending with 6?

Wait, let me check 24² is 576, which ends with 6, but that's three-digit. 26² is 676, also three-digit. 14² is 196, three-digit. 16² is 256, three-digit. So no, there are no other two-digit numbers that are squares ending with 6. Therefore, only 16 and 36. Therefore, B can be 1 or 3. Therefore, two possible values. Therefore, the answer is 2? But I feel like maybe I'm missing something here. Let me check again.

Wait, maybe there are numbers where the square is two-digit, ends with 6, but the square root is not between 4 and 9? Wait, but if the square is two-digit, then the square root must be between 4 and 9, since 4²=16 and 9²=81. So numbers from 4 to 9. So square roots are 4,5,6,7,8,9. As we saw, only 4 and 6 have squares ending with 6. Therefore, only 16 and 36. Therefore, B is 1 and 3. Therefore, two values. Therefore, answer is 2? But maybe the problem is in another base? No, the problem says "two-digit number", so base 10. So I think that's correct.

Wait, but let me check once more. Let me list all two-digit numbers ending with 6 and check if they are perfect squares.

Numbers ending with 6 between 10 and 99:

16, 26, 36, 46, 56, 66, 76, 86, 96.

Now check which of these are perfect squares.

16: 4², yes.

26: Not a square.

36: 6², yes.

46: Not a square.

56: Not a square.

66: Not a square.

76: Not a square.

86: Not a square.

96: Not a square.

So only 16 and 36. Therefore, B is 1 and 3. Therefore, two possible values. Therefore, the answer is 2. But the problem says "how many distinct possible values are there for B?" So 2? But maybe I made a mistake here?

Wait, but let me check if there are any other squares. Wait, 0² is 0, but that's not two-digit. 1² is 1, same. 2² is 4, same. 3² is 9, same. 4² is 16, which is two-digit. So yes, only 16 and 36. Therefore, answer is 2. But maybe the problem is trickier?

Wait, but let me check if there are numbers where B is a digit, but the square is two-digit. For example, if B is 0, then the number is 06, which is 6, not two-digit. So B can't be 0. So only 1 and 3. Therefore, answer is 2. But maybe the problem is in another interpretation?

Wait, the problem says "the two-digit number B6, where B is the tens digit, is the square of a positive integer." So maybe there are other squares where the units digit is 6, but the number is two-digit. But as we saw, only 16 and 36. Therefore, answer is 2. But maybe I need to check if there are other squares?

Wait, let me check 16, 36. What about 100? No, that's three-digit. 121 is three-digit. So no. Therefore, only 16 and 36. Therefore, answer is 2. But maybe the problem is expecting more? Maybe I need to check numbers where the square is two-digit, but the square root is not between 4 and 9? Wait, but if the square is two-digit, then square root is between 4 and 9. Because sqrt(10) is approx 3.16, so 4 is the first integer. So numbers from 4 to 9. Therefore, only 4 and 6. Therefore, answer is 2. Therefore, the number of distinct possible values for B is 2. Therefore, answer is \boxed{2}?

But wait, let me check once again. Maybe I made a mistake in thinking that only numbers ending with 4 or 6 can have squares ending with 6. Let me check.

Let me recall that the last digit of a square number depends on the last digit of the original number. Let me check all digits 0-9:

0² = 0 → ends with 0

1² = 1 → ends with 1

2² = 4 → ends with 4

3² = 9 → ends with 9

4² = 16 → ends with 6

5² = 25 → ends with 5

6² = 36 → ends with 6

7² = 49 → ends with 9

8² = 64 → ends with 4

9² = 81 → ends with 1

Therefore, only numbers ending with 4 or 6 have squares ending with 6. Therefore, square roots must end with 4 or 6. Therefore, in the two-digit numbers, the square roots are numbers from 4 to 9, so only 4 and 6. Therefore, their squares are 16 and 36. Therefore, B is 1 and 3. Therefore, answer is 2. Therefore, I think that is correct.

But maybe there's something else? Wait, maybe numbers like 14, 24, etc., but their squares are three-digit numbers. For example, 14² is 196, which is three-digit. So even though 14 ends with 4, its square is three-digit. Therefore, not applicable here. Therefore, only 4 and 6. Therefore, answer is 2. Therefore, the number of distinct possible values for B is 2. Therefore, answer is \boxed{2}.

But wait, hold on. Wait, let me check if there are any other numbers. For example, if the square is 16, which is 4², so B is 1. If the square is 36, which is 6², so B is 3. Are there any other squares? Let me check 25, which is 5², ends with 5. 49 is 7², ends with 9. 64 is 8², ends with 4. 81 is 9², ends with 1. So none of these end with 6. Therefore, only 16 and 36. Therefore, answer is 2. Therefore, I think that is correct.

But maybe the problem is in another base? But the problem didn't specify, so it's base 10. Therefore, answer is 2. Therefore, I think the answer is 2. But maybe I need to check once again.

Alternatively, maybe there are numbers where B is a digit, but the square is two-digit, but the square root is not an integer? No, the problem says it's the square of a positive integer. Therefore, the square must be a perfect square. Therefore, only 16 and 36. Therefore, answer is 2. Therefore, I think that is correct.

But maybe I need to check if there are other squares. Let me think. For example, 16, 36. What about 100? No, three-digit. 121? Three-digit. 144? Three-digit. So no. Therefore, only 16 and 36. Therefore, answer is 2. Therefore, the number of distinct possible values for B is 2. Therefore, answer is \boxed{2}.

Wait, but I just thought of something. What about numbers like 25? Wait, no, ends with 5. 49 ends with 9. 64 ends with 4. 81 ends with 1. So no. Therefore, only 16 and 36. Therefore, answer is 2. Therefore, I think that is correct.

But maybe the problem is in the way the number is written? For example, if B is a digit, but maybe there are numbers where the square is two-digit, but the square root is a two-digit number? Wait, but if the square is two-digit, then the square root is between 4 and 9, as we saw. Therefore, single-digit numbers. Therefore, no. Therefore, answer is 2. Therefore, I think that is correct.

But maybe the problem is from a different source, and the answer is different? Wait, but according to my reasoning, it's 2. Let me check once again.

List of two-digit numbers ending with 6: 16, 26, 36, 46, 56, 66, 76, 86, 96.

Check which are perfect squares:

- 16: 4² ✔
- 26: Not a square
- 36: 6² ✔
- 46: Not a square
- 56: Not a square
- 66: Not a square
- 76: Not a square
- 86: Not a square
- 96: Not a square

Yes, only 16 and 36. Therefore, B is 1 and 3. Therefore, two values. Therefore, answer is 2. Therefore, I think that is correct.

But maybe the problem is in another interpretation? For example, maybe B is allowed to be 0? If B is 0, then the number is 06, which is 6, which is a single-digit number. But the problem says "two-digit number", so 06 is not considered a two-digit number. Therefore, B cannot be 0. Therefore, only 1 and 3. Therefore, answer is 2. Therefore, I think that is correct.

Therefore, I think the answer is 2. Therefore, the number of distinct possible values for B is 2. Therefore, the answer is \boxed{2}.

**Final Answer**
\boxed{2}

To solve the problem, we are given a two-digit number of the form **B6**, where **B** is the tens digit and **6** is the units digit. We are told that this number is a perfect square of a positive integer. Our goal is to determine how many **distinct possible values** there are for the digit **B**.

---

### Step 1: Understand the Structure of the Number

The number **B6** can be written as:

$$
10B + 6
$$

We are told that this number is a **perfect square**. So, we are looking for all integers $ n $ such that:

$$
n^2 = 10B + 6
$$

This implies that $ n^2 $ must be a two-digit number ending in **6**.

---

### Step 2: Determine the Range of Possible Squares

Since $ n^2 $ is a two-digit number, it must be between 10 and 99. Therefore, $ n $ must be an integer such that:

$$
\sqrt{10} \leq n \leq \sqrt{99}
$$

$$
3.16 \leq n \leq 9.95
$$

So, the possible integer values of $ n $ are:

$$
n = 4, 5, 6, 7, 8, 9
$$

Now, we compute the squares of these values and check which ones end in **6**:

- $ 4^2 = 16 $ → ends in 6 ✅
- $ 5^2 = 25 $ → ends in 5 ❌
- $ 6^2 = 36 $ → ends in 6 ✅
- $ 7^2 = 49 $ → ends in 9 ❌
- $ 8^2 = 64 $ → ends in 4 ❌
- $ 9^2 = 81 $ → ends in 1 ❌

So, the only two-digit perfect squares that end in **6** are:

- $ 16 = 4^2 $ → B = 1
- $ 36 = 6^2 $ → B = 3

---

### Step 3: Count the Distinct Values of B

From the above, the possible values of **B** are:

- 1 (from 16)
- 3 (from 36)

These are the only two distinct values of **B** for which the number **B6** is a perfect square.

---

### Final Answer

$$
\boxed{2}
$$

Solution: Of the two-digit perfect squares, only $4^2=16$ and $6^2=36$ end in $6$. Thus, there are $\boxed{2}$ distinct possible values for $B$.


Explanation:
Okay, so I need to find tan A for triangle ABC, given that the area of the triangle is equal to a² - (b - c)². Hmm, let me start by recalling some formulas related to the area of a triangle and trigonometric identities.

