Let `A={a in R}` the equation `(1+2i)x^3-2(3+i)x^2+(5-4i)x+a^2=0` has at least one real root. Then the value of `(suma^2)/2` is_______.

Explanation

To solve the problem, we need to analyze the given polynomial equation and determine the conditions under which it has at least one real root. The equation is:

\[ (1 + 2i)x^3 - 2(3 + i)x^2 + (5 - 4i)x + a^2 = 0 \]

### Step 1: Assume a real root
Let \( \alpha \) be a real root of the polynomial. This means that substituting \( x = \alpha \) into the equation will yield zero:

\[ (1 + 2i)\alpha^3 - 2(3 + i)\alpha^2 + (5 - 4i)\alpha + a^2 = 0 \]

### Step 2: Separate real and imaginary parts
We can separate the equation into real and imaginary parts. The equation can be rewritten as:

\[ (1\alpha^3 - 6\alpha^2 + 5\alpha + a^2) + i(2\alpha^3 - 2\alpha^2 - 4\alpha) = 0 \]

For the equation to hold, both the real part and the imaginary part must equal zero:

1. Real part:
\[ \alpha^3 - 6\alpha^2 + 5\alpha + a^2 = 0 \tag{1} \]

2. Imaginary part:
\[ 2\alpha^3 - 2\alpha^2 - 4\alpha = 0 \tag{2} \]

### Step 3: Solve the imaginary part equation
From equation (2), we can factor out \( 2\alpha \):

\[ 2\alpha(\alpha^2 - \alpha - 2) = 0 \]

This gives us two cases:

1. \( \alpha = 0 \)
2. \( \alpha^2 - \alpha - 2 = 0 \)

Solving the quadratic equation \( \alpha^2 - \alpha - 2 = 0 \) using the quadratic formula:

\[ \alpha = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm 3}{2} \]

This yields:

\[ \alpha = 2 \quad \text{or} \quad \alpha = -1 \]

### Step 4: Substitute values of \( \alpha \) into the real part equation
Now we substitute each value of \( \alpha \) back into equation (1) to find \( a^2 \).

1. **For \( \alpha = 0 \)**:
\[ 0^3 - 6(0^2) + 5(0) + a^2 = 0 \implies a^2 = 0 \implies a = 0 \]

2. **For \( \alpha = 2 \)**:
\[ 2^3 - 6(2^2) + 5(2) + a^2 = 0 \implies 8 - 24 + 10 + a^2 = 0 \]
\[ -6 + a^2 = 0 \implies a^2 = 6 \implies a = \pm \sqrt{6} \]

3. **For \( \alpha = -1 \)**:
\[ (-1)^3 - 6(-1)^2 + 5(-1) + a^2 = 0 \implies -1 - 6 - 5 + a^2 = 0 \]
\[ -12 + a^2 = 0 \implies a^2 = 12 \implies a = \pm \sqrt{12} \]

### Step 5: Calculate the sum of squares of \( a \)
Now we have the possible values of \( a \):

- \( a = 0 \)
- \( a = \sqrt{6} \)
- \( a = -\sqrt{6} \)
- \( a = \sqrt{12} \)
- \( a = -\sqrt{12} \)

Calculating \( a^2 \):

- For \( a = 0 \): \( a^2 = 0 \)
- For \( a = \sqrt{6} \): \( a^2 = 6 \)
- For \( a = -\sqrt{6} \): \( a^2 = 6 \)
- For \( a = \sqrt{12} \): \( a^2 = 12 \)
- For \( a = -\sqrt{12} \): \( a^2 = 12 \)

Thus, the sum of squares is:

\[ 0 + 6 + 6 + 12 + 12 = 36 \]

### Step 6: Find the final answer
The problem asks for \( \frac{\sum a^2}{2} \):

\[ \frac{36}{2} = 18 \]

Thus, the final answer is:

\[ \boxed{18} \]