What is the condition that one root of the equation `ax^(2) + bx + c = 0,a !=0` should be doube the other ?

Options

A. `2a^(2) = 9bc`
B. `2b^(2) = 9ac`
C. `2c^(2) = 9ab`
D. None of these

Answer

To find the condition that one root of the quadratic equation \( ax^2 + bx + c = 0 \) (where \( a \neq 0 \)) is double the other root, we can follow these steps:

### Step-by-Step Solution:

1. **Assume the Roots**:
Let the roots of the quadratic equation be \( \alpha \) and \( \beta \). According to the problem, we can assume:
\[ \beta = 2\alpha \]

2. **Use Vieta's Formulas**:
According to Vieta's formulas, the sum of the roots \( \alpha + \beta \) is given by:
\[ \alpha + \beta = -\frac{b}{a} \]
Substituting \( \beta = 2\alpha \) into this equation:
\[ \alpha + 2\alpha = -\frac{b}{a} \]
This simplifies to:
\[ 3\alpha = -\frac{b}{a} \]

3. **Solve for \( \alpha \)**:
From the equation \( 3\alpha = -\frac{b}{a} \), we can solve for \( \alpha \):
\[ \alpha = -\frac{b}{3a} \]

4. **Calculate the Product of the Roots**:
The product of the roots \( \alpha \cdot \beta \) is given by:
\[ \alpha \cdot \beta = \frac{c}{a} \]
Substituting \( \beta = 2\alpha \):
\[ \alpha \cdot (2\alpha) = \frac{c}{a} \]
This simplifies to:
\[ 2\alpha^2 = \frac{c}{a} \]

5. **Substitute \( \alpha \)**:
Now substitute \( \alpha = -\frac{b}{3a} \) into the product equation:
\[ 2\left(-\frac{b}{3a}\right)^2 = \frac{c}{a} \]
Simplifying this gives:
\[ 2 \cdot \frac{b^2}{9a^2} = \frac{c}{a} \]

6. **Clear the Denominator**:
Multiply both sides by \( 9a^2 \):
\[ 2b^2 = 9ac \]

### Final Condition:
Thus, the condition that one root of the equation \( ax^2 + bx + c = 0 \) is double the other root is:
\[ 2b^2 = 9ac \]