First, the area of a triangle can be expressed in terms of two sides and the sine of the included angle. The formula is:

Area = (1/2) * b * c * sin A

But in this problem, the area is given as a² - (b - c)². So maybe I can set these two expressions for the area equal to each other? Let me write that down:

(1/2) * b * c * sin A = a² - (b - c)²

Okay, so I need to simplify the right-hand side. Let me expand (b - c)²:

(b - c)² = b² - 2bc + c²

Therefore, the area becomes:

a² - (b² - 2bc + c²) = a² - b² + 2bc - c²

So, simplifying:

a² - b² - c² + 2bc

Hmm, let me rearrange terms:

(a² - b² - c²) + 2bc

Wait, but I also know from the Law of Cosines that in any triangle:

a² = b² + c² - 2bc cos A

So maybe I can substitute that into the expression for the area. Let me try that.

Substituting a² from the Law of Cosines:

Area = (b² + c² - 2bc cos A) - b² - c² + 2bc

Simplify term by term:

b² + c² - 2bc cos A - b² - c² + 2bc

The b² and -b² cancel, same with c² and -c². So we have:

-2bc cos A + 2bc

Factor out 2bc:

2bc(1 - cos A)

So the area is 2bc(1 - cos A). But earlier, we had another expression for the area: (1/2)bc sin A. Therefore, setting them equal:

(1/2)bc sin A = 2bc(1 - cos A)

Hmm, okay, so let me write that equation:

(1/2) bc sin A = 2 bc (1 - cos A)

First, I can divide both sides by bc, assuming bc ≠ 0, which they aren't in a triangle. So:

(1/2) sin A = 2(1 - cos A)

Multiply both sides by 2:

sin A = 4(1 - cos A)

So, sin A = 4 - 4 cos A

Hmm, so now I have an equation involving sin A and cos A. I need to find tan A. Maybe I can use some trigonometric identities here. Let me recall that tan A = sin A / cos A. So if I can express sin A in terms of cos A or vice versa, I can find tan A.

Alternatively, I can use the identity sin² A + cos² A = 1. Let me try that. Let me rearrange the equation sin A = 4(1 - cos A) to express sin A in terms of cos A, then substitute into the identity.

Let me write:

sin A = 4 - 4 cos A

Let me square both sides to use the Pythagorean identity:

sin² A = (4 - 4 cos A)² = 16(1 - cos A)²

But sin² A = 1 - cos² A, so:

1 - cos² A = 16(1 - cos A)²

Let me expand the right-hand side:

16(1 - 2 cos A + cos² A)

So:

1 - cos² A = 16 - 32 cos A + 16 cos² A

Bring all terms to one side:

1 - cos² A - 16 + 32 cos A - 16 cos² A = 0

Combine like terms:

(1 - 16) + 32 cos A + (-cos² A - 16 cos² A) = 0

Which is:

-15 + 32 cos A - 17 cos² A = 0

Let me rearrange terms:

-17 cos² A + 32 cos A - 15 = 0

Multiply both sides by -1 to make the quadratic coefficient positive:

17 cos² A - 32 cos A + 15 = 0

So now we have a quadratic equation in terms of cos A. Let me denote x = cos A for simplicity:

17x² - 32x + 15 = 0

Let me try to solve this quadratic equation. The quadratic formula is x = [32 ± sqrt( (-32)^2 - 4*17*15 )]/(2*17)

Compute discriminant D:

D = 1024 - 4*17*15

Calculate 4*17 = 68, 68*15 = 1020

So D = 1024 - 1020 = 4

So sqrt(D) = 2

Therefore, x = [32 ± 2]/34

So two solutions:

x = (32 + 2)/34 = 34/34 = 1

x = (32 - 2)/34 = 30/34 = 15/17

So cos A = 1 or cos A = 15/17

But cos A = 1 would imply that angle A is 0 degrees, which is impossible in a triangle. Therefore, we discard cos A = 1, and take cos A = 15/17.

Therefore, cos A = 15/17. Now, we can find sin A from the earlier equation:

sin A = 4(1 - cos A) = 4(1 - 15/17) = 4*(2/17) = 8/17

Therefore, sin A = 8/17 and cos A = 15/17. Therefore, tan A = sin A / cos A = (8/17)/(15/17) = 8/15.

Wait, so tan A is 8/15? Let me check if this makes sense.

Let me verify the steps again to make sure I didn't make a mistake. Starting from the area expressions:

Original area given: a² - (b - c)^2. Expanded to a² - b² + 2bc - c². Then using Law of Cosines, substituted a² = b² + c² - 2bc cos A. Then substituted into the area expression:

(b² + c² - 2bc cos A) - b² - c² + 2bc = -2bc cos A + 2bc = 2bc(1 - cos A). Then equate to (1/2) bc sin A. So:

(1/2) bc sin A = 2 bc (1 - cos A). Dividing by bc gives (1/2) sin A = 2(1 - cos A). Multiply by 2: sin A = 4(1 - cos A). Then squared both sides: sin² A = 16(1 - 2 cos A + cos² A). Then replaced sin² A with 1 - cos² A:

1 - cos² A = 16 - 32 cos A + 16 cos² A. Then rearranged terms:

-15 + 32 cos A -17 cos² A = 0. Multiply by -1: 17 cos² A - 32 cos A + 15 = 0. Solved quadratic, got cos A = 1 or 15/17. Discarded 1, so cos A = 15/17. Then sin A = 4(1 - 15/17) = 4*(2/17) = 8/17. Therefore tan A = 8/15. Seems correct.

But let me check if the area expressions are consistent. Let me take a triangle with angle A such that cos A = 15/17, sin A = 8/17. Then area is (1/2) bc sin A = (1/2) bc*(8/17) = (4/17) bc. On the other hand, the given area is a² - (b - c)^2. Let me compute that. From Law of Cosines, a² = b² + c² - 2bc*(15/17). Therefore:

a² - (b - c)^2 = [b² + c² - (30/17) bc] - [b² - 2bc + c²] = b² + c² - (30/17) bc - b² + 2bc - c² = (-30/17 bc + 2 bc) = (-30/17 + 34/17) bc = (4/17) bc. Which matches the area from the other formula. So yes, that checks out. Therefore, tan A = 8/15 is correct.

Wait, but let me think again. The problem says "the area of triangle ABC is equal to a² - (b - c)^2". But in my calculation, I used the formula for area as (1/2) bc sin A. But is there another way to check?

Alternatively, maybe I can use Heron's formula? But that might be more complicated. Alternatively, maybe check with specific values. Let me try to imagine a triangle where cos A = 15/17, sin A = 8/17. Let me pick specific values for b and c. Let me choose b = 17, c = 17. Then, since cos A = 15/17, then a² = b² + c² - 2bc cos A = 17² + 17² - 2*17*17*(15/17) = 2*289 - 2*17*15 = 578 - 510 = 68. Therefore, a = sqrt(68) = 2*sqrt(17). Then compute the area as a² - (b - c)^2. Since b = c = 17, (b - c)^2 = 0, so area is a² = 68. On the other hand, using (1/2) bc sin A: (1/2)*17*17*(8/17) = (1/2)*17*8 = 68. Which matches. So that works. So tan A = 8/15 is correct.

Alternatively, if I take different values for b and c. Let me take b = 17, c = 1. Then compute a² = 17² + 1² - 2*17*1*(15/17) = 289 + 1 - 2*17*(15/17) = 290 - 30 = 260. Therefore, a = sqrt(260). Then compute area as a² - (b - c)^2 = 260 - (16)^2 = 260 - 256 = 4. On the other hand, (1/2)*b*c*sin A = 0.5*17*1*(8/17) = 0.5*8 = 4. Which also matches. So that works too. Therefore, tan A = 8/15 is correct.

Therefore, the answer is 8/15. Wait, but let me check if there's another way to approach this problem, maybe using other area formulas or relations?

Alternatively, since we have the area expressed as a² - (b - c)^2, which we simplified to 2bc(1 - cos A). But also, area is (1/2) bc sin A. Therefore, equating them gives (1/2) bc sin A = 2 bc (1 - cos A), leading to sin A = 4(1 - cos A). Then, using the identity sin A = 4(1 - cos A). Let me think if there's another identity that can be used here. For example, using tan(A/2). Because there's an identity that relates sin A and 1 - cos A.

Recall that 1 - cos A = 2 sin²(A/2), and sin A = 2 sin(A/2) cos(A/2). Therefore, substituting into sin A = 4(1 - cos A):

2 sin(A/2) cos(A/2) = 4 * 2 sin²(A/2)

Simplify:

2 sin(A/2) cos(A/2) = 8 sin²(A/2)

Divide both sides by 2 sin(A/2), assuming sin(A/2) ≠ 0 (which it isn't because A is between 0 and π in a triangle):

cos(A/2) = 4 sin(A/2)

Therefore, tan(A/2) = cos(A/2) / sin(A/2) = 1/4

Wait, no, if cos(A/2) = 4 sin(A/2), then tan(A/2) = sin(A/2)/cos(A/2) = 1/4. So tan(A/2) = 1/4. Then, using the double-angle formula for tan A:

tan A = 2 tan(A/2) / (1 - tan²(A/2)) = 2*(1/4) / (1 - (1/4)^2) = (1/2) / (1 - 1/16) = (1/2) / (15/16) = (1/2)*(16/15) = 8/15.

So that's another way to get tan A = 8/15. So that confirms the result again. Therefore, regardless of the method, the answer is 8/15.

Therefore, I think that's solid. So the answer is 8/15.

**Final Answer**
\boxed{\dfrac{8}{15}}

To find $\tan A$ in triangle $ABC$, we are given that the area of the triangle is:

$$
\text{Area} = a^2 - (b - c)^2
$$

We are also told that $a$, $b$, and $c$ are the sides of triangle $ABC$, and we are to compute $\tan A$.

