Explanation:
Okay, so I need to find the class average for this test. Let me think about how to approach this. The class has 30 students total. Some scored 80, some 90, and some 100. The average is the sum of all the scores divided by the number of students, right? So I need to figure out the total points scored by all the students and then divide that by 30.

Let me break it down. First, there are 20 students who scored 80. So if I multiply 20 by 80, that should give me the total points from those students. Let me do that calculation: 20 times 80. Hmm, 20 times 80 is 1600. Okay, so 1600 points from the first group.

Next, there are 8 students who scored 90. So I need to calculate 8 times 90. Let me do that. 8 times 90... Well, 8 times 9 is 72, so adding a zero makes it 720. So that's 720 points from the second group.

Then there are 2 students who scored 100. Multiplying 2 by 100 is straightforward. That's 200 points.

Now, to get the total points for the whole class, I need to add up those three amounts: 1600, 720, and 200. Let me add them step by step. First, 1600 plus 720. Let me think: 1600 + 700 is 2300, and then plus 20 more is 2320. Then add the 200. So 2320 + 200 is 2520. So the total points scored by all students is 2520.

Now, the average is total points divided by number of students. There are 30 students. So I need to divide 2520 by 30. Let me do that division. Hmm, 2520 divided by 30. Well, 30 times 80 is 2400. So subtract 2400 from 2520, which leaves 120. Then 30 goes into 120 exactly 4 times. So 80 + 4 is 84. Therefore, the average is 84.

Wait, let me check that again to make sure I didn't make a mistake. So total points: 20*80=1600, 8*90=720, 2*100=200. Adding them up: 1600+720 is 2320, plus 200 is 2520. Divided by 30. 2520 divided by 30. Let me think of another way. Dividing both numerator and denominator by 10 first: 2520/10=252, 30/10=3. So now it's 252 divided by 3. 3 goes into 252 how many times? 3*80=240, so 252-240=12, and 3 goes into 12 four times. So 80+4=84. Yep, same result. So that seems right.

Alternatively, maybe I can check with another method. Let me think. If all 30 students scored 80, the total would be 30*80=2400. But some students scored higher. The 8 students who scored 90 are 10 points above 80, so that's 8*10=80 extra points. The 2 students who scored 100 are 20 points above 80, so that's 2*20=40 extra points. So total extra points are 80+40=120. Therefore, total points would be 2400 + 120 = 2520, which matches what I had before. Then average is 2520/30=84. Yep, same answer. So that seems to confirm it.

Alternatively, maybe I can think of the average as a weighted average. The average would be (20/30)*80 + (8/30)*90 + (2/30)*100. Let me compute that.

First, 20/30 is 2/3. 2/3 of 80 is (2*80)/3 = 160/3 ≈ 53.333...

Then 8/30 is 4/15. 4/15 of 90 is (4*90)/15 = 360/15 = 24.

Then 2/30 is 1/15. 1/15 of 100 is 100/15 ≈ 6.666...

Adding those together: 53.333... + 24 + 6.666... Let me add 53.333 and 6.666 first. That's 60. Then 60 +24 is 84. So that also gives 84. So that's another way to get the same answer.

So all methods lead to 84. Therefore, I think the class average is 84.

But just to make sure I didn't make any arithmetic errors. Let me verify once again:

Total points:

20 students * 80 = 1600

8 students * 90 = 720

2 students * 100 = 200

Total: 1600 + 720 = 2320; 2320 + 200 = 2520. Correct.

Number of students: 20 + 8 + 2 = 30. Correct.

Average: 2520 / 30. Let me divide 2520 by 30:

Divide numerator and denominator by 10: 252 / 3 = 84. Correct.

Yes, that's straightforward. So I think that's solid.

Alternatively, if I use decimal division: 2520 divided by 30. 30 goes into 2520 how many times? Well, 30*80=2400, as before. 2520-2400=120. 30*4=120. So total is 84. Yep.

So all methods confirm that the average is 84. Therefore, I can be confident that the answer is 84.

**Final Answer**
The class average on the test was \boxed{84}.

To find the **class average** (mean) on the test, we need to calculate the **total score** of all students and then divide it by the **total number of students**.

---

### Step 1: Calculate the total score

We are given the following:

- 20 students scored **80**
- 8 students scored **90**
- 2 students scored **100**

We compute the total score by multiplying the number of students by their respective scores:

$$
\text{Total score} = (20 \times 80) + (8 \times 90) + (2 \times 100)
$$

$$
= 1600 + 720 + 200 = 2520
$$

---

### Step 2: Calculate the class average

The class has **30 students**, so the average is:

$$
\text{Average} = \frac{\text{Total score}}{\text{Number of students}} = \frac{2520}{30}
$$

$$
= 84
$$

---

### Final Answer

$$
\boxed{84}
$$

[Assistant]
Explanation:
Okay, so I need to find the class average for this test. Let me think about how to approach this. The class has 30 students total. Some scored 80, some 90, and some 100. The average is the sum of all the scores divided by the number of students, right? So I need to figure out the total points scored by all the students and then divide that by 30.

Let me break it down. First, there are 20 students who scored 80. So if I multiply 20 by 80, that should give me the total points from those students. Let me do that calculation: 20 times 80. Hmm, 20 times 80 is 1600. Okay, so 1600 points from the first group.

Next, there are 8 students who scored 90. So I need to calculate 8 times 90. Let me do that. 8 times 90... Well, 8 times 9 is 72, so adding a zero makes it 720. So that's 720 points from the second group.

Then there are 2 students who scored 100. Multiplying 2 by 100 is straightforward. That's 200 points.

Now, to get the total points for the whole class, I need to add up those three amounts: 1600, 720, and 200. Let me add them step by step. First, 1600 plus 720. Let me think: 1600 + 700 is 2300, and then plus 20 more is 2320. Then add the 200. So 2320 + 200 is 2520. So the total points scored by all students is 2520.

Now, the average is total points divided by number of students. There are 30 students. So I need to divide 2520 by 30. Let me do that division. Hmm, 2520 divided by 30. Well, 30 times 80 is 2400. So subtract 2400 from 2520, which leaves 120. Then 30 goes into 120 exactly 4 times. So 80 + 4 is 84. Therefore, the average is 84.

