Solve the equation `x^4 -2x^3 + 4x^2 + 6x -21=0`
the sum of two of roots being zero.

Answer

To solve the equation \( x^4 - 2x^3 + 4x^2 + 6x - 21 = 0 \) given that the sum of two of its roots is zero, we can follow these steps:

### Step 1: Identify the roots
Let the roots of the polynomial be \( \alpha, \beta, \gamma, \delta \). Given that \( \alpha + \beta = 0 \), we can express \( \beta \) as \( \beta = -\alpha \).

### Step 2: Use Vieta's formulas
According to Vieta's formulas, the sum of the roots \( \alpha + \beta + \gamma + \delta \) is equal to the coefficient of \( x^3 \) divided by the leading coefficient (which is 1 here). Therefore, we have:
\[ \alpha + \beta + \gamma + \delta = -(-2) = 2 \]
Substituting \( \beta = -\alpha \):
\[ \alpha - \alpha + \gamma + \delta = 2 \implies \gamma + \delta = 2 \]

### Step 3: Define products of the roots
Let \( \alpha \beta = p \) and \( \gamma \delta = q \). We can form two quadratic equations based on the roots:
1. For roots \( \alpha \) and \( \beta \):
\[ x^2 - (\alpha + \beta)x + \alpha \beta = 0 \implies x^2 + p = 0 \quad \text{(since } \alpha + \beta = 0\text{)} \]
This gives us:
\[ x^2 + p = 0 \quad \text{(Equation 1)} \]

2. For roots \( \gamma \) and \( \delta \):
\[ x^2 - (\gamma + \delta)x + \gamma \delta = 0 \implies x^2 - 2x + q = 0 \quad \text{(since } \gamma + \delta = 2\text{)} \]
This gives us:
\[ x^2 - 2x + q = 0 \quad \text{(Equation 2)} \]

### Step 4: Multiply the quadratic equations
The product of these two quadratic equations will give us the original polynomial:
\[ (x^2 + p)(x^2 - 2x + q) = 0 \]
Expanding this:
\[ x^4 - 2x^3 + (q + p)x^2 - 2px + pq = 0 \]

### Step 5: Compare coefficients
Now, we can compare coefficients with the original polynomial \( x^4 - 2x^3 + 4x^2 + 6x - 21 = 0 \):
1. Coefficient of \( x^3 \): \(-2 = -2\) (matches)
2. Coefficient of \( x^2 \): \(q + p = 4\) (Equation 3)
3. Coefficient of \( x \): \(-2p = 6\) (Equation 4)
4. Constant term: \(pq = -21\) (Equation 5)

### Step 6: Solve the equations
From Equation 4:
\[ -2p = 6 \implies p = -3 \]
Substituting \( p = -3 \) into Equation 3:
\[ q - 3 = 4 \implies q = 7 \]
Now substituting \( p \) and \( q \) into Equation 5:
\[ (-3)(7) = -21 \quad \text{(this holds true)} \]

### Step 7: Form the quadratic equations
Now we can write the two quadratic equations:
1. From Equation 1:
\[ x^2 + (-3) = 0 \implies x^2 = 3 \implies x = \pm \sqrt{3} \]
2. From Equation 2:
\[ x^2 - 2x + 7 = 0 \]
Using the quadratic formula:
\[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 7}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 - 28}}{2} = \frac{2 \pm \sqrt{-24}}{2} = 1 \pm \sqrt{6}i \]

### Final Roots
Thus, the roots of the equation \( x^4 - 2x^3 + 4x^2 + 6x - 21 = 0 \) are:
\[ x = \sqrt{3}, -\sqrt{3}, 1 + \sqrt{6}i, 1 - \sqrt{6}i \]