---

### Step 1: Use the standard area formula

The area of a triangle can also be expressed using two sides and the sine of the included angle:

$$
\text{Area} = \frac{1}{2}bc \sin A
$$

Equating this with the given expression for the area:

$$
\frac{1}{2}bc \sin A = a^2 - (b - c)^2
$$

---

### Step 2: Expand and simplify the right-hand side

$$
(b - c)^2 = b^2 - 2bc + c^2
$$

So,

$$
a^2 - (b - c)^2 = a^2 - b^2 + 2bc - c^2
$$

Now, recall the **Law of Cosines**:

$$
a^2 = b^2 + c^2 - 2bc \cos A
$$

Substitute this into the expression:

$$
a^2 - b^2 - c^2 + 2bc = (b^2 + c^2 - 2bc \cos A) - b^2 - c^2 + 2bc = -2bc \cos A + 2bc
$$

So the area becomes:

$$
\text{Area} = 2bc(1 - \cos A)
$$

Now equate this with the earlier expression:

$$
\frac{1}{2}bc \sin A = 2bc(1 - \cos A)
$$

Divide both sides by $bc$ (since $b, c > 0$):

$$
\frac{1}{2} \sin A = 2(1 - \cos A)
$$

Multiply both sides by 2:

$$
\sin A = 4(1 - \cos A)
$$

---

### Step 3: Use trigonometric identities

We now have:

$$
\sin A = 4(1 - \cos A)
$$

Let’s square both sides:

$$
\sin^2 A = 16(1 - \cos A)^2
$$

Use the identity $\sin^2 A = 1 - \cos^2 A$:

$$
1 - \cos^2 A = 16(1 - 2\cos A + \cos^2 A)
$$

Expand the right-hand side:

$$
1 - \cos^2 A = 16 - 32\cos A + 16\cos^2 A
$$

Bring all terms to one side:

$$
1 - \cos^2 A - 16 + 32\cos A - 16\cos^2 A = 0
$$

Simplify:

$$
-15 + 32\cos A - 17\cos^2 A = 0
$$

Multiply through by $-1$:

$$
17\cos^2 A - 32\cos A + 15 = 0
$$

---

### Step 4: Solve the quadratic

Let $x = \cos A$, then:

$$
17x^2 - 32x + 15 = 0
$$

Use the quadratic formula:

$$
x = \frac{32 \pm \sqrt{(-32)^2 - 4(17)(15)}}{2(17)} = \frac{32 \pm \sqrt{1024 - 1020}}{34} = \frac{32 \pm 2}{34}
$$

So:

$$
x = \frac{34}{34} = 1 \quad \text{or} \quad x = \frac{30}{34} = \frac{15}{17}
$$

Since $\cos A = 1$ would imply angle $A = 0^\circ$, which is not possible in a triangle, we take:

$$
\cos A = \frac{15}{17}
$$

Now use the earlier equation:

$$
\sin A = 4(1 - \cos A) = 4\left(1 - \frac{15}{17}\right) = 4 \cdot \frac{2}{17} = \frac{8}{17}
$$

---

### Step 5: Compute $\tan A$

$$
\tan A = \frac{\sin A}{\cos A} = \frac{8/17}{15/17} = \frac{8}{15}
$$

---

### Final Answer

$$
\boxed{\dfrac{8}{15}}
$$

[Assistant]
Explanation:
Okay, so I need to find tan A for triangle ABC, given that the area of the triangle is equal to a² - (b - c)². Hmm, let me start by recalling some formulas related to the area of a triangle and trigonometric identities.

First, the area of a triangle can be expressed in terms of two sides and the sine of the included angle. The formula is:

Area = (1/2) * b * c * sin A

But in this problem, the area is given as a² - (b - c)². So maybe I can set these two expressions for the area equal to each other? Let me write that down:

(1/2) * b * c * sin A = a² - (b - c)²

Okay, so I need to simplify the right-hand side. Let me expand (b - c)²:

(b - c)² = b² - 2bc + c²

Therefore, the area becomes:

a² - (b² - 2bc + c²) = a² - b² + 2bc - c²

So, simplifying:

a² - b² - c² + 2bc

Hmm, let me rearrange terms:

(a² - b² - c²) + 2bc

Wait, but I also know from the Law of Cosines that in any triangle:

a² = b² + c² - 2bc cos A

So maybe I can substitute that into the expression for the area. Let me try that.

Substituting a² from the Law of Cosines:

Area = (b² + c² - 2bc cos A) - b² - c² + 2bc

Simplify term by term:

b² + c² - 2bc cos A - b² - c² + 2bc

The b² and -b² cancel, same with c² and -c². So we have:

-2bc cos A + 2bc

Factor out 2bc:

2bc(1 - cos A)

So the area is 2bc(1 - cos A). But earlier, we had another expression for the area: (1/2)bc sin A. Therefore, setting them equal:

(1/2)bc sin A = 2bc(1 - cos A)

Hmm, okay, so let me write that equation:

(1/2) bc sin A = 2 bc (1 - cos A)

First, I can divide both sides by bc, assuming bc ≠ 0, which they aren't in a triangle. So:

(1/2) sin A = 2(1 - cos A)

Multiply both sides by 2:

sin A = 4(1 - cos A)

So, sin A = 4 - 4 cos A

Hmm, so now I have an equation involving sin A and cos A. I need to find tan A. Maybe I can use some trigonometric identities here. Let me recall that tan A = sin A / cos A. So if I can express sin A in terms of cos A or vice versa, I can find tan A.

Alternatively, I can use the identity sin² A + cos² A = 1. Let me try that. Let me rearrange the equation sin A = 4(1 - cos A) to express sin A in terms of cos A, then substitute into the identity.

Let me write:

sin A = 4 - 4 cos A

Let me square both sides to use the Pythagorean identity:

sin² A = (4 - 4 cos A)² = 16(1 - cos A)²

But sin² A = 1 - cos² A, so:

1 - cos² A = 16(1 - cos A)²

Let me expand the right-hand side:

16(1 - 2 cos A + cos² A)

So:

1 - cos² A = 16 - 32 cos A + 16 cos² A

Bring all terms to one side:

1 - cos² A - 16 + 32 cos A - 16 cos² A = 0

Combine like terms:

(1 - 16) + 32 cos A + (-cos² A - 16 cos² A) = 0

Which is:

-15 + 32 cos A - 17 cos² A = 0

Let me rearrange terms:

-17 cos² A + 32 cos A - 15 = 0

Multiply both sides by -1 to make the quadratic coefficient positive:

17 cos² A - 32 cos A + 15 = 0

So now we have a quadratic equation in terms of cos A. Let me denote x = cos A for simplicity:

17x² - 32x + 15 = 0

Let me try to solve this quadratic equation. The quadratic formula is x = [32 ± sqrt( (-32)^2 - 4*17*15 )]/(2*17)

Compute discriminant D:

D = 1024 - 4*17*15

Calculate 4*17 = 68, 68*15 = 1020

So D = 1024 - 1020 = 4

So sqrt(D) = 2

Therefore, x = [32 ± 2]/34

So two solutions:

x = (32 + 2)/34 = 34/34 = 1

x = (32 - 2)/34 = 30/34 = 15/17

So cos A = 1 or cos A = 15/17

But cos A = 1 would imply that angle A is 0 degrees, which is impossible in a triangle. Therefore, we discard cos A = 1, and take cos A = 15/17.

Therefore, cos A = 15/17. Now, we can find sin A from the earlier equation:

sin A = 4(1 - cos A) = 4(1 - 15/17) = 4*(2/17) = 8/17

Therefore, sin A = 8/17 and cos A = 15/17. Therefore, tan A = sin A / cos A = (8/17)/(15/17) = 8/15.

Wait, so tan A is 8/15? Let me check if this makes sense.

Let me verify the steps again to make sure I didn't make a mistake. Starting from the area expressions:

Original area given: a² - (b - c)^2. Expanded to a² - b² + 2bc - c². Then using Law of Cosines, substituted a² = b² + c² - 2bc cos A. Then substituted into the area expression:

(b² + c² - 2bc cos A) - b² - c² + 2bc = -2bc cos A + 2bc = 2bc(1 - cos A). Then equate to (1/2) bc sin A. So:

(1/2) bc sin A = 2 bc (1 - cos A). Dividing by bc gives (1/2) sin A = 2(1 - cos A). Multiply by 2: sin A = 4(1 - cos A). Then squared both sides: sin² A = 16(1 - 2 cos A + cos² A). Then replaced sin² A with 1 - cos² A:

1 - cos² A = 16 - 32 cos A + 16 cos² A. Then rearranged terms:

-15 + 32 cos A -17 cos² A = 0. Multiply by -1: 17 cos² A - 32 cos A + 15 = 0. Solved quadratic, got cos A = 1 or 15/17. Discarded 1, so cos A = 15/17. Then sin A = 4(1 - 15/17) = 4*(2/17) = 8/17. Therefore tan A = 8/15. Seems correct.

But let me check if the area expressions are consistent. Let me take a triangle with angle A such that cos A = 15/17, sin A = 8/17. Then area is (1/2) bc sin A = (1/2) bc*(8/17) = (4/17) bc. On the other hand, the given area is a² - (b - c)^2. Let me compute that. From Law of Cosines, a² = b² + c² - 2bc*(15/17). Therefore:

a² - (b - c)^2 = [b² + c² - (30/17) bc] - [b² - 2bc + c²] = b² + c² - (30/17) bc - b² + 2bc - c² = (-30/17 bc + 2 bc) = (-30/17 + 34/17) bc = (4/17) bc. Which matches the area from the other formula. So yes, that checks out. Therefore, tan A = 8/15 is correct.

