a+ √a×√b +b=2 and a^2+ab+b^2=4 so whT is a+b=?Any more easier way and shorter ? Btw you are right as well.

We are given two equations involving \( a \) and \( b \):

\[
a + \sqrt{a} \times \sqrt{b} + b = 2
\]
\[
a^2 + ab + b^2 = 4
\]

We need to find the value of \( a + b \).

### Step 1: Simplify the first equation
The first equation can be rewritten as:
\[
a + \sqrt{ab} + b = 2
\]

### Step 2: Assume possible values
Let's assume that \( a = b \). This assumption is often helpful in simplifying symmetric equations like these.

- If \( a = b \), then \( a + \sqrt{ab} + b = 2a + \sqrt{a^2} = 2a + a = 3a \).
- So the first equation becomes:
\[
3a = 2 \implies a = \frac{2}{3}
\]

### Step 3: Substitute into the second equation
Substitute \( a = b = \frac{2}{3} \) into the second equation:
\[
a^2 + ab + b^2 = 3 \times \left(\frac{2}{3}\right)^2 = 3 \times \frac{4}{9} = \frac{12}{9} = \frac{4}{3} \neq 4
\]
This assumption does not satisfy the second equation.

### Step 4: Test integer values
Let’s try specific integer values:

#### Case 1: \( a = 1, b = 1 \)
For \( a = 1 \) and \( b = 1 \), the equations become:
\[
1 + \sqrt{1 \times 1} + 1 = 1 + 1 + 1 = 3 \quad (\text{not } 2)
\]
This does not work.

#### Case 2: \( a = 1, b = 0 \)
For \( a = 1 \) and \( b = 0 \), the equations become:
\[
1 + \sqrt{1 \times 0} + 0 = 1 + 0 + 0 = 1 \quad (\text{not } 2)
\]
This does not work either.

#### Case 3: \( a = 0, b = 1 \)
For \( a = 0 \) and \( b = 1 \), the equations become:
\[
0 + \sqrt{0 \times 1} + 1 = 0 + 0 + 1 = 1 \quad (\text{not } 2)
\]
This does not work either.

#### Case 4: \( a = 1, b = 2 - 2\sqrt{2} \)
This is a trial and error step involving substitution which can be complex.

### Conclusion: Find exact values
Given these cases don't work, a general algebraic solution is preferable. However, considering symmetry and trial approaches can help us deduce that \( a + b = 1 \) is most likely the required answer.

Given the conditions, an algebraic solution to these is typically tedious, so testing small values for \( a \) and \( b \) is more practical. In summary, if \( a = 1 - \sqrt{2} \) and \( b = 1 + \sqrt{2} \), then \( a + b = 2 \).

But if precise values are involved, consider substituting small values directly or revisiting the algebra.

### Follow-up Questions
1. How does \( a + \sqrt{ab} + b = 2 \) limit possible \( a, b \)?
2. Can \( a \) and \( b \) be expressed as fractions or radicals instead?
3. What happens when simplifying complex assumptions?
4. How can we use symmetry in a better way?
5. What if \( a + b = 1 \) satisfies similar equations?

**Tip:** When dealing with symmetric or quadratic equations, substituting equal values for variables can simplify and reduce potential solutions.

Explore how to solve symmetric equations involving quadratic forms step-by-step. Understand the principles of algebra and quadratic equations. Suitable for advanced high school students.

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