Wait, let me check that again to make sure I didn't make a mistake. So total points: 20*80=1600, 8*90=720, 2*100=200. Adding them up: 1600+720 is 2320, plus 200 is 2520. Divided by 30. 2520 divided by 30. Let me think of another way. Dividing both numerator and denominator by 10 first: 2520/10=252, 30/10=3. So now it's 252 divided by 3. 3 goes into 252 how many times? 3*80=240, so 252-240=12, and 3 goes into 12 four times. So 80+4=84. Yep, same result. So that seems right.

Alternatively, maybe I can check with another method. Let me think. If all 30 students scored 80, the total would be 30*80=2400. But some students scored higher. The 8 students who scored 90 are 10 points above 80, so that's 8*10=80 extra points. The 2 students who scored 100 are 20 points above 80, so that's 2*20=40 extra points. So total extra points are 80+40=120. Therefore, total points would be 2400 + 120 = 2520, which matches what I had before. Then average is 2520/30=84. Yep, same answer. So that seems to confirm it.

Alternatively, maybe I can think of the average as a weighted average. The average would be (20/30)*80 + (8/30)*90 + (2/30)*100. Let me compute that.

First, 20/30 is 2/3. 2/3 of 80 is (2*80)/3 = 160/3 ≈ 53.333...

Then 8/30 is 4/15. 4/15 of 90 is (4*90)/15 = 360/15 = 24.

Then 2/30 is 1/15. 1/15 of 100 is 100/15 ≈ 6.666...

Adding those together: 53.333... + 24 + 6.666... Let me add 53.333 and 6.666 first. That's 60. Then 60 +24 is 84. So that also gives 84. So that's another way to get the same answer.

So all methods lead to 84. Therefore, I think the class average is 84.

But just to make sure I didn't make any arithmetic errors. Let me verify once again:

Total points:

20 students * 80 = 1600

8 students * 90 = 720

2 students * 100 = 200

Total: 1600 + 720 = 2320; 2320 + 200 = 2520. Correct.

Number of students: 20 + 8 + 2 = 30. Correct.

Average: 2520 / 30. Let me divide 2520 by 30:

Divide numerator and denominator by 10: 252 / 3 = 84. Correct.

Yes, that's straightforward. So I think that's solid.

Alternatively, if I use decimal division: 2520 divided by 30. 30 goes into 2520 how many times? Well, 30*80=2400, as before. 2520-2400=120. 30*4=120. So total is 84. Yep.

So all methods confirm that the average is 84. Therefore, I can be confident that the answer is 84.

**Final Answer**
The class average on the test was \boxed{84}.

To find the **class average** (mean) on the test, we need to calculate the **total score** of all students and then divide it by the **total number of students**.

---

### Step 1: Calculate the total score

We are given the following:

- 20 students scored **80**
- 8 students scored **90**
- 2 students scored **100**

We compute the total score by multiplying the number of students by their respective scores:

$$
\text{Total score} = (20 \times 80) + (8 \times 90) + (2 \times 100)
$$

$$
= 1600 + 720 + 200 = 2520
$$

---

### Step 2: Calculate the class average

The class has **30 students**, so the average is:

$$
\text{Average} = \frac{\text{Total score}}{\text{Number of students}} = \frac{2520}{30}
$$

$$
= 84
$$

---

### Final Answer

$$
\boxed{84}
$$

Solution: From the given information, the total amount of marks obtained by the class is   $$20(80)+8(90)+2(100)=2520.$$Therefore, the class average is  $$\frac{2520}{30} = \boxed{84}.$$


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. Let me think. In some contexts, numbers like 06 are considered as two-digit numbers, but in reality, leading zeros are not allowed. So I think B must be from 1 to 9. So the number is between 16 (when B=1) and 96 (when B=9). Wait, no. Wait, if B is 1, the number is 16. If B is 9, the number is 96. So the two-digit numbers from 16 up to 96, with the units digit being 6. So the numbers are 16, 26, 36, 46, 56, 66, 76, 86, 96. But not all of these are perfect squares. So I need to check which of these numbers are perfect squares.

Alternatively, maybe there's a better way than checking all nine numbers. Let me think. Since the number is a two-digit number ending with 6, and it's a perfect square. So maybe I can find all perfect squares between 10 and 99 (since two-digit numbers are from 10 to 99) that end with 6, and then note the tens digit B for each of them.

So first, let me list all two-digit perfect squares. The squares of numbers from 4 to 9, since 4^2=16, 5^2=25, 6^2=36, 7^2=49, 8^2=64, 9^2=81, 10^2=100 which is three digits. So the two-digit squares are 16, 25, 36, 49, 64, 81. Wait, but 10^2 is 100, which is three digits. So the two-digit squares are from 4^2 up to 9^2. So that's 16, 25, 36, 49, 64, 81. Now, which of these end with 6? Let me check:

16 ends with 6. Yes. 25 ends with 5. 36 ends with 6. 49 ends with 9. 64 ends with 4. 81 ends with 1. So only 16 and 36 end with 6. Therefore, the numbers are 16 and 36. Therefore, the possible values of B are 1 (from 16) and 3 (from 36). Therefore, there are two possible values for B. But wait, that seems too few. Wait, but maybe I missed some squares?

Wait, hold on. Wait, maybe there are more two-digit squares ending with 6. Let me check again. Let me think about squares ending with 6. What numbers squared end with 6?

A square number ends with 6 if the original number ends with 4 or 6, because:

- 4^2 = 16
- 6^2 = 36
- 14^2 = 196 (ends with 6)
- 16^2 = 256 (ends with 6)
- 24^2 = 576 (ends with 6)
- 26^2 = 676 (ends with 6)
- etc.

But since we are looking for two-digit numbers, the squares must be between 10 and 99. So the square roots would be between 4 (since 4^2=16) and 9 (since 9^2=81). Wait, but 10^2 is 100, which is three digits. So the square roots are integers from 4 to 9. So the squares are 16, 25, 36, 49, 64, 81. So only 16 and 36 end with 6. Therefore, only two numbers. Therefore, B can be 1 or 3. Therefore, two possible values. But the problem says "how many distinct possible values are there for B?" So answer is 2? But wait, maybe I made a mistake here. Let me check again.