Wait, but let me think again. The problem says "the area of triangle ABC is equal to a² - (b - c)^2". But in my calculation, I used the formula for area as (1/2) bc sin A. But is there another way to check?

Alternatively, maybe I can use Heron's formula? But that might be more complicated. Alternatively, maybe check with specific values. Let me try to imagine a triangle where cos A = 15/17, sin A = 8/17. Let me pick specific values for b and c. Let me choose b = 17, c = 17. Then, since cos A = 15/17, then a² = b² + c² - 2bc cos A = 17² + 17² - 2*17*17*(15/17) = 2*289 - 2*17*15 = 578 - 510 = 68. Therefore, a = sqrt(68) = 2*sqrt(17). Then compute the area as a² - (b - c)^2. Since b = c = 17, (b - c)^2 = 0, so area is a² = 68. On the other hand, using (1/2) bc sin A: (1/2)*17*17*(8/17) = (1/2)*17*8 = 68. Which matches. So that works. So tan A = 8/15 is correct.

Alternatively, if I take different values for b and c. Let me take b = 17, c = 1. Then compute a² = 17² + 1² - 2*17*1*(15/17) = 289 + 1 - 2*17*(15/17) = 290 - 30 = 260. Therefore, a = sqrt(260). Then compute area as a² - (b - c)^2 = 260 - (16)^2 = 260 - 256 = 4. On the other hand, (1/2)*b*c*sin A = 0.5*17*1*(8/17) = 0.5*8 = 4. Which also matches. So that works too. Therefore, tan A = 8/15 is correct.

Therefore, the answer is 8/15. Wait, but let me check if there's another way to approach this problem, maybe using other area formulas or relations?

Alternatively, since we have the area expressed as a² - (b - c)^2, which we simplified to 2bc(1 - cos A). But also, area is (1/2) bc sin A. Therefore, equating them gives (1/2) bc sin A = 2 bc (1 - cos A), leading to sin A = 4(1 - cos A). Then, using the identity sin A = 4(1 - cos A). Let me think if there's another identity that can be used here. For example, using tan(A/2). Because there's an identity that relates sin A and 1 - cos A.

Recall that 1 - cos A = 2 sin²(A/2), and sin A = 2 sin(A/2) cos(A/2). Therefore, substituting into sin A = 4(1 - cos A):

2 sin(A/2) cos(A/2) = 4 * 2 sin²(A/2)

Simplify:

2 sin(A/2) cos(A/2) = 8 sin²(A/2)

Divide both sides by 2 sin(A/2), assuming sin(A/2) ≠ 0 (which it isn't because A is between 0 and π in a triangle):

cos(A/2) = 4 sin(A/2)

Therefore, tan(A/2) = cos(A/2) / sin(A/2) = 1/4

Wait, no, if cos(A/2) = 4 sin(A/2), then tan(A/2) = sin(A/2)/cos(A/2) = 1/4. So tan(A/2) = 1/4. Then, using the double-angle formula for tan A:

tan A = 2 tan(A/2) / (1 - tan²(A/2)) = 2*(1/4) / (1 - (1/4)^2) = (1/2) / (1 - 1/16) = (1/2) / (15/16) = (1/2)*(16/15) = 8/15.

So that's another way to get tan A = 8/15. So that confirms the result again. Therefore, regardless of the method, the answer is 8/15.

Therefore, I think that's solid. So the answer is 8/15.

**Final Answer**
\boxed{\dfrac{8}{15}}

To find $\tan A$ in triangle $ABC$, we are given that the area of the triangle is:

$$
\text{Area} = a^2 - (b - c)^2
$$

We are also told that $a$, $b$, and $c$ are the sides of triangle $ABC$, and we are to compute $\tan A$.

---

### Step 1: Use the standard area formula

The area of a triangle can also be expressed using two sides and the sine of the included angle:

$$
\text{Area} = \frac{1}{2}bc \sin A
$$

Equating this with the given expression for the area:

$$
\frac{1}{2}bc \sin A = a^2 - (b - c)^2
$$

---

### Step 2: Expand and simplify the right-hand side

$$
(b - c)^2 = b^2 - 2bc + c^2
$$

So,

$$
a^2 - (b - c)^2 = a^2 - b^2 + 2bc - c^2
$$

Now, recall the **Law of Cosines**:

$$
a^2 = b^2 + c^2 - 2bc \cos A
$$

Substitute this into the expression:

$$
a^2 - b^2 - c^2 + 2bc = (b^2 + c^2 - 2bc \cos A) - b^2 - c^2 + 2bc = -2bc \cos A + 2bc
$$

So the area becomes:

$$
\text{Area} = 2bc(1 - \cos A)
$$

Now equate this with the earlier expression:

$$
\frac{1}{2}bc \sin A = 2bc(1 - \cos A)
$$

Divide both sides by $bc$ (since $b, c > 0$):

$$
\frac{1}{2} \sin A = 2(1 - \cos A)
$$

Multiply both sides by 2:

$$
\sin A = 4(1 - \cos A)
$$

---

### Step 3: Use trigonometric identities

We now have:

$$
\sin A = 4(1 - \cos A)
$$

Let’s square both sides:

$$
\sin^2 A = 16(1 - \cos A)^2
$$

Use the identity $\sin^2 A = 1 - \cos^2 A$:

$$
1 - \cos^2 A = 16(1 - 2\cos A + \cos^2 A)
$$

Expand the right-hand side:

$$
1 - \cos^2 A = 16 - 32\cos A + 16\cos^2 A
$$

Bring all terms to one side:

$$
1 - \cos^2 A - 16 + 32\cos A - 16\cos^2 A = 0
$$

Simplify:

$$
-15 + 32\cos A - 17\cos^2 A = 0
$$

Multiply through by $-1$:

$$
17\cos^2 A - 32\cos A + 15 = 0
$$

---

### Step 4: Solve the quadratic

Let $x = \cos A$, then:

$$
17x^2 - 32x + 15 = 0
$$

Use the quadratic formula:

$$
x = \frac{32 \pm \sqrt{(-32)^2 - 4(17)(15)}}{2(17)} = \frac{32 \pm \sqrt{1024 - 1020}}{34} = \frac{32 \pm 2}{34}
$$

So:

$$
x = \frac{34}{34} = 1 \quad \text{or} \quad x = \frac{30}{34} = \frac{15}{17}
$$

Since $\cos A = 1$ would imply angle $A = 0^\circ$, which is not possible in a triangle, we take:

$$
\cos A = \frac{15}{17}
$$

Now use the earlier equation:

$$
\sin A = 4(1 - \cos A) = 4\left(1 - \frac{15}{17}\right) = 4 \cdot \frac{2}{17} = \frac{8}{17}
$$

---

### Step 5: Compute $\tan A$

$$
\tan A = \frac{\sin A}{\cos A} = \frac{8/17}{15/17} = \frac{8}{15}
$$

---

### Final Answer

$$
\boxed{\dfrac{8}{15}}
$$

Solution:
The area of triangle $ABC$ is given by
\[\frac{1}{2} bc \sin A.\]Hence,
\[\frac{1}{2} bc \sin A = a^2 - (b - c)^2 = a^2 - b^2 + 2bc - c^2.\]By the Law of Cosines, $b^2 + c^2 - 2bc \cos A = a^2,$ so
\[\frac{1}{2} bc \sin A = 2bc - 2bc \cos A.\]This simplifies to $\sin A = 4 - 4 \cos A.$  Squaring both sides, we get
\[\sin^2 A = 16 - 32 \cos A + 16 \cos^2 A,\]so $1 - \cos^2 A = 16 - 32 \cos A + 16 \cos^2 A.$  This simplifies to
\[17 \cos^2 A - 32 \cos A + 15 = 0.\]This factors as $(\cos A - 1)(17 \cos A - 15) = 0.$  Since $\cos A$ cannot be equal to 1, $\cos A = \frac{15}{17}.$

Then $\sin A = 4 - 4 \cos A = \frac{8}{17},$ so
\[\tan A = \frac{\sin A}{\cos A} = \boxed{\frac{8}{15}}.\]


Explanation:
Okay, so I need to solve this equation: √(x + √(3x + 6)) + √(x - √(3x + 6)) = 6. Hmm, that looks a bit complicated with the nested square roots. Let me try to break it down step by step.

First, maybe I can simplify the equation by letting some substitution. Let me think... The equation has two square roots added together. Let me denote the entire expression inside the square roots. Let me set y = √(3x + 6). Then maybe I can rewrite the original equation in terms of y?

Wait, let me check. The original equation is √(x + y) + √(x - y) = 6, where y = √(3x + 6). Hmm, that might be helpful. Let me write that down:

Let y = √(3x + 6). Then the equation becomes √(x + y) + √(x - y) = 6.

