The sum of all the real roots of equation `x^4-3x^3-2x^2-3x+1=0` is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To find the sum of all the real roots of the equation \(x^4 - 3x^3 - 2x^2 - 3x + 1 = 0\), we can follow these steps:

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

### Step 2: Factor the polynomial
We can factor the polynomial into two quadratic equations. We can express it as:
\[ (x^2 - 4x + 1)(x^2 + x + 1) = 0 \]

### Step 3: Solve the first quadratic equation
Now, we solve the first quadratic equation:
\[ x^2 - 4x + 1 = 0 \]
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
- Here, \(a = 1\), \(b = -4\), and \(c = 1\).

Calculating the discriminant:
\[ b^2 - 4ac = (-4)^2 - 4 \cdot 1 \cdot 1 = 16 - 4 = 12 \]
Now, applying the quadratic formula:
\[ x = \frac{4 \pm \sqrt{12}}{2} \]
Simplifying \(\sqrt{12} = 2\sqrt{3}\):
\[ x = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3} \]

### Step 4: Solve the second quadratic equation
Next, we solve the second quadratic equation:
\[ x^2 + x + 1 = 0 \]
Using the quadratic formula again:
- Here, \(a = 1\), \(b = 1\), and \(c = 1\).

Calculating the discriminant:
\[ b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 \]
Since the discriminant is negative, this quadratic has no real roots.

### Step 5: Sum of the real roots
The real roots we found from the first quadratic are:
\[ 2 + \sqrt{3} \quad \text{and} \quad 2 - \sqrt{3} \]
To find the sum of the real roots:
\[ (2 + \sqrt{3}) + (2 - \sqrt{3}) = 2 + \sqrt{3} + 2 - \sqrt{3} = 4 \]

### Final Answer
The sum of all the real roots of the equation is:
\[ \boxed{4} \]