Wait, maybe there are more squares? Let me check numbers from 1 to 9 squared:

1^2 = 1 (single-digit)
2^2 = 4 (single-digit)
3^2 = 9 (single-digit)
4^2 = 16 (two-digit)
5^2 = 25
6^2 = 36
7^2 = 49
8^2 = 64
9^2 = 81
10^2 = 100 (three-digit)

So yes, only 16, 25, 36, 49, 64, 81. So only 16 and 36 end with 6. Therefore, B is 1 and 3. Therefore, two possible values. But the problem says "the two-digit number B6", so maybe there are more possibilities? Wait, maybe I need to check numbers beyond 9? Wait, but if the square is two-digit, then the square root must be between 4 and 9. So numbers from 4 to 9. So their squares are as above. Therefore, only 16 and 36. Therefore, B is 1 and 3. Therefore, answer is 2. But maybe I need to check if there are other squares with two digits ending with 6?

Wait, let me think again. For example, 14^2 is 196, which is three-digit. So not applicable. 16^2 is 256, also three-digit. So no. So in two-digit numbers, 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 made a mistake here. Wait, let me check if there are other squares. Wait, maybe numbers like 26? Wait, 26 is not a square. 25 is 5^2, 26 is not. 36 is 6^2. 46? Not a square. 56? Not. 66? Not. 76? Not. 86? Not. 96? Not. So only 16 and 36. Therefore, B is 1 and 3. Therefore, two values. But maybe I need to check if there are other squares with two digits ending with 6. Wait, let me think about the last digit. For a square number to end with 6, the original number must end with 4 or 6. So let me check numbers ending with 4 or 6 in the range of square roots that produce two-digit numbers.

So square roots between 4 and 9. So numbers from 4 to 9. Which of these end with 4 or 6? 4, 6. So 4 and 6. So 4^2=16, 6^2=36. So that's the only ones. Therefore, only 16 and 36. Therefore, B is 1 and 3. Therefore, two possible values. Therefore, answer is 2. But maybe the problem is trickier? Wait, maybe there are other squares with two digits ending with 6? Let me check again. Let me think of squares ending with 6. For example, 14^2 is 196, which ends with 6, but it's three-digit. Similarly, 16^2 is 256, three-digit. 24^2 is 576, three-digit. 26^2 is 676, three-digit. So no, in two-digit numbers, only 16 and 36. Therefore, answer is 2. But maybe I need to check if there are other numbers?

Wait, let me think again. Let me check all two-digit numbers ending with 6 and see if they are squares. The numbers are 16, 26, 36, 46, 56, 66, 76, 86, 96. Let me check each:

- 16: 4^2, yes.
- 26: Not a square. 5^2 is 25, 6^2 is 36. So no.
- 36: 6^2, yes.
- 46: Not a square. 7^2 is 49, 6^2 is 36. No.
- 56: Not a square. 7^2 is 49, 8^2 is 64. No.
- 66: Not a square. 8^2 is 64, 9^2 is 81. No.
- 76: Not a square. 8^2 is 64, 9^2 is 81. No.
- 86: Not a square. 9^2 is 81, 10^2 is 100. No.
- 96: Not a square. 9^2 is 81, 10^2 is 100. No.

So only 16 and 36. Therefore, B is 1 and 3. Therefore, two possible values. Therefore, the answer is 2. But wait, the problem says "the two-digit number B6", so maybe there's something I'm missing here? Let me check again. Wait, maybe there are squares with leading zeros? But no, leading zeros aren't allowed. So B can't be zero. Therefore, only 1 and 3. Therefore, answer is 2. But maybe the problem is in another base? No, the problem doesn't mention that. It's in base 10. So I think my answer is correct. But maybe I need to check again. Wait, let me think of another approach.

Let me think algebraically. Let me denote the number as N = 10B + 6, which is a perfect square. So N = k^2, where k is a positive integer. Since N is a two-digit number, 10 ≤ N ≤ 99. Therefore, k^2 is between 10 and 99. Therefore, k is between 4 (since 4^2=16) and 9 (since 9^2=81). So k can be 4,5,6,7,8,9. Let me compute k^2 for each:

- 4^2 = 16 → B=1
- 5^2 = 25 → ends with 5, not 6
- 6^2 = 36 → B=3
- 7^2 = 49 → ends with 9
- 8^2 = 64 → ends with 4
- 9^2 = 81 → ends with 1

So only 16 and 36. Therefore, B can be 1 or 3. Therefore, two possible values. Therefore, the answer is 2. Therefore, the number of distinct possible values for B is 2. Therefore, the answer is \boxed{2}.

Wait, but I feel like maybe there's something wrong here. Let me check once more. Maybe there are other squares outside of this range? Wait, no. Because if k is 10, then k^2 is 100, which is three digits. So no. So only k from 4 to 9. Therefore, only 16 and 36. Therefore, answer is 2. But maybe the problem is in another interpretation? For example, maybe B is allowed to be zero? If B is zero, then the number is 06, which is 6, which is a single-digit number. But the problem says "two-digit number B6", so leading zeros are not allowed. Therefore, B cannot be zero. Therefore, only 1 and 3. Therefore, answer is 2. So I think that's correct.

But let me check once again. Maybe there's a square number with two digits ending with 6 that I haven't considered? For example, 16, 36. Let me check 16: 4^2. 36: 6^2. What about 25? Ends with 5. 49: 9. 64: 4. 81: 1. So no. So only 16 and 36. Therefore, answer is 2. Therefore, I think the answer is 2. But maybe the problem is expecting more? Wait, let me check if there are other numbers. For example, 10B + 6 = k^2. Let me think of solving for B. So B = (k^2 - 6)/10. Since B must be an integer between 1 and 9, inclusive. So (k^2 - 6) must be divisible by 10, and the result must be between 1 and 9. So let's check for k from 4 to 9:

k=4: 16 - 6 = 10 → 10/10 = 1 → B=1. Valid.

k=5: 25 - 6 = 19 → 19/10 = 1.9 → Not integer. Invalid.

k=6: 36 - 6 = 30 → 30/10 = 3 → B=3. Valid.

k=7: 49 - 6 = 43 → 43/10 = 4.3 → Not integer. Invalid.

k=8: 64 - 6 = 58 → 58/10 = 5.8 → Not integer. Invalid.

k=9: 81 - 6 = 75 → 75/10 = 7.5 → Not integer. Invalid.

