Given that one root of `2x^3+3x^2-8x+3=0 ` is double of another root , find the roots of the equation.

Answer

To solve the equation \(2x^3 + 3x^2 - 8x + 3 = 0\) given that one root is double another root, we will follow these steps:

### Step 1: Define the Roots
Let the roots of the polynomial be:
- \( \alpha \) (first root)
- \( 2\alpha \) (second root, which is double the first)
- \( \beta \) (third root)

### Step 2: Use Vieta's Formulas
According to Vieta's formulas for a cubic equation \(ax^3 + bx^2 + cx + d = 0\):
1. The sum of the roots \( \alpha + 2\alpha + \beta = -\frac{b}{a} \)
2. The product of the roots \( \alpha \cdot 2\alpha \cdot \beta = -\frac{d}{a} \)
3. The sum of the products of the roots taken two at a time \( \alpha \cdot 2\alpha + 2\alpha \cdot \beta + \alpha \cdot \beta = \frac{c}{a} \)

### Step 3: Apply Vieta's Formulas
For our polynomial \(2x^3 + 3x^2 - 8x + 3\):
- \(a = 2\), \(b = 3\), \(c = -8\), \(d = 3\)

Using Vieta’s formulas:
1. **Sum of the Roots**:
\[ \alpha + 2\alpha + \beta = -\frac{3}{2} \implies 3\alpha + \beta = -\frac{3}{2} \quad \text{(Equation 1)} \]

2. **Product of the Roots**:
\[ \alpha \cdot 2\alpha \cdot \beta = -\frac{3}{2} \implies 2\alpha^2 \beta = -\frac{3}{2} \quad \text{(Equation 2)} \]

3. **Sum of Products of Roots**:
\[ \alpha \cdot 2\alpha + 2\alpha \cdot \beta + \alpha \cdot \beta = -\frac{-8}{2} \implies 2\alpha^2 + 3\alpha\beta = -4 \quad \text{(Equation 3)} \]

### Step 4: Solve the Equations
From Equation 1, we can express \(\beta\):
\[ \beta = -\frac{3}{2} - 3\alpha \quad \text{(Substituting into Equation 2)} \]

Substituting \(\beta\) into Equation 2:
\[ 2\alpha^2 \left(-\frac{3}{2} - 3\alpha\right) = -\frac{3}{2} \]
Expanding:
\[ -3\alpha^2 - 6\alpha^3 = -\frac{3}{2} \]
Multiplying through by -2 to eliminate the fraction:
\[ 6\alpha^2 + 12\alpha^3 = 3 \quad \text{(Rearranging gives)} \quad 12\alpha^3 + 6\alpha^2 - 3 = 0 \]
Dividing through by 3:
\[ 4\alpha^3 + 2\alpha^2 - 1 = 0 \quad \text{(Equation 4)} \]

### Step 5: Factor or Solve Equation 4
To solve \(4\alpha^3 + 2\alpha^2 - 1 = 0\), we can use the Rational Root Theorem or numerical methods. Testing \( \alpha = \frac{1}{2} \):
\[ 4\left(\frac{1}{2}\right)^3 + 2\left(\frac{1}{2}\right)^2 - 1 = 4 \cdot \frac{1}{8} + 2 \cdot \frac{1}{4} - 1 = \frac{1}{2} + \frac{1}{2} - 1 = 0 \]
Thus, \( \alpha = \frac{1}{2} \) is a root.

### Step 6: Find \(\beta\)
Substituting \(\alpha = \frac{1}{2}\) back into Equation 1:
\[ \beta = -\frac{3}{2} - 3\left(\frac{1}{2}\right) = -\frac{3}{2} - \frac{3}{2} = -3 \]

### Step 7: Find the Roots
Now we have:
- \( \alpha = \frac{1}{2} \)
- \( 2\alpha = 1 \)
- \( \beta = -3 \)

### Final Roots
The roots of the equation \(2x^3 + 3x^2 - 8x + 3 = 0\) are:
\[ \frac{1}{2}, 1, -3 \]