Now, maybe I can square both sides to eliminate the square roots? Let me try that. If I square both sides, I get:

[√(x + y) + √(x - y)]² = 6²

Expanding the left side: (√(x + y))² + 2√(x + y)√(x - y) + (√(x - y))² = 36

Simplify each term:

(x + y) + 2√[(x + y)(x - y)] + (x - y) = 36

Combine like terms: x + y + x - y + 2√(x² - y²) = 36

So that simplifies to 2x + 2√(x² - y²) = 36

Divide both sides by 2: x + √(x² - y²) = 18

Hmm, okay. Now, remember that y = √(3x + 6). Let me substitute that back in:

x + √[x² - (√(3x + 6))²] = 18

Compute the square inside the square root:

(√(3x + 6))² is just 3x + 6. So:

x + √[x² - (3x + 6)] = 18

Simplify inside the square root:

x² - 3x - 6. So:

x + √(x² - 3x - 6) = 18

Now, let me isolate the square root term. Subtract x from both sides:

√(x² - 3x - 6) = 18 - x

Now, square both sides again to eliminate the square root:

[√(x² - 3x - 6)]² = (18 - x)²

Which simplifies to:

x² - 3x - 6 = 324 - 36x + x²

Wait, let me check that expansion. (18 - x)² is 18² - 2*18*x + x², which is 324 - 36x + x². Yes, that's correct.

Now, subtract x² from both sides:

-3x - 6 = 324 - 36x

Now, let me bring all terms to one side. Let me add 36x to both sides and add 6 to both sides:

-3x - 6 + 36x + 6 = 324

Simplify:

33x = 324

Therefore, x = 324 / 33. Let me simplify that fraction. Both numerator and denominator are divisible by 3:

324 ÷ 3 = 108; 33 ÷ 3 = 11. So x = 108/11. Let me check if that reduces further. 108 and 11 have no common factors since 11 is prime and doesn't divide 108. So x = 108/11.

But wait, I need to check if this solution is valid because we squared both sides twice, which can introduce extraneous solutions. So I need to verify this solution in the original equation.

First, let me compute 3x + 6. If x = 108/11, then 3x = 324/11, so 3x + 6 = 324/11 + 66/11 = 390/11. Therefore, √(3x + 6) = √(390/11). Let me compute that value numerically to check if it's real. Since 390/11 is approximately 35.45, so square root is about 5.95. Let me keep it symbolic for now.

Now, let me check the original equation:

√(x + √(3x + 6)) + √(x - √(3x + 6)) = 6.

Let me compute x + √(3x + 6) and x - √(3x + 6). Let me denote √(3x + 6) as y again. So y = √(390/11). Then x + y = 108/11 + √(390/11) and x - y = 108/11 - √(390/11). Let me compute these terms.

But maybe instead of calculating numerically, I can check if the equation holds. Let me recall that earlier steps led us to x = 108/11. Let me check if when we squared both sides, we didn't introduce any inconsistencies.

Wait, when we had √(x² - 3x - 6) = 18 - x, we squared both sides. So we need to make sure that 18 - x is non-negative because the square root is non-negative. So 18 - x ≥ 0 ⇒ x ≤ 18. Since x = 108/11 ≈ 9.818, which is less than 18, so that's okay.

Also, the expression under the square root in the original equation must be non-negative. Let me check the domains.

First, for √(3x + 6) to be real, 3x + 6 ≥ 0 ⇒ x ≥ -2.

Then, for √(x + √(3x + 6)) to be real, x + √(3x + 6) ≥ 0. Since x ≥ -2 and √(3x + 6) is non-negative, x + √(3x + 6) is at least -2 + 0 = -2. But we need it to be non-negative. Let me check for x ≥ -2:

Let me suppose x ≥ -2. Let me check if x + √(3x + 6) ≥ 0. Let me consider x ≥ -2. Let me check for x = -2: √(3*(-2) + 6) = √(0) = 0. Then x + √(...) = -2 + 0 = -2 < 0. So that's invalid. Therefore, x must be such that x + √(3x + 6) ≥ 0. Similarly, for the other square root, x - √(3x + 6) ≥ 0. So both expressions under the square roots must be non-negative.

So, for √(x + √(3x + 6)) to be real: x + √(3x + 6) ≥ 0.

For √(x - √(3x + 6)) to be real: x - √(3x + 6) ≥ 0 ⇒ x ≥ √(3x + 6).

So, combining these two conditions:

First, x ≥ √(3x + 6). Let me solve this inequality.

Let me square both sides (since both sides are non-negative? Wait, x must be ≥ √(3x + 6). Let me check if x is non-negative. Let me think. If x ≥ √(3x + 6), then since √(3x + 6) is non-negative, x must be non-negative. Therefore, x ≥ 0. So, squaring both sides:

x² ≥ 3x + 6 ⇒ x² - 3x - 6 ≥ 0.

Solving x² - 3x - 6 ≥ 0. The quadratic equation x² - 3x - 6 = 0 has solutions x = [3 ± √(9 + 24)] / 2 = [3 ± √33]/2. Since √33 ≈ 5.744, so the positive root is (3 + 5.744)/2 ≈ 8.744 / 2 ≈ 4.372. Therefore, the inequality x² - 3x - 6 ≥ 0 holds when x ≤ [3 - √33]/2 or x ≥ [3 + √33]/2. But since x ≥ 0, and [3 - √33]/2 is negative (since √33 ≈ 5.744, so 3 - 5.744 ≈ -2.744, divided by 2 is ≈ -1.372). Therefore, the inequality holds for x ≥ [3 + √33]/2 ≈ 4.372. Therefore, our solution x = 108/11 ≈ 9.818 is in this interval, so it satisfies x ≥ √(3x + 6). Therefore, the domain is okay.

Also, check x + √(3x + 6) ≥ 0. Since x ≥ [3 + √33]/2 ≈ 4.372, and √(3x + 6) is positive, so x + √(3x + 6) is definitely positive. Therefore, the domain conditions are satisfied.

So, x = 108/11 is a valid solution? Let me check in the original equation.

Let me compute the left-hand side (LHS) of the original equation:

√(x + √(3x + 6)) + √(x - √(3x + 6)).

Let me compute each term step by step.

First, compute 3x + 6 when x = 108/11:

3*(108/11) + 6 = 324/11 + 66/11 = 390/11. So √(3x + 6) = √(390/11). Let me compute that value numerically:

√(390/11) ≈ √(35.4545) ≈ 5.954.

Then, compute x + √(3x + 6) ≈ 108/11 + 5.954 ≈ 9.818 + 5.954 ≈ 15.772. Then √(15.772) ≈ 3.971.

Similarly, compute x - √(3x + 6) ≈ 9.818 - 5.954 ≈ 3.864. Then √(3.864) ≈ 1.966.

Adding them together: 3.971 + 1.966 ≈ 5.937. Wait, but the original equation says it should equal 6. Hmm, there's a discrepancy here. So my approximate calculation suggests that the left-hand side is approximately 5.937, which is close to 6 but not exactly. Maybe my approximations are off? Let me check more accurately.

Alternatively, maybe I made a mistake in algebra. Let me check my steps again.

Starting from the original equation:

√(x + √(3x + 6)) + √(x - √(3x + 6)) = 6.

Let me denote A = √(x + √(3x + 6)) and B = √(x - √(3x + 6)). Then A + B = 6.

Earlier, I squared both sides and got 2x + 2√(x² - y²) = 36, where y = √(3x + 6). Then simplified to x + √(x² - 3x - 6) = 18. Then solved for x = 108/11. But when I plug back in, the approximate value isn't exactly 6. Maybe my calculation was wrong?

Alternatively, maybe I need to do exact calculations. Let me try that.

Let me compute x = 108/11. Let me compute √(3x + 6):

3x + 6 = 3*(108/11) + 6 = 324/11 + 66/11 = 390/11. So √(390/11). Let me denote this as y = √(390/11).

Then compute x + y = 108/11 + √(390/11). Let me compute √(x + y) and √(x - y). Wait, but maybe there's a better way. Let me recall that earlier steps led us to x = 108/11. Let me check if when we squared the equation, we might have introduced an error.

Wait, let me check the step where I squared √(x² - 3x - 6) = 18 - x. Let me compute x² - 3x - 6 when x = 108/11.

First, x² = (108/11)^2 = 11664/121.

Then, 3x = 324/11, so 3x + 6 = 324/11 + 66/11 = 390/11. Therefore, x² - 3x - 6 = 11664/121 - 390/11.

Convert 390/11 to 121 denominator: 390/11 = (390 * 11)/121 = 4290/121. Wait, no. Wait, 390/11 is equal to (390 * 11)/121? No, that's not correct. To convert 390/11 to denominator 121, multiply numerator and denominator by 11: 390*11 / 121. Let me compute that:

390 * 11 = 4290. Therefore, x² - 3x - 6 = 11664/121 - 4290/121 = (11664 - 4290)/121 = 7374 / 121.

Now, compute 18 - x = 18 - 108/11 = (198 - 108)/11 = 90/11. Then (18 - x)^2 = (90/11)^2 = 8100 / 121.

Now, check if x² - 3x - 6 equals (18 - x)^2:

x² - 3x - 6 = 7374 / 121 ≈ 60.94

(18 - x)^2 = 8100 / 121 ≈ 66.94

Wait, these are not equal! That means there's a mistake in my algebra. Wait, this is a problem. So according to my previous steps, I had:

After squaring both sides, I had x² - 3x - 6 = 324 - 36x + x². Then subtract x²: -3x -6 = 324 - 36x. Then adding 36x to both sides: 33x -6 = 324. Then adding 6: 33x = 330 ⇒ x = 10. Wait, wait! Wait, earlier I said 33x = 324? Wait, no! Wait, let me check again.

Wait, let me go back. After squaring both sides, I had:

x² - 3x - 6 = 324 - 36x + x².

