The minimum number of real roots of `(x^(2)+3x+a)(x^(2)+ax+1)=0` equals

Explanation

To solve the problem, we need to analyze the quadratic equations given in the expression \((x^2 + 3x + a)(x^2 + ax + 1) = 0\). We will find the minimum number of real roots of this equation.

### Step 1: Identify the two quadratic equations
The equation can be split into two separate quadratic equations:
1. \(x^2 + 3x + a = 0\)
2. \(x^2 + ax + 1 = 0\)

### Step 2: Find the discriminant of the first quadratic
The discriminant \(D_1\) of the first quadratic equation \(x^2 + 3x + a = 0\) is given by:
\[ D_1 = b^2 - 4ac = 3^2 - 4(1)(a) = 9 - 4a \]
For this equation to have real roots, the discriminant must be non-negative:
\[ 9 - 4a \geq 0 \]
This simplifies to:
\[ a \leq \frac{9}{4} \]

### Step 3: Find the discriminant of the second quadratic
The discriminant \(D_2\) of the second quadratic equation \(x^2 + ax + 1 = 0\) is given by:
\[ D_2 = b^2 - 4ac = a^2 - 4(1)(1) = a^2 - 4 \]
For this equation to have real roots, the discriminant must also be non-negative:
\[ a^2 - 4 \geq 0 \]
This factors to:
\[ (a - 2)(a + 2) \geq 0 \]
The solutions to this inequality are:
\[ a \leq -2 \quad \text{or} \quad a \geq 2 \]

### Step 4: Combine the conditions
Now we have two conditions:
1. \(a \leq \frac{9}{4}\)
2. \(a \leq -2\) or \(a \geq 2\)

We will analyze the intervals:
- From \(a \leq -2\), we have real roots from the second equation, and since \(a \leq \frac{9}{4}\), we can have:
- If \(a < -2\), both quadratics can have real roots.
- From \(2 \leq a \leq \frac{9}{4}\), we can have:
- The first quadratic has real roots, and the second quadratic also has real roots.

### Step 5: Count the minimum number of real roots
- If \(a < -2\): Both quadratics have 2 real roots each, giving a total of 4 real roots.
- If \(-2 \leq a < 2\): The first quadratic has 2 real roots, and the second quadratic has no real roots, giving a total of 2 real roots.
- If \(2 \leq a \leq \frac{9}{4}\): Both quadratics have 2 real roots each, giving a total of 4 real roots.
- If \(a > \frac{9}{4}\): The first quadratic has no real roots, and the second quadratic has real roots, giving a total of 2 real roots.

### Conclusion
The minimum number of real roots occurs in the interval \(-2 \leq a < 2\) or when \(a > \frac{9}{4}\), which gives us a minimum of **2 real roots**.

Thus, the answer is:
\[ \text{Minimum number of real roots} = 2 \]