So only k=4 and k=6 give integer values for B. Therefore, B=1 and B=3. Therefore, two possible values. Therefore, answer is 2. Therefore, I think that's correct. Therefore, the answer is \boxed{2}.

But wait, just to be thorough, let me check if there are any other k outside of 4-9 that could result in a two-digit number. For example, if k is 3, then k^2 is 9, which is single-digit. If k is 10, then 100, which is three-digit. So no. Therefore, no other possibilities. Therefore, answer is 2. Therefore, I think that's correct.

**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. This number is a perfect square, and we are to determine how many **distinct possible values** of **B** exist.

---

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

The number **B6** can be expressed algebraically as:

$$
N = 10B + 6
$$

We are told that this number is a **perfect square**, so:

$$
N = k^2 \quad \text{for some integer } k
$$

Also, since **B6** is a two-digit number, the value of **N** must be between 10 and 99 (inclusive). Therefore:

$$
10 \leq k^2 \leq 99
$$

Taking square roots:

$$
\sqrt{10} \leq k \leq \sqrt{99} \Rightarrow 3.16 \leq k \leq 9.95
$$

So **k** must be an integer between **4** and **9** (inclusive), since **k** must be an integer and **k = 3** gives a single-digit square (9), and **k = 10** gives a three-digit square (100).

---

### Step 2: List the Squares of Integers from 4 to 9

We now compute the squares of integers from 4 to 9:

- $4^2 = 16$
- $5^2 = 25$
- $6^2 = 36$
- $7^2 = 49$
- $8^2 = 64$
- $9^2 = 81$

Now, we check which of these numbers end in **6** (i.e., have the form **B6**):

- $16$ → ends in 6 → valid
- $25$ → ends in 5 → invalid
- $36$ → ends in 6 → valid
- $49$ → ends in 9 → invalid
- $64$ → ends in 4 → invalid
- $81$ → ends in 1 → invalid

So the only valid numbers are **16** and **36**.

---

### Step 3: Determine the Corresponding Values of B

From the valid numbers:

- $16$ → B = 1
- $36$ → B = 3

Thus, the possible values of **B** are **1** and **3**.

---

### Step 4: Count the Distinct Values of B

The distinct values of **B** are:

$$
\{1, 3\}
$$

So, there are **2** distinct possible values for **B**.

---

### 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. Let me think. In some contexts, numbers like 06 are considered as two-digit numbers, but in reality, leading zeros are not allowed. So I think B must be from 1 to 9. So the number is between 16 (when B=1) and 96 (when B=9). Wait, no. Wait, if B is 1, the number is 16. If B is 9, the number is 96. So the two-digit numbers from 16 up to 96, with the units digit being 6. So the numbers are 16, 26, 36, 46, 56, 66, 76, 86, 96. But not all of these are perfect squares. So I need to check which of these numbers are perfect squares.

Alternatively, maybe there's a better way than checking all nine numbers. Let me think. Since the number is a two-digit number ending with 6, and it's a perfect square. So maybe I can find all perfect squares between 10 and 99 (since two-digit numbers are from 10 to 99) that end with 6, and then note the tens digit B for each of them.

So first, let me list all two-digit perfect squares. The squares of numbers from 4 to 9, since 4^2=16, 5^2=25, 6^2=36, 7^2=49, 8^2=64, 9^2=81, 10^2=100 which is three digits. So the two-digit squares are 16, 25, 36, 49, 64, 81. Wait, but 10^2 is 100, which is three digits. So the two-digit squares are from 4^2 up to 9^2. So that's 16, 25, 36, 49, 64, 81. Now, which of these end with 6? Let me check:

16 ends with 6. Yes. 25 ends with 5. 36 ends with 6. 49 ends with 9. 64 ends with 4. 81 ends with 1. So only 16 and 36 end with 6. Therefore, the numbers are 16 and 36. Therefore, the possible values of B are 1 (from 16) and 3 (from 36). Therefore, there are two possible values for B. But wait, that seems too few. Wait, but maybe I missed some squares?

Wait, hold on. Wait, maybe there are more two-digit squares ending with 6. Let me check again. Let me think about squares ending with 6. What numbers squared end with 6?

A square number ends with 6 if the original number ends with 4 or 6, because:

- 4^2 = 16
- 6^2 = 36
- 14^2 = 196 (ends with 6)
- 16^2 = 256 (ends with 6)
- 24^2 = 576 (ends with 6)
- 26^2 = 676 (ends with 6)
- etc.

But since we are looking for two-digit numbers, the squares must be between 10 and 99. So the square roots would be between 4 (since 4^2=16) and 9 (since 9^2=81). Wait, but 10^2 is 100, which is three digits. So the square roots are integers from 4 to 9. So the squares are 16, 25, 36, 49, 64, 81. So only 16 and 36 end with 6. Therefore, only two numbers. Therefore, B can be 1 or 3. Therefore, two possible values. But the problem says "how many distinct possible values are there for B?" So answer is 2? But wait, maybe I made a mistake here. Let me check again.

Wait, maybe there are more squares? Let me check numbers from 1 to 9 squared:

1^2 = 1 (single-digit)
2^2 = 4 (single-digit)
3^2 = 9 (single-digit)
4^2 = 16 (two-digit)
5^2 = 25
6^2 = 36
7^2 = 49
8^2 = 64
9^2 = 81
10^2 = 100 (three-digit)

So yes, only 16, 25, 36, 49, 64, 81. So only 16 and 36 end with 6. Therefore, B is 1 and 3. Therefore, two possible values. But the problem says "the two-digit number B6", so maybe there are more possibilities? Wait, maybe I need to check numbers beyond 9? Wait, but if the square is two-digit, then the square root must be between 4 and 9. So numbers from 4 to 9. So their squares are as above. Therefore, only 16 and 36. Therefore, B is 1 and 3. Therefore, answer is 2. But maybe I need to check if there are other squares with two digits ending with 6?

