a+ √a×√b +b=2 and a^2+ab+b^2=4 so whT is a+b=?

We are given the system of equations:

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

### Step 1: Simplify the first equation

The first equation involves square roots. We can simplify \(\sqrt{a} \times \sqrt{b}\) to \(\sqrt{ab}\). Thus, the first equation becomes:

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

### Step 2: Work with the second equation

The second equation is a quadratic form:

\[
a^2 + ab + b^2 = 4
\]

### Step 3: Solve by trial or substitution

We will try some simple integer values of \(a\) and \(b\) to see if we can solve the system.

#### Trying \(a = 1\) and \(b = 1\):

For the first equation:
\[
1 + \sqrt{1 \times 1} + 1 = 1 + 1 + 1 = 3 \neq 2
\]
This doesn't work.

#### Trying \(a = 1\) and \(b = 0\):

For the first equation:
\[
1 + \sqrt{1 \times 0} + 0 = 1 + 0 + 0 = 1 \neq 2
\]
This doesn't work.

#### Trying \(a = 1\) and \(b = 1\):

Now, let's test \(a = 2, b = -2. 

In method 

Learn how to solve the system of equations involving quadratic and square root terms: a + √a × √b + b = 2 and a^2 + ab + b^2 = 4. Explore step-by-step solutions and trial methods. Suitable for advanced high school students.

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