Subtract x² from both sides:

-3x - 6 = 324 - 36x.

Then, adding 36x to both sides:

(-3x + 36x) - 6 = 324 ⇒ 33x - 6 = 324.

Then adding 6 to both sides:

33x = 324 + 6 = 330 ⇒ x = 330 / 33 = 10. Wait! Earlier I said 33x = 324, but that was a mistake. It's 33x = 330. Therefore, x = 10. Oh no! I must have miscalculated earlier. So x = 10, not 108/11. That explains why my approximate calculation didn't match. I had a calculation error earlier. So x = 10. Let me check that again.

So, let me redo that step. After squaring both sides:

x² - 3x - 6 = 324 - 36x + x².

Subtract x²: -3x - 6 = 324 - 36x.

Then, adding 36x to both sides: 33x - 6 = 324.

Adding 6: 33x = 330 ⇒ x = 10. Yes, that's correct. So my previous calculation had a mistake where I thought 33x = 324, but it's actually 33x = 330. Therefore, x = 10. That was a critical error. So x = 10. Let me check this solution.

So x = 10. Let me check if this satisfies the original equation.

First, check the domain conditions. For √(3x + 6): 3*10 + 6 = 36, so √36 = 6. Then check x - √(3x + 6) = 10 - 6 = 4 ≥ 0. So both square roots are defined.

Now compute the original equation:

√(10 + √(3*10 + 6)) + √(10 - √(3*10 + 6)) = √(10 + √36) + √(10 - √36) = √(10 + 6) + √(10 - 6) = √16 + √4 = 4 + 2 = 6. Which matches the right-hand side. Perfect! So x = 10 is the correct solution.

But why did I get 108/11 earlier? Because I made a mistake in arithmetic when solving for x. I had written 33x = 324, but it's actually 33x = 330. So that was a critical error. So the correct solution is x = 10.

But let me check if there are other solutions. When we squared both sides, sometimes extraneous solutions can be introduced, but in this case, we only had one solution. Let me check if there could be another solution.

Looking back at the equation after squaring:

√(x² - 3x - 6) = 18 - x. We had to have 18 - x ≥ 0 ⇒ x ≤ 18. Also, the expression under the square root must be non-negative: x² - 3x - 6 ≥ 0. Which, as we saw earlier, requires x ≥ [3 + √33]/2 ≈ 4.372. So x is between approximately 4.372 and 18. But when we solved, we found x = 10. Is there another solution?

Let me check. Suppose there is another solution. Let me consider the equation √(x² - 3x - 6) = 18 - x. Let me square both sides again:

x² - 3x - 6 = (18 - x)^2 ⇒ x² - 3x - 6 = 324 - 36x + x² ⇒ -3x -6 = 324 - 36x ⇒ 33x = 330 ⇒ x = 10. So only one solution. Therefore, x = 10 is the only solution. Therefore, the answer is x = 10.

But let me check if there's another way to approach the problem, just to confirm.

Alternative approach: Let me denote the original equation as:

√(x + √(3x + 6)) + √(x - √(3x + 6)) = 6.

Let me let’s denote √(3x + 6) = t. Then, t ≥ 0, and 3x + 6 = t² ⇒ x = (t² - 6)/3.

Substituting into the original equation:

√[( (t² - 6)/3 ) + t] + √[( (t² - 6)/3 ) - t] = 6.

Simplify each term inside the square roots:

First term inside first sqrt: (t² - 6)/3 + t = (t² - 6 + 3t)/3 = (t² + 3t - 6)/3.

Second term inside second sqrt: (t² - 6)/3 - t = (t² - 6 - 3t)/3 = (t² - 3t - 6)/3.

Therefore, the equation becomes:

√[(t² + 3t - 6)/3] + √[(t² - 3t - 6)/3] = 6.

Let me factor out 1/√3:

[1/√3][√(t² + 3t - 6) + √(t² - 3t - 6)] = 6.

Multiply both sides by √3:

√(t² + 3t - 6) + √(t² - 3t - 6) = 6√3.

Hmm, this seems more complicated. Maybe not helpful. Let me check if this approach can lead to something. Alternatively, maybe square both sides again.

Let me denote A = √(t² + 3t - 6) and B = √(t² - 3t - 6). Then A + B = 6√3. Squaring both sides:

A² + 2AB + B² = 36*3 = 108.

Compute A² + B²:

(t² + 3t - 6) + (t² - 3t - 6) = 2t² - 12.

Therefore:

2t² - 12 + 2AB = 108 ⇒ 2AB = 108 + 12 - 2t² ⇒ 2AB = 120 - 2t² ⇒ AB = 60 - t².

But AB = √(t² + 3t - 6) * √(t² - 3t - 6) = √[(t² + 3t - 6)(t² - 3t - 6)].

Let me compute the product inside the square root:

(t² + 3t - 6)(t² - 3t - 6) = [t² - 6 + 3t][t² - 6 - 3t] = (t² - 6)^2 - (3t)^2 = t^4 - 12t² + 36 - 9t² = t^4 - 21t² + 36.

Therefore, AB = √(t^4 - 21t² + 36).

So we have:

√(t^4 - 21t² + 36) = 60 - t².

Square both sides:

t^4 - 21t² + 36 = (60 - t²)^2 = 3600 - 120t² + t^4.

Subtract t^4 from both sides:

-21t² + 36 = 3600 - 120t².

Bring all terms to left side:

-21t² + 36 - 3600 + 120t² = 0 ⇒ 99t² - 3564 = 0 ⇒ 99t² = 3564 ⇒ t² = 3564 / 99.

Simplify:

Divide numerator and denominator by 9: 3564 ÷ 9 = 396; 99 ÷ 9 = 11. So t² = 396 / 11 = 36. Therefore, t² = 36 ⇒ t = 6 or t = -6. But since t = √(3x + 6) ≥ 0, so t = 6.

Therefore, t = 6. Then, since t = √(3x + 6) = 6 ⇒ 3x + 6 = 36 ⇒ 3x = 30 ⇒ x = 10. Which matches our previous result. So this confirms that x = 10 is the solution. Therefore, the answer is x = 10.

Therefore, my initial mistake was in the arithmetic when solving for x, but after correcting that, both methods lead to x = 10, which checks out in the original equation. Therefore, the solution is x = 10.

**Final Answer**
The solution is \boxed{10}.

To solve the equation:

$$
\sqrt{x + \sqrt{3x + 6}} + \sqrt{x - \sqrt{3x + 6}} = 6,
$$

we begin by simplifying the structure of the equation. Let us define:

$$
y = \sqrt{3x + 6}.
$$

Then the original equation becomes:

$$
\sqrt{x + y} + \sqrt{x - y} = 6.
$$

---

### Step 1: Square Both Sides

Squaring both sides:

$$
\left( \sqrt{x + y} + \sqrt{x - y} \right)^2 = 6^2 = 36.
$$

Expanding the left-hand side:

$$
(x + y) + (x - y) + 2\sqrt{(x + y)(x - y)} = 36.
$$

Simplify:

$$
2x + 2\sqrt{x^2 - y^2} = 36.
$$

Divide both sides by 2:

$$
x + \sqrt{x^2 - y^2} = 18.
$$

---

### Step 2: Substitute Back $ y = \sqrt{3x + 6} $

Recall that $ y^2 = 3x + 6 $, so:

$$
x + \sqrt{x^2 - (3x + 6)} = 18.
$$

Simplify the expression under the square root:

$$
x + \sqrt{x^2 - 3x - 6} = 18.
$$

---

### Step 3: Isolate the Square Root

Subtract $ x $ from both sides:

$$
\sqrt{x^2 - 3x - 6} = 18 - x.
$$

Now square both sides again:

$$
x^2 - 3x - 6 = (18 - x)^2.
$$

Expand the right-hand side:

$$
x^2 - 3x - 6 = 324 - 36x + x^2.
$$

Subtract $ x^2 $ from both sides:

$$
-3x - 6 = 324 - 36x.
$$

Add $ 36x $ to both sides:

$$
33x - 6 = 324.
$$

Add 6 to both sides:

$$
33x = 330.
$$

Solve for $ x $:

$$
x = \frac{330}{33} = 10.
$$

---

### Step 4: Verify the Solution

Let’s verify that $ x = 10 $ satisfies the original equation.

Compute $ \sqrt{3x + 6} = \sqrt{3(10) + 6} = \sqrt{36} = 6 $.

Now compute the original expression:

$$
\sqrt{10 + 6} + \sqrt{10 - 6} = \sqrt{16} + \sqrt{4} = 4 + 2 = 6.
$$

This matches the right-hand side of the equation, so the solution is valid.

---

### Final Answer

$$
\boxed{10}
$$

[Assistant]
Explanation:
Okay, so I need to solve this equation: √(x + √(3x + 6)) + √(x - √(3x + 6)) = 6. Hmm, that looks a bit complicated with the nested square roots. Let me try to break it down step by step.

First, maybe I can simplify the equation by letting some substitution. Let me think... The equation has two square roots added together. Let me denote the entire expression inside the square roots. Let me set y = √(3x + 6). Then maybe I can rewrite the original equation in terms of y?

Wait, let me check. The original equation is √(x + y) + √(x - y) = 6, where y = √(3x + 6). Hmm, that might be helpful. Let me write that down:

Let y = √(3x + 6). Then the equation becomes √(x + y) + √(x - y) = 6.

