If the equation `x^(3)+ax^(2)+b=0 (b ne 0)` has a double root then

Options

A. `4a +27b^(3)=0`
B. `4a^(3)+27b=0`
C. `27 a +4b^(3)=0`
D. none

Answer

To solve the problem where the equation \( x^3 + ax^2 + b = 0 \) has a double root, we will follow these steps:

### Step 1: Understand the concept of double roots
A double root means that one of the roots of the polynomial is repeated. Let's denote the double root as \( \alpha \) and the other root as \( \beta \). Thus, the roots can be expressed as \( \alpha, \alpha, \beta \).

### Step 2: Form the polynomial using the roots
The polynomial can be expressed in terms of its roots:
\[ (x - \alpha)^2 (x - \beta) = 0 \]
Expanding this gives:
\[ (x^2 - 2\alpha x + \alpha^2)(x - \beta) = 0 \]
Expanding further:
\[ x^3 - \beta x^2 - 2\alpha x^2 + 2\alpha \beta x + \alpha^2 x - \alpha^2 \beta = 0 \]
Combining like terms:
\[ x^3 + (-\beta - 2\alpha)x^2 + (2\alpha \beta + \alpha^2)x - \alpha^2 \beta = 0 \]

### Step 3: Compare coefficients
Now, we compare the coefficients of the expanded polynomial with the given polynomial \( x^3 + ax^2 + b = 0 \):
- Coefficient of \( x^2 \):
\[ -\beta - 2\alpha = a \tag{1} \]
- Coefficient of \( x \):
\[ 2\alpha \beta + \alpha^2 = 0 \tag{2} \]
- Constant term:
\[ -\alpha^2 \beta = b \tag{3} \]

### Step 4: Solve the equations
From equation (2):
\[ 2\alpha \beta + \alpha^2 = 0 \implies \alpha(2\beta + \alpha) = 0 \]
Since \( \alpha \) cannot be zero (as it is a double root), we have:
\[ 2\beta + \alpha = 0 \implies \alpha = -2\beta \tag{4} \]

### Step 5: Substitute \( \alpha \) into the other equations
Substituting equation (4) into equation (1):
\[ -\beta - 2(-2\beta) = a \implies -\beta + 4\beta = a \implies 3\beta = a \tag{5} \]

Now substituting equation (4) into equation (3):
\[ -\alpha^2 \beta = b \implies -(-2\beta)^2 \beta = b \implies -4\beta^2 = b \tag{6} \]

### Step 6: Express \( a \) and \( b \) in terms of \( \beta \)
From equations (5) and (6), we have:
\[ a = 3\beta \quad \text{and} \quad b = -4\beta^2 \]

### Step 7: Find the relationship between \( a \) and \( b \)
Now, we can express \( 4a + 27b = 0 \):
\[ 4(3\beta) + 27(-4\beta^2) = 0 \implies 12\beta - 108\beta^2 = 0 \]
Factoring out \( \beta \):
\[ \beta(12 - 108\beta) = 0 \]
This gives us:
\[ \beta = 0 \quad \text{or} \quad 12 - 108\beta = 0 \implies \beta = \frac{1}{9} \]

### Step 8: Substitute back to find \( a \) and \( b \)
If \( \beta = \frac{1}{9} \):
\[ a = 3\left(\frac{1}{9}\right) = \frac{1}{3} \]
\[ b = -4\left(\frac{1}{9}\right)^2 = -4 \cdot \frac{1}{81} = -\frac{4}{81} \]

### Conclusion
Thus, the values of \( a \) and \( b \) that satisfy the condition of the cubic equation having a double root are:
\[ a = \frac{1}{3}, \quad b = -\frac{4}{81} \]