Wait, let me think again. For example, 14^2 is 196, which is three-digit. So not applicable. 16^2 is 256, also three-digit. So no. So in two-digit numbers, 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 made a mistake here. Wait, let me check if there are other squares. Wait, maybe numbers like 26? Wait, 26 is not a square. 25 is 5^2, 26 is not. 36 is 6^2. 46? Not a square. 56? Not. 66? Not. 76? Not. 86? Not. 96? Not. So only 16 and 36. Therefore, B is 1 and 3. Therefore, two values. But maybe I need to check if there are other squares with two digits ending with 6. Wait, let me think about the last digit. For a square number to end with 6, the original number must end with 4 or 6. So let me check numbers ending with 4 or 6 in the range of square roots that produce two-digit numbers.

So square roots between 4 and 9. So numbers from 4 to 9. Which of these end with 4 or 6? 4, 6. So 4 and 6. So 4^2=16, 6^2=36. So that's the only ones. Therefore, only 16 and 36. Therefore, B is 1 and 3. Therefore, two possible values. Therefore, answer is 2. But maybe the problem is trickier? Wait, maybe there are other squares with two digits ending with 6? Let me check again. Let me think of squares ending with 6. For example, 14^2 is 196, which ends with 6, but it's three-digit. Similarly, 16^2 is 256, three-digit. 24^2 is 576, three-digit. 26^2 is 676, three-digit. So no, in two-digit numbers, only 16 and 36. Therefore, answer is 2. But maybe I need to check if there are other numbers?

Wait, let me think again. Let me check all two-digit numbers ending with 6 and see if they are squares. The numbers are 16, 26, 36, 46, 56, 66, 76, 86, 96. Let me check each:

- 16: 4^2, yes.
- 26: Not a square. 5^2 is 25, 6^2 is 36. So no.
- 36: 6^2, yes.
- 46: Not a square. 7^2 is 49, 6^2 is 36. No.
- 56: Not a square. 7^2 is 49, 8^2 is 64. No.
- 66: Not a square. 8^2 is 64, 9^2 is 81. No.
- 76: Not a square. 8^2 is 64, 9^2 is 81. No.
- 86: Not a square. 9^2 is 81, 10^2 is 100. No.
- 96: Not a square. 9^2 is 81, 10^2 is 100. No.

So only 16 and 36. Therefore, B is 1 and 3. Therefore, two possible values. Therefore, the answer is 2. But wait, the problem says "the two-digit number B6", so maybe there's something I'm missing here? Let me check again. Wait, maybe there are squares with leading zeros? But no, leading zeros aren't allowed. So B can't be zero. Therefore, only 1 and 3. Therefore, answer is 2. But maybe the problem is in another base? No, the problem doesn't mention that. It's in base 10. So I think my answer is correct. But maybe I need to check again. Wait, let me think of another approach.

Let me think algebraically. Let me denote the number as N = 10B + 6, which is a perfect square. So N = k^2, where k is a positive integer. Since N is a two-digit number, 10 ≤ N ≤ 99. Therefore, k^2 is between 10 and 99. Therefore, k is between 4 (since 4^2=16) and 9 (since 9^2=81). So k can be 4,5,6,7,8,9. Let me compute k^2 for each:

- 4^2 = 16 → B=1
- 5^2 = 25 → ends with 5, not 6
- 6^2 = 36 → B=3
- 7^2 = 49 → ends with 9
- 8^2 = 64 → ends with 4
- 9^2 = 81 → ends with 1

So only 16 and 36. Therefore, B can be 1 or 3. Therefore, two possible values. Therefore, the answer is 2. Therefore, the number of distinct possible values for B is 2. Therefore, the answer is \boxed{2}.

Wait, but I feel like maybe there's something wrong here. Let me check once more. Maybe there are other squares outside of this range? Wait, no. Because if k is 10, then k^2 is 100, which is three digits. So no. So only k from 4 to 9. Therefore, only 16 and 36. Therefore, answer is 2. But maybe the problem is in another interpretation? For example, maybe B is allowed to be zero? If B is zero, then the number is 06, which is 6, which is a single-digit number. But the problem says "two-digit number B6", so leading zeros are not allowed. Therefore, B cannot be zero. Therefore, only 1 and 3. Therefore, answer is 2. So I think that's correct.

But let me check once again. Maybe there's a square number with two digits ending with 6 that I haven't considered? For example, 16, 36. Let me check 16: 4^2. 36: 6^2. What about 25? Ends with 5. 49: 9. 64: 4. 81: 1. So no. So only 16 and 36. Therefore, answer is 2. Therefore, I think the answer is 2. But maybe the problem is expecting more? Wait, let me check if there are other numbers. For example, 10B + 6 = k^2. Let me think of solving for B. So B = (k^2 - 6)/10. Since B must be an integer between 1 and 9, inclusive. So (k^2 - 6) must be divisible by 10, and the result must be between 1 and 9. So let's check for k from 4 to 9:

k=4: 16 - 6 = 10 → 10/10 = 1 → B=1. Valid.

k=5: 25 - 6 = 19 → 19/10 = 1.9 → Not integer. Invalid.

k=6: 36 - 6 = 30 → 30/10 = 3 → B=3. Valid.

k=7: 49 - 6 = 43 → 43/10 = 4.3 → Not integer. Invalid.

k=8: 64 - 6 = 58 → 58/10 = 5.8 → Not integer. Invalid.

k=9: 81 - 6 = 75 → 75/10 = 7.5 → Not integer. Invalid.

So only k=4 and k=6 give integer values for B. Therefore, B=1 and B=3. Therefore, two possible values. Therefore, answer is 2. Therefore, I think that's correct. Therefore, the answer is \boxed{2}.

But wait, just to be thorough, let me check if there are any other k outside of 4-9 that could result in a two-digit number. For example, if k is 3, then k^2 is 9, which is single-digit. If k is 10, then 100, which is three-digit. So no. Therefore, no other possibilities. Therefore, answer is 2. Therefore, I think that's correct.