Now, maybe I can square both sides to eliminate the square roots? Let me try that. If I square both sides, I get:

[√(x + y) + √(x - y)]² = 6²

Expanding the left side: (√(x + y))² + 2√(x + y)√(x - y) + (√(x - y))² = 36

Simplify each term:

(x + y) + 2√[(x + y)(x - y)] + (x - y) = 36

Combine like terms: x + y + x - y + 2√(x² - y²) = 36

So that simplifies to 2x + 2√(x² - y²) = 36

Divide both sides by 2: x + √(x² - y²) = 18

Hmm, okay. Now, remember that y = √(3x + 6). Let me substitute that back in:

x + √[x² - (√(3x + 6))²] = 18

Compute the square inside the square root:

(√(3x + 6))² is just 3x + 6. So:

x + √[x² - (3x + 6)] = 18

Simplify inside the square root:

x² - 3x - 6. So:

x + √(x² - 3x - 6) = 18

Now, let me isolate the square root term. Subtract x from both sides:

√(x² - 3x - 6) = 18 - x

Now, square both sides again to eliminate the square root:

[√(x² - 3x - 6)]² = (18 - x)²

Which simplifies to:

x² - 3x - 6 = 324 - 36x + x²

Wait, let me check that expansion. (18 - x)² is 18² - 2*18*x + x², which is 324 - 36x + x². Yes, that's correct.

Now, subtract x² from both sides:

-3x - 6 = 324 - 36x

Now, let me bring all terms to one side. Let me add 36x to both sides and add 6 to both sides:

-3x - 6 + 36x + 6 = 324

Simplify:

33x = 324

Therefore, x = 324 / 33. Let me simplify that fraction. Both numerator and denominator are divisible by 3:

324 ÷ 3 = 108; 33 ÷ 3 = 11. So x = 108/11. Let me check if that reduces further. 108 and 11 have no common factors since 11 is prime and doesn't divide 108. So x = 108/11.

But wait, I need to check if this solution is valid because we squared both sides twice, which can introduce extraneous solutions. So I need to verify this solution in the original equation.

First, let me compute 3x + 6. If x = 108/11, then 3x = 324/11, so 3x + 6 = 324/11 + 66/11 = 390/11. Therefore, √(3x + 6) = √(390/11). Let me compute that value numerically to check if it's real. Since 390/11 is approximately 35.45, so square root is about 5.95. Let me keep it symbolic for now.

Now, let me check the original equation:

√(x + √(3x + 6)) + √(x - √(3x + 6)) = 6.

Let me compute x + √(3x + 6) and x - √(3x + 6). Let me denote √(3x + 6) as y again. So y = √(390/11). Then x + y = 108/11 + √(390/11) and x - y = 108/11 - √(390/11). Let me compute these terms.

But maybe instead of calculating numerically, I can check if the equation holds. Let me recall that earlier steps led us to x = 108/11. Let me check if when we squared both sides, we didn't introduce any inconsistencies.

Wait, when we had √(x² - 3x - 6) = 18 - x, we squared both sides. So we need to make sure that 18 - x is non-negative because the square root is non-negative. So 18 - x ≥ 0 ⇒ x ≤ 18. Since x = 108/11 ≈ 9.818, which is less than 18, so that's okay.

Also, the expression under the square root in the original equation must be non-negative. Let me check the domains.

First, for √(3x + 6) to be real, 3x + 6 ≥ 0 ⇒ x ≥ -2.

Then, for √(x + √(3x + 6)) to be real, x + √(3x + 6) ≥ 0. Since x ≥ -2 and √(3x + 6) is non-negative, x + √(3x + 6) is at least -2 + 0 = -2. But we need it to be non-negative. Let me check for x ≥ -2:

Let me suppose x ≥ -2. Let me check if x + √(3x + 6) ≥ 0. Let me consider x ≥ -2. Let me check for x = -2: √(3*(-2) + 6) = √(0) = 0. Then x + √(...) = -2 + 0 = -2 < 0. So that's invalid. Therefore, x must be such that x + √(3x + 6) ≥ 0. Similarly, for the other square root, x - √(3x + 6) ≥ 0. So both expressions under the square roots must be non-negative.

So, for √(x + √(3x + 6)) to be real: x + √(3x + 6) ≥ 0.

For √(x - √(3x + 6)) to be real: x - √(3x + 6) ≥ 0 ⇒ x ≥ √(3x + 6).

So, combining these two conditions:

First, x ≥ √(3x + 6). Let me solve this inequality.

Let me square both sides (since both sides are non-negative? Wait, x must be ≥ √(3x + 6). Let me check if x is non-negative. Let me think. If x ≥ √(3x + 6), then since √(3x + 6) is non-negative, x must be non-negative. Therefore, x ≥ 0. So, squaring both sides:

x² ≥ 3x + 6 ⇒ x² - 3x - 6 ≥ 0.

Solving x² - 3x - 6 ≥ 0. The quadratic equation x² - 3x - 6 = 0 has solutions x = [3 ± √(9 + 24)] / 2 = [3 ± √33]/2. Since √33 ≈ 5.744, so the positive root is (3 + 5.744)/2 ≈ 8.744 / 2 ≈ 4.372. Therefore, the inequality x² - 3x - 6 ≥ 0 holds when x ≤ [3 - √33]/2 or x ≥ [3 + √33]/2. But since x ≥ 0, and [3 - √33]/2 is negative (since √33 ≈ 5.744, so 3 - 5.744 ≈ -2.744, divided by 2 is ≈ -1.372). Therefore, the inequality holds for x ≥ [3 + √33]/2 ≈ 4.372. Therefore, our solution x = 108/11 ≈ 9.818 is in this interval, so it satisfies x ≥ √(3x + 6). Therefore, the domain is okay.

Also, check x + √(3x + 6) ≥ 0. Since x ≥ [3 + √33]/2 ≈ 4.372, and √(3x + 6) is positive, so x + √(3x + 6) is definitely positive. Therefore, the domain conditions are satisfied.

So, x = 108/11 is a valid solution? Let me check in the original equation.

Let me compute the left-hand side (LHS) of the original equation:

√(x + √(3x + 6)) + √(x - √(3x + 6)).

Let me compute each term step by step.

First, compute 3x + 6 when x = 108/11:

3*(108/11) + 6 = 324/11 + 66/11 = 390/11. So √(3x + 6) = √(390/11). Let me compute that value numerically:

√(390/11) ≈ √(35.4545) ≈ 5.954.

Then, compute x + √(3x + 6) ≈ 108/11 + 5.954 ≈ 9.818 + 5.954 ≈ 15.772. Then √(15.772) ≈ 3.971.

Similarly, compute x - √(3x + 6) ≈ 9.818 - 5.954 ≈ 3.864. Then √(3.864) ≈ 1.966.

Adding them together: 3.971 + 1.966 ≈ 5.937. Wait, but the original equation says it should equal 6. Hmm, there's a discrepancy here. So my approximate calculation suggests that the left-hand side is approximately 5.937, which is close to 6 but not exactly. Maybe my approximations are off? Let me check more accurately.

Alternatively, maybe I made a mistake in algebra. Let me check my steps again.

Starting from the original equation:

√(x + √(3x + 6)) + √(x - √(3x + 6)) = 6.

Let me denote A = √(x + √(3x + 6)) and B = √(x - √(3x + 6)). Then A + B = 6.

Earlier, I squared both sides and got 2x + 2√(x² - y²) = 36, where y = √(3x + 6). Then simplified to x + √(x² - 3x - 6) = 18. Then solved for x = 108/11. But when I plug back in, the approximate value isn't exactly 6. Maybe my calculation was wrong?

Alternatively, maybe I need to do exact calculations. Let me try that.

Let me compute x = 108/11. Let me compute √(3x + 6):

3x + 6 = 3*(108/11) + 6 = 324/11 + 66/11 = 390/11. So √(390/11). Let me denote this as y = √(390/11).

Then compute x + y = 108/11 + √(390/11). Let me compute √(x + y) and √(x - y). Wait, but maybe there's a better way. Let me recall that earlier steps led us to x = 108/11. Let me check if when we squared the equation, we might have introduced an error.

Wait, let me check the step where I squared √(x² - 3x - 6) = 18 - x. Let me compute x² - 3x - 6 when x = 108/11.

First, x² = (108/11)^2 = 11664/121.

Then, 3x = 324/11, so 3x + 6 = 324/11 + 66/11 = 390/11. Therefore, x² - 3x - 6 = 11664/121 - 390/11.

Convert 390/11 to 121 denominator: 390/11 = (390 * 11)/121 = 4290/121. Wait, no. Wait, 390/11 is equal to (390 * 11)/121? No, that's not correct. To convert 390/11 to denominator 121, multiply numerator and denominator by 11: 390*11 / 121. Let me compute that:

390 * 11 = 4290. Therefore, x² - 3x - 6 = 11664/121 - 4290/121 = (11664 - 4290)/121 = 7374 / 121.

Now, compute 18 - x = 18 - 108/11 = (198 - 108)/11 = 90/11. Then (18 - x)^2 = (90/11)^2 = 8100 / 121.

Now, check if x² - 3x - 6 equals (18 - x)^2:

x² - 3x - 6 = 7374 / 121 ≈ 60.94

(18 - x)^2 = 8100 / 121 ≈ 66.94

Wait, these are not equal! That means there's a mistake in my algebra. Wait, this is a problem. So according to my previous steps, I had:

After squaring both sides, I had x² - 3x - 6 = 324 - 36x + x². Then subtract x²: -3x -6 = 324 - 36x. Then adding 36x to both sides: 33x -6 = 324. Then adding 6: 33x = 330 ⇒ x = 10. Wait, wait! Wait, earlier I said 33x = 324? Wait, no! Wait, let me check again.

