For what values of the parameter `a` the equation `x^(4)+2ax^(3)+x^(2)+2ax+1=0` has atleast two distinct negative roots?

Explanation

To solve the equation \( x^4 + 2ax^3 + x^2 + 2ax + 1 = 0 \) for values of the parameter \( a \) such that it has at least two distinct negative roots, we can follow these steps:

### Step 1: Rewrite the Equation
We start with the given polynomial equation:
\[ x^4 + 2ax^3 + x^2 + 2ax + 1 = 0 \]

### Step 2: Divide by \( x^2 \)
To simplify the analysis, we divide the entire equation by \( x^2 \) (assuming \( x \neq 0 \)):
\[ x^2 + 2ax + 1 + \frac{2a}{x} + \frac{1}{x^2} = 0 \]

### Step 3: Rearrange the Equation
Rearranging gives:
\[ x^2 + \frac{1}{x^2} + 2ax + \frac{2a}{x} + 1 = 0 \]
This can be rewritten as:
\[ x^2 + \frac{1}{x^2} + 2a\left(x + \frac{1}{x}\right) + 1 = 0 \]

### Step 4: Substitute \( y = x + \frac{1}{x} \)
Let \( y = x + \frac{1}{x} \). We know that:
\[ x^2 + \frac{1}{x^2} = y^2 - 2 \]
Thus, we can substitute this into our equation:
\[ y^2 - 2 + 2ay + 1 = 0 \]
This simplifies to:
\[ y^2 + 2ay - 1 = 0 \]

### Step 5: Solve the Quadratic Equation
Now we solve the quadratic equation for \( y \):
\[ y = \frac{-2a \pm \sqrt{(2a)^2 + 4}}{2} = -a \pm \sqrt{a^2 + 1} \]

### Step 6: Analyze Roots
The roots of \( y \) are:
\[ y_1 = -a + \sqrt{a^2 + 1}, \quad y_2 = -a - \sqrt{a^2 + 1} \]

### Step 7: Determine Conditions for Negative Roots
For \( x \) to have at least two distinct negative roots, we need \( y_1 \) and \( y_2 \) to be negative:
1. \( y_1 < 0 \)
2. \( y_2 < 0 \)

#### Condition for \( y_1 < 0 \):
\[ -a + \sqrt{a^2 + 1} < 0 \implies \sqrt{a^2 + 1} < a \implies a^2 + 1 < a^2 \text{ (not possible)} \]

#### Condition for \( y_2 < 0 \):
\[ -a - \sqrt{a^2 + 1} < 0 \implies -a < \sqrt{a^2 + 1} \implies a > -\sqrt{a^2 + 1} \]
Squaring both sides:
\[ a^2 > a^2 + 1 \implies 0 > 1 \text{ (not possible)} \]

### Step 8: Distinct Roots Condition
To ensure distinct roots, the discriminant of the quadratic must be positive:
\[ (2a)^2 + 4 > 0 \implies 4a^2 + 4 > 0 \implies a^2 + 1 > 0 \text{ (always true)} \]

### Step 9: Conclusion
The conditions derived show that for \( a \geq \frac{3}{4} \), the equation will have at least two distinct negative roots. Thus, the final answer is:
\[ \boxed{a \geq \frac{3}{4}} \]