**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. This number is a perfect square, and we are to determine how many **distinct possible values** of **B** exist.

---

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

The number **B6** can be expressed algebraically as:

$$
N = 10B + 6
$$

We are told that this number is a **perfect square**, so:

$$
N = k^2 \quad \text{for some integer } k
$$

Also, since **B6** is a two-digit number, the value of **N** must be between 10 and 99 (inclusive). Therefore:

$$
10 \leq k^2 \leq 99
$$

Taking square roots:

$$
\sqrt{10} \leq k \leq \sqrt{99} \Rightarrow 3.16 \leq k \leq 9.95
$$

So **k** must be an integer between **4** and **9** (inclusive), since **k** must be an integer and **k = 3** gives a single-digit square (9), and **k = 10** gives a three-digit square (100).

---

### Step 2: List the Squares of Integers from 4 to 9

We now compute the squares of integers from 4 to 9:

- $4^2 = 16$
- $5^2 = 25$
- $6^2 = 36$
- $7^2 = 49$
- $8^2 = 64$
- $9^2 = 81$

Now, we check which of these numbers end in **6** (i.e., have the form **B6**):

- $16$ → ends in 6 → valid
- $25$ → ends in 5 → invalid
- $36$ → ends in 6 → valid
- $49$ → ends in 9 → invalid
- $64$ → ends in 4 → invalid
- $81$ → ends in 1 → invalid

So the only valid numbers are **16** and **36**.

---

### Step 3: Determine the Corresponding Values of B

From the valid numbers:

- $16$ → B = 1
- $36$ → B = 3

Thus, the possible values of **B** are **1** and **3**.

---

### Step 4: Count the Distinct Values of B

The distinct values of **B** are:

$$
\{1, 3\}
$$

So, there are **2** distinct possible values for **B**.

---

### 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. 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 formulas?

Alternatively, since we have the area expressed as a² - (b - c)^2, which we simplified to 2bc(1 - cos A). Also, area is (1/2) bc sin A. Therefore, equate them:

(1/2) bc sin A = 2 bc (1 - cos A) => 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 relate sin A and cos A here. For example, using tan(A/2). Because there's an identity that sin A = 2 tan(A/2) / (1 + tan²(A/2)) and 1 - cos A = 2 tan²(A/2) / (1 + tan²(A/2)). Let me try that.

Let me denote t = tan(A/2). Then:

sin A = 2t / (1 + t²)

and

1 - cos A = 2t² / (1 + t²)

Therefore, substituting into sin A = 4(1 - cos A):

2t / (1 + t²) = 4*(2t² / (1 + t²))

Multiply both sides by (1 + t²):

2t = 8t²

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

1 = 4t => t = 1/4

Therefore, tan(A/2) = 1/4. Then, tan A can be found using the double-angle formula:

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

Same result! So that confirms tan A = 8/15. Therefore, regardless of the method, the answer is 8/15.

Therefore, I think that's the correct answer. Let me just check once more if there's any possible mistake in the algebra. For example, when I squared both sides of sin A = 4(1 - cos A), could that introduce extraneous solutions? Well, we checked that cos A = 1 is invalid, and the other solution worked. So no problem there.

Alternatively, if I use the values of sin A and cos A, tan A is 8/15, which is positive, and since angle A is between 0 and π, tan A is positive, so angle A is acute or obtuse? Since cos A = 15/17 is positive, angle A is acute. Therefore, tan A is positive, which matches.

Therefore, I think I can be confident that tan A 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:

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

Simplify:

$$
-2bc \cos A + 2bc = 2bc(1 - \cos A)
$$

So the area becomes:

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

---

### Step 3: Equating both expressions for area

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

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 4: Use trigonometric identities to solve for $\tan A$

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 5: 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 $\cos A = 1$ implies angle $A = 0^\circ$, which is not possible in a triangle, we discard it.

Thus:

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

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 6: 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. 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 formulas?

Alternatively, since we have the area expressed as a² - (b - c)^2, which we simplified to 2bc(1 - cos A). Also, area is (1/2) bc sin A. Therefore, equate them:

(1/2) bc sin A = 2 bc (1 - cos A) => 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 relate sin A and cos A here. For example, using tan(A/2). Because there's an identity that sin A = 2 tan(A/2) / (1 + tan²(A/2)) and 1 - cos A = 2 tan²(A/2) / (1 + tan²(A/2)). Let me try that.

Let me denote t = tan(A/2). Then:

sin A = 2t / (1 + t²)

and

1 - cos A = 2t² / (1 + t²)

Therefore, substituting into sin A = 4(1 - cos A):

2t / (1 + t²) = 4*(2t² / (1 + t²))

Multiply both sides by (1 + t²):

2t = 8t²

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

1 = 4t => t = 1/4

Therefore, tan(A/2) = 1/4. Then, tan A can be found using the double-angle formula:

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

Same result! So that confirms tan A = 8/15. Therefore, regardless of the method, the answer is 8/15.

Therefore, I think that's the correct answer. Let me just check once more if there's any possible mistake in the algebra. For example, when I squared both sides of sin A = 4(1 - cos A), could that introduce extraneous solutions? Well, we checked that cos A = 1 is invalid, and the other solution worked. So no problem there.

Alternatively, if I use the values of sin A and cos A, tan A is 8/15, which is positive, and since angle A is between 0 and π, tan A is positive, so angle A is acute or obtuse? Since cos A = 15/17 is positive, angle A is acute. Therefore, tan A is positive, which matches.

Therefore, I think I can be confident that tan A 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:

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

Simplify:

$$
-2bc \cos A + 2bc = 2bc(1 - \cos A)
$$

So the area becomes:

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

---

### Step 3: Equating both expressions for area

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

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 4: Use trigonometric identities to solve for $\tan A$

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 5: 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 $\cos A = 1$ implies angle $A = 0^\circ$, which is not possible in a triangle, we discard it.

Thus:

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

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 6: 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 these two approximated square roots: 3.971 + 1.966 ≈ 5.937. Wait, but the original equation says it should equal 6. Hmm, there's a discrepancy here. The approximate value is about 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 calculation. Let me try to compute more precisely.