Wait, let me go back. After squaring both sides, I had:

x² - 3x - 6 = 324 - 36x + x².

Subtract x² from both sides:

-3x - 6 = 324 - 36x.

Then, adding 36x to both sides:

(-3x + 36x) - 6 = 324 ⇒ 33x - 6 = 324.

Then adding 6 to both sides:

33x = 324 + 6 = 330 ⇒ x = 330 / 33 = 10. Wait! Earlier I said 33x = 324, but that was a mistake. It's 33x = 330. Therefore, x = 10. Oh no! I must have miscalculated earlier. So x = 10, not 108/11. That explains why my approximate calculation didn't match. I had a calculation error earlier. So x = 10. Let me check that again.

So, let me redo that step. After squaring both sides:

x² - 3x - 6 = 324 - 36x + x².

Subtract x²: -3x - 6 = 324 - 36x.

Then, adding 36x to both sides: 33x - 6 = 324.

Adding 6: 33x = 330 ⇒ x = 10. Yes, that's correct. So my previous calculation had a mistake where I thought 33x = 324, but it's actually 33x = 330. Therefore, x = 10. That was a critical error. So x = 10. Let me check this solution.

So x = 10. Let me check if this satisfies the original equation.

First, check the domain conditions. For √(3x + 6): 3*10 + 6 = 36, so √36 = 6. Then check x - √(3x + 6) = 10 - 6 = 4 ≥ 0. So both square roots are defined.

Now compute the original equation:

√(10 + √(3*10 + 6)) + √(10 - √(3*10 + 6)) = √(10 + √36) + √(10 - √36) = √(10 + 6) + √(10 - 6) = √16 + √4 = 4 + 2 = 6. Which matches the right-hand side. Perfect! So x = 10 is the correct solution.

But why did I get 108/11 earlier? Because I made a mistake in arithmetic when solving for x. I had written 33x = 324, but it's actually 33x = 330. So that was a critical error. So the correct solution is x = 10.

But let me check if there are other solutions. When we squared both sides, sometimes extraneous solutions can be introduced, but in this case, we only had one solution. Let me check if there could be another solution.

Looking back at the equation after squaring:

√(x² - 3x - 6) = 18 - x. We had to have 18 - x ≥ 0 ⇒ x ≤ 18. Also, the expression under the square root must be non-negative: x² - 3x - 6 ≥ 0. Which, as we saw earlier, requires x ≥ [3 + √33]/2 ≈ 4.372. So x is between approximately 4.372 and 18. But when we solved, we found x = 10. Is there another solution?

Let me check. Suppose there is another solution. Let me consider the equation √(x² - 3x - 6) = 18 - x. Let me square both sides again:

x² - 3x - 6 = (18 - x)^2 ⇒ x² - 3x - 6 = 324 - 36x + x² ⇒ -3x -6 = 324 - 36x ⇒ 33x = 330 ⇒ x = 10. So only one solution. Therefore, x = 10 is the only solution. Therefore, the answer is x = 10.

But let me check if there's another way to approach the problem, just to confirm.

Alternative approach: Let me denote the original equation as:

√(x + √(3x + 6)) + √(x - √(3x + 6)) = 6.

Let me let’s denote √(3x + 6) = t. Then, t ≥ 0, and 3x + 6 = t² ⇒ x = (t² - 6)/3.

Substituting into the original equation:

√[( (t² - 6)/3 ) + t] + √[( (t² - 6)/3 ) - t] = 6.

Simplify each term inside the square roots:

First term inside first sqrt: (t² - 6)/3 + t = (t² - 6 + 3t)/3 = (t² + 3t - 6)/3.

Second term inside second sqrt: (t² - 6)/3 - t = (t² - 6 - 3t)/3 = (t² - 3t - 6)/3.

Therefore, the equation becomes:

√[(t² + 3t - 6)/3] + √[(t² - 3t - 6)/3] = 6.

Let me factor out 1/√3:

[1/√3][√(t² + 3t - 6) + √(t² - 3t - 6)] = 6.

Multiply both sides by √3:

√(t² + 3t - 6) + √(t² - 3t - 6) = 6√3.

Hmm, this seems more complicated. Maybe not helpful. Let me check if this approach can lead to something. Alternatively, maybe square both sides again.

Let me denote A = √(t² + 3t - 6) and B = √(t² - 3t - 6). Then A + B = 6√3. Squaring both sides:

A² + 2AB + B² = 36*3 = 108.

Compute A² + B²:

(t² + 3t - 6) + (t² - 3t - 6) = 2t² - 12.

Therefore:

2t² - 12 + 2AB = 108 ⇒ 2AB = 108 + 12 - 2t² ⇒ 2AB = 120 - 2t² ⇒ AB = 60 - t².

But AB = √(t² + 3t - 6) * √(t² - 3t - 6) = √[(t² + 3t - 6)(t² - 3t - 6)].

Let me compute the product inside the square root:

(t² + 3t - 6)(t² - 3t - 6) = [t² - 6 + 3t][t² - 6 - 3t] = (t² - 6)^2 - (3t)^2 = t^4 - 12t² + 36 - 9t² = t^4 - 21t² + 36.

Therefore, AB = √(t^4 - 21t² + 36).

So we have:

√(t^4 - 21t² + 36) = 60 - t².

Square both sides:

t^4 - 21t² + 36 = (60 - t²)^2 = 3600 - 120t² + t^4.

Subtract t^4 from both sides:

-21t² + 36 = 3600 - 120t².

Bring all terms to left side:

-21t² + 36 - 3600 + 120t² = 0 ⇒ 99t² - 3564 = 0 ⇒ 99t² = 3564 ⇒ t² = 3564 / 99.

Simplify:

Divide numerator and denominator by 9: 3564 ÷ 9 = 396; 99 ÷ 9 = 11. So t² = 396 / 11 = 36. Therefore, t² = 36 ⇒ t = 6 or t = -6. But since t = √(3x + 6) ≥ 0, so t = 6.

Therefore, t = 6. Then, since t = √(3x + 6) = 6 ⇒ 3x + 6 = 36 ⇒ 3x = 30 ⇒ x = 10. Which matches our previous result. So this confirms that x = 10 is the solution. Therefore, the answer is x = 10.

Therefore, my initial mistake was in the arithmetic when solving for x, but after correcting that, both methods lead to x = 10, which checks out in the original equation. Therefore, the solution is x = 10.

**Final Answer**
The solution is \boxed{10}.

To solve the equation:

$$
\sqrt{x + \sqrt{3x + 6}} + \sqrt{x - \sqrt{3x + 6}} = 6,
$$

we begin by simplifying the structure of the equation. Let us define:

$$
y = \sqrt{3x + 6}.
$$

Then the original equation becomes:

$$
\sqrt{x + y} + \sqrt{x - y} = 6.
$$

---

### Step 1: Square Both Sides

Squaring both sides:

$$
\left( \sqrt{x + y} + \sqrt{x - y} \right)^2 = 6^2 = 36.
$$

Expanding the left-hand side:

$$
(x + y) + (x - y) + 2\sqrt{(x + y)(x - y)} = 36.
$$

Simplify:

$$
2x + 2\sqrt{x^2 - y^2} = 36.
$$

Divide both sides by 2:

$$
x + \sqrt{x^2 - y^2} = 18.
$$

---

### Step 2: Substitute Back $ y = \sqrt{3x + 6} $

Recall that $ y^2 = 3x + 6 $, so:

$$
x + \sqrt{x^2 - (3x + 6)} = 18.
$$

Simplify the expression under the square root:

$$
x + \sqrt{x^2 - 3x - 6} = 18.
$$

---

### Step 3: Isolate the Square Root

Subtract $ x $ from both sides:

$$
\sqrt{x^2 - 3x - 6} = 18 - x.
$$

Now square both sides again:

$$
x^2 - 3x - 6 = (18 - x)^2.
$$

Expand the right-hand side:

$$
x^2 - 3x - 6 = 324 - 36x + x^2.
$$

Subtract $ x^2 $ from both sides:

$$
-3x - 6 = 324 - 36x.
$$

Add $ 36x $ to both sides:

$$
33x - 6 = 324.
$$

Add 6 to both sides:

$$
33x = 330.
$$

Solve for $ x $:

$$
x = \frac{330}{33} = 10.
$$

---

### Step 4: Verify the Solution

Let’s verify that $ x = 10 $ satisfies the original equation.

Compute $ \sqrt{3x + 6} = \sqrt{3(10) + 6} = \sqrt{36} = 6 $.

Now compute the original expression:

$$
\sqrt{10 + 6} + \sqrt{10 - 6} = \sqrt{16} + \sqrt{4} = 4 + 2 = 6.
$$

This matches the right-hand side of the equation, so the solution is valid.

---

### Final Answer

$$
\boxed{10}
$$

Solution:
Squaring both sides, we get
\[x + \sqrt{3x + 6} + 2 \sqrt{x + \sqrt{3x + 6}} \sqrt{x - \sqrt{3x + 6}} + x - \sqrt{3x + 6} = 36.\]Then
\[2x + 2 \sqrt{x^2 - 3x - 6} = 36,\]so
\[\sqrt{x^2 - 3x - 6} = 18 - x.\]Squaring both sides, we get $x^2 - 3x - 6 = 324 - 36x + x^2.$  Hence, $x = \boxed{10}.$  We check that this solution works.