Find the square root of
`a+x+sqrt(2ax+x^(2))`.

Answer

To find the square root of the expression \( a + x + \sqrt{2ax + x^2} \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ n = \sqrt{a + x + \sqrt{2ax + x^2}} \]

### Step 2: Simplify the square root
Notice that \( \sqrt{2ax + x^2} \) can be rewritten:
\[ \sqrt{2ax + x^2} = \sqrt{x^2 + 2ax} = \sqrt{x(x + 2a)} \]
So we can rewrite \( n \) as:
\[ n = \sqrt{a + x + \sqrt{x(x + 2a)}} \]

### Step 3: Combine terms
Now, we can express \( n \) in a more manageable form:
\[ n = \sqrt{a + x + \sqrt{x(x + 2a)}} \]

### Step 4: Assume a form for \( n \)
Let’s assume that \( n \) can be expressed in the form:
\[ n = \sqrt{p} + \sqrt{q} \]
for some \( p \) and \( q \).

### Step 5: Square both sides
Squaring both sides gives:
\[ n^2 = p + q + 2\sqrt{pq} \]
We want this to equal \( a + x + \sqrt{2ax + x^2} \).

### Step 6: Identify \( p \) and \( q \)
By comparing terms, we can set:
\[ p + q = a + x \]
and
\[ 2\sqrt{pq} = \sqrt{2ax + x^2} \]

### Step 7: Solve for \( p \) and \( q \)
From the second equation, squaring gives:
\[ 4pq = 2ax + x^2 \]
Thus:
\[ pq = \frac{1}{4}(2ax + x^2) \]

### Step 8: Solve the system of equations
Now we have a system:
1. \( p + q = a + x \)
2. \( pq = \frac{1}{4}(2ax + x^2) \)

Let \( p \) and \( q \) be the roots of the quadratic:
\[ t^2 - (a+x)t + \frac{1}{4}(2ax + x^2) = 0 \]

### Step 9: Calculate the discriminant
The discriminant \( D \) of this quadratic is:
\[ D = (a+x)^2 - 4 \cdot \frac{1}{4}(2ax + x^2) = (a+x)^2 - (2ax + x^2) \]
Simplifying gives:
\[ D = a^2 + 2ax + x^2 - 2ax - x^2 = a^2 \]

### Step 10: Find \( p \) and \( q \)
Since \( D = a^2 \), the roots are:
\[ p, q = \frac{(a+x) \pm a}{2} \]
This gives:
\[ p = x \quad \text{and} \quad q = a \]
or vice versa.

### Final Step: Write the final answer
Thus, we have:
\[ n = \sqrt{x} + \sqrt{a} \]

### Conclusion
The square root of \( a + x + \sqrt{2ax + x^2} \) is:
\[ \sqrt{a} + \sqrt{x} \]

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