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

3x = 3*(108)/11 = 324/11 ≈ 29.4545

Then 3x + 6 = 29.4545 + 6 = 35.4545. So √(35.4545) ≈ 5.9544.

Then x + √(3x + 6) = 108/11 + √(35.4545). Let me compute 108/11 ≈ 9.8181818...

So 9.8181818 + 5.9544 ≈ 15.77258. Then √(15.77258) ≈ 3.9714.

Similarly, x - √(3x + 6) ≈ 9.81818 - 5.9544 ≈ 3.86378. Then √(3.86378) ≈ 1.9656.

Adding them together: 3.9714 + 1.9656 ≈ 5.937. Hmm, which is about 5.937, not 6. So that suggests that maybe x = 108/11 is not a solution? But according to our algebra, it should be. Maybe there was an error in the algebra?

Let me check my steps again.

Original equation: √(x + √(3x + 6)) + √(x - √(3x + 6)) = 6.

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

Squaring both sides: (x + y) + (x - y) + 2√(x² - y²) = 36 ⇒ 2x + 2√(x² - y²) = 36 ⇒ x + √(x² - y²) = 18.

Then, since y² = 3x + 6, so x² - y² = x² - 3x - 6. Therefore, equation becomes x + √(x² - 3x - 6) = 18.

Then, moving x: √(x² - 3x - 6) = 18 - x.

Squaring again: x² - 3x - 6 = 324 - 36x + x² ⇒ -3x -6 = 324 - 36x ⇒ 33x = 330 ⇒ x = 10. Wait, wait! Wait, earlier I said 33x = 324? Wait, let me check again.

Wait, let me redo this step:

After squaring both sides:

x² - 3x - 6 = (18 - x)² = 324 - 36x + x².

Subtract x² from both sides:

-3x - 6 = 324 - 36x.

Then, adding 36x to both sides:

33x - 6 = 324.

Then, adding 6 to both sides:

33x = 330 ⇒ x = 330 / 33 = 10. Oh! Wait, earlier I had written 33x = 324, but that was a mistake. It should be 33x = 330. So x = 10. That was my mistake earlier! I must have miscalculated when I was doing the arithmetic. So x = 10, not 108/11. That explains why my approximate calculation didn't match. So I need to correct that.

So, x = 10. Let me check this value.

First, check if x = 10 satisfies the domain conditions. Earlier, we had that x must be ≥ [3 + √33]/2 ≈ 4.372. 10 is greater than that, so okay.

Now, check the original equation with x = 10.

Compute 3x + 6 = 3*10 + 6 = 36. Therefore, √(3x + 6) = √36 = 6.

Then, compute √(x + √(3x + 6)) + √(x - √(3x + 6)) = √(10 + 6) + √(10 - 6) = √16 + √4 = 4 + 2 = 6. Which matches the right-hand side. Perfect! So x = 10 is the correct solution. My earlier mistake was in the algebra step where I incorrectly calculated 33x = 324 instead of 33x = 330. So that was a critical error. Let me recap the correct steps.

So, after squaring both sides the second time, we had:

-3x - 6 = 324 - 36x

Adding 36x to both sides:

33x - 6 = 324

Adding 6:

33x = 330 ⇒ x = 10. Correct.

So, x = 10 is the solution. Let me check if there are other solutions. When we squared both sides, sometimes extraneous solutions can be introduced, but in this case, since we had only one solution from the algebra, and it checks out, so that's the only solution.

But just to be thorough, let me check if there could be another solution. Suppose when we squared the equation √(x² - 3x - 6) = 18 - x, we might have introduced an extraneous solution, but since we only had one solution, and it checks out, so that's the only one.

Alternatively, let me check if there's another solution when solving √(x² - 3x - 6) = 18 - x. The equation √(x² - 3x - 6) = 18 - x implies that 18 - x ≥ 0 ⇒ x ≤ 18, which is already considered. Also, the expression under the square root must be non-negative: x² - 3x - 6 ≥ 0. Which we had earlier, leading to x ≥ [3 + √33]/2 ≈ 4.372. So x is between approx 4.372 and 18. But solving √(x² - 3x - 6) = 18 - x gives only one solution x = 10. Let me check if there's another solution.

Suppose we 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 once again with exact values. Let me compute the original equation with x = 10:

√(10 + √(3*10 + 6)) + √(10 - √(3*10 + 6)) = √(10 + √36) + √(10 - √36) = √(10 + 6) + √(10 - 6) = √16 + √4 = 4 + 2 = 6. Perfect, that's exactly the right-hand side. So x = 10 is correct.

Therefore, my initial mistake was a calculation error when I thought 33x = 324, but it's actually 33x = 330, leading to x = 10. So the correct solution is 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:

$$
x^2 - y^2 = x^2 - (3x + 6) = x^2 - 3x - 6.
$$

Substitute into the equation:

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

---

### Step 3: Isolate the Square Root

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

Now square both sides again:

$$
x^2 - 3x - 6 = (18 - x)^2 = 324 - 36x + x^2.
$$

Subtract $ x^2 $ from both sides:

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

Solve for $ x $:

$$
33x = 330 \Rightarrow x = 10.
$$

---

### Step 4: Verify the Solution

We now check whether $ x = 10 $ satisfies the original equation.

Compute:

- $ 3x + 6 = 3(10) + 6 = 36 $, so $ \sqrt{3x + 6} = \sqrt{36} = 6 $.
- Then:
  $$
  \sqrt{x + \sqrt{3x + 6}} = \sqrt{10 + 6} = \sqrt{16} = 4,
  $$
  $$
  \sqrt{x - \sqrt{3x + 6}} = \sqrt{10 - 6} = \sqrt{4} = 2.
  $$

Add them:

$$
4 + 2 = 6.
$$

This matches the right-hand side of the original equation, confirming that $ x = 10 $ is a valid solution.

---

### Step 5: Check for Extraneous Solutions

We squared the equation twice, which can introduce extraneous solutions. However, the algebra led to a single solution, and it checks out in the original equation. Therefore, there are no other valid solutions.

---

### 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 these two approximated square roots: 3.971 + 1.966 ≈ 5.937. Wait, but the original equation says it should equal 6. Hmm, there's a discrepancy here. The approximate value is about 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 calculation. Let me try to compute more precisely.

