Let `r_1, r_2, r_3` be the three (not necessarily distinct) solution to the equation `x^3 + 4x^2 - ax +1 = 0`. If a can be any real number, then find the minimum value of `(r_1+1/r_1)^2+(r_2+1/r_2)^2+(r_3+1/r_3)^2`.

Options

A. even
B. odd
C. prime
D. divisible by `'13'`

Explanation

To find the minimum value of \( (r_1 + \frac{1}{r_1})^2 + (r_2 + \frac{1}{r_2})^2 + (r_3 + \frac{1}{r_3})^2 \), we start with the roots \( r_1, r_2, r_3 \) of the polynomial \( x^3 + 4x^2 - ax + 1 = 0 \).

### Step 1: Use Vieta's Formulas
From Vieta's formulas, we know:
- \( r_1 + r_2 + r_3 = -4 \)
- \( r_1 r_2 + r_2 r_3 + r_3 r_1 = -a \)
- \( r_1 r_2 r_3 = -1 \)

### Step 2: Rewrite the Expression
We can rewrite the expression we want to minimize:
\[ (r_i + \frac{1}{r_i})^2 = r_i^2 + 2 + \frac{1}{r_i^2} \]
Thus, we have:
\[ A = (r_1 + \frac{1}{r_1})^2 + (r_2 + \frac{1}{r_2})^2 + (r_3 + \frac{1}{r_3})^2 = r_1^2 + r_2^2 + r_3^2 + \frac{1}{r_1^2} + \frac{1}{r_2^2} + \frac{1}{r_3^2} + 6 \]

### Step 3: Find \( r_1^2 + r_2^2 + r_3^2 \)
Using the identity:
\[ r_1^2 + r_2^2 + r_3^2 = (r_1 + r_2 + r_3)^2 - 2(r_1 r_2 + r_2 r_3 + r_3 r_1) \]
Substituting the values from Vieta's:
\[ r_1^2 + r_2^2 + r_3^2 = (-4)^2 - 2(-a) = 16 + 2a \]

### Step 4: Find \( \frac{1}{r_1^2} + \frac{1}{r_2^2} + \frac{1}{r_3^2} \)
Using the identity:
\[ \frac{1}{r_1^2} + \frac{1}{r_2^2} + \frac{1}{r_3^2} = \frac{(r_2 r_3)^2 + (r_1 r_3)^2 + (r_1 r_2)^2}{(r_1 r_2 r_3)^2} \]
Substituting \( r_1 r_2 r_3 = -1 \):
\[ \frac{1}{r_1^2} + \frac{1}{r_2^2} + \frac{1}{r_3^2} = (r_2 r_3)^2 + (r_1 r_3)^2 + (r_1 r_2)^2 \]

### Step 5: Use the Square of Sums
Using the identity:
\[ (r_1 r_2 + r_2 r_3 + r_3 r_1)^2 = r_1^2 r_2^2 + r_2^2 r_3^2 + r_3^2 r_1^2 + 2r_1 r_2 r_3(r_1 + r_2 + r_3) \]
Substituting the values:
\[ (-a)^2 = r_1^2 r_2^2 + r_2^2 r_3^2 + r_3^2 r_1^2 + 2(-1)(-4) \]
This gives:
\[ a^2 = r_1^2 r_2^2 + r_2^2 r_3^2 + r_3^2 r_1^2 + 8 \]
Thus:
\[ r_1^2 r_2^2 + r_2^2 r_3^2 + r_3^2 r_1^2 = a^2 - 8 \]

### Step 6: Substitute Back into \( A \)
Now we can substitute back into \( A \):
\[ A = (16 + 2a) + (a^2 - 8) + 6 = a^2 + 2a + 14 \]

### Step 7: Find the Minimum Value
To find the minimum value of \( A = a^2 + 2a + 14 \), we complete the square:
\[ A = (a + 1)^2 + 13 \]
The minimum value occurs when \( (a + 1)^2 = 0 \), which gives \( a = -1 \). Thus:
\[ A_{\text{min}} = 0 + 13 = 13 \]

### Final Answer
The minimum value of \( (r_1 + \frac{1}{r_1})^2 + (r_2 + \frac{1}{r_2})^2 + (r_3 + \frac{1}{r_3})^2 \) is \( \boxed{13} \).