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

3x = 3*(108)/11 = 324/11 ≈ 29.4545

Then 3x + 6 = 29.4545 + 6 = 35.4545. So √(35.4545) ≈ 5.9544.

Then x + √(3x + 6) = 108/11 + √(35.4545). Let me compute 108/11 ≈ 9.8181818...

So 9.8181818 + 5.9544 ≈ 15.77258. Then √(15.77258) ≈ 3.9714.

Similarly, x - √(3x + 6) ≈ 9.81818 - 5.9544 ≈ 3.86378. Then √(3.86378) ≈ 1.9656.

Adding them together: 3.9714 + 1.9656 ≈ 5.937. Hmm, which is about 5.937, not 6. So that suggests that maybe x = 108/11 is not a solution? But according to our algebra, it should be. Maybe there was an error in the algebra?

Let me check my steps again.

Original equation: √(x + √(3x + 6)) + √(x - √(3x + 6)) = 6.

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

Squaring both sides: (x + y) + (x - y) + 2√(x² - y²) = 36 ⇒ 2x + 2√(x² - y²) = 36 ⇒ x + √(x² - y²) = 18.

Then, since y² = 3x + 6, so x² - y² = x² - 3x - 6. Therefore, equation becomes x + √(x² - 3x - 6) = 18.

Then, moving x: √(x² - 3x - 6) = 18 - x.

Squaring again: x² - 3x - 6 = 324 - 36x + x² ⇒ -3x -6 = 324 - 36x ⇒ 33x = 330 ⇒ x = 10. Wait, wait! Wait, earlier I said 33x = 324? Wait, let me check again.

Wait, let me redo this step:

After squaring both sides:

x² - 3x - 6 = (18 - x)² = 324 - 36x + x².

Subtract x² from both sides:

-3x - 6 = 324 - 36x.

Then, adding 36x to both sides:

33x - 6 = 324.

Then, adding 6 to both sides:

33x = 330 ⇒ x = 330 / 33 = 10. Oh! Wait, earlier I had written 33x = 324, but that was a mistake. It should be 33x = 330. So x = 10. That was my mistake earlier! I must have miscalculated when I was doing the arithmetic. So x = 10, not 108/11. That explains why my approximate calculation didn't match. So I need to correct that.

So, x = 10. Let me check this value.

First, check if x = 10 satisfies the domain conditions. Earlier, we had that x must be ≥ [3 + √33]/2 ≈ 4.372. 10 is greater than that, so okay.

Now, check the original equation with x = 10.

Compute 3x + 6 = 3*10 + 6 = 36. Therefore, √(3x + 6) = √36 = 6.

Then, compute √(x + √(3x + 6)) + √(x - √(3x + 6)) = √(10 + 6) + √(10 - 6) = √16 + √4 = 4 + 2 = 6. Which matches the right-hand side. Perfect! So x = 10 is the correct solution. My earlier mistake was in the algebra step where I incorrectly calculated 33x = 324 instead of 33x = 330. So that was a critical error. Let me recap the correct steps.

So, after squaring both sides the second time, we had:

-3x - 6 = 324 - 36x

Adding 36x to both sides:

33x - 6 = 324

Adding 6:

33x = 330 ⇒ x = 10. Correct.

So, x = 10 is the solution. Let me check if there are other solutions. When we squared both sides, sometimes extraneous solutions can be introduced, but in this case, since we had only one solution from the algebra, and it checks out, so that's the only solution.

But just to be thorough, let me check if there could be another solution. Suppose when we squared the equation √(x² - 3x - 6) = 18 - x, we might have introduced an extraneous solution, but since we only had one solution, and it checks out, so that's the only one.

Alternatively, let me check if there's another solution when solving √(x² - 3x - 6) = 18 - x. The equation √(x² - 3x - 6) = 18 - x implies that 18 - x ≥ 0 ⇒ x ≤ 18, which is already considered. Also, the expression under the square root must be non-negative: x² - 3x - 6 ≥ 0. Which we had earlier, leading to x ≥ [3 + √33]/2 ≈ 4.372. So x is between approx 4.372 and 18. But solving √(x² - 3x - 6) = 18 - x gives only one solution x = 10. Let me check if there's another solution.

Suppose we 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 once again with exact values. Let me compute the original equation with x = 10:

√(10 + √(3*10 + 6)) + √(10 - √(3*10 + 6)) = √(10 + √36) + √(10 - √36) = √(10 + 6) + √(10 - 6) = √16 + √4 = 4 + 2 = 6. Perfect, that's exactly the right-hand side. So x = 10 is correct.

Therefore, my initial mistake was a calculation error when I thought 33x = 324, but it's actually 33x = 330, leading to x = 10. So the correct solution is 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:

$$
x^2 - y^2 = x^2 - (3x + 6) = x^2 - 3x - 6.
$$

Substitute into the equation:

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

---

### Step 3: Isolate the Square Root

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

Now square both sides again:

$$
x^2 - 3x - 6 = (18 - x)^2 = 324 - 36x + x^2.
$$

Subtract $ x^2 $ from both sides:

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

Solve for $ x $:

$$
33x = 330 \Rightarrow x = 10.
$$

---

### Step 4: Verify the Solution

We now check whether $ x = 10 $ satisfies the original equation.

Compute:

- $ 3x + 6 = 3(10) + 6 = 36 $, so $ \sqrt{3x + 6} = \sqrt{36} = 6 $.
- Then:
  $$
  \sqrt{x + \sqrt{3x + 6}} = \sqrt{10 + 6} = \sqrt{16} = 4,
  $$
  $$
  \sqrt{x - \sqrt{3x + 6}} = \sqrt{10 - 6} = \sqrt{4} = 2.
  $$

Add them:

$$
4 + 2 = 6.
$$

This matches the right-hand side of the original equation, confirming that $ x = 10 $ is a valid solution.

---

### Step 5: Check for Extraneous Solutions

We squared the equation twice, which can introduce extraneous solutions. However, the algebra led to a single solution, and it checks out in the original equation. Therefore, there are no other valid solutions.

---

### 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.