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

Options

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

Explanation

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

### Step 1: Define the Roots
Let the roots of the equation be \( \alpha \) and \( 2\alpha \). This means one root is \( \alpha \) and the other root is \( 2\alpha \).

### Step 2: Use the Sum of Roots
According to Vieta's formulas, the sum of the roots of the quadratic equation is given by:
\[ \alpha + 2\alpha = -\frac{b}{a} \]
This simplifies to:
\[ 3\alpha = -\frac{b}{a} \]
From this, we can express \( \alpha \):
\[ \alpha = -\frac{b}{3a} \]

### Step 3: Use the Product of Roots
The product of the roots is also given by Vieta's formulas:
\[ \alpha \cdot 2\alpha = \frac{c}{a} \]
This simplifies to:
\[ 2\alpha^2 = \frac{c}{a} \]

### Step 4: Substitute \( \alpha \) into the Product Equation
Now, substitute \( \alpha = -\frac{b}{3a} \) into the product equation:
\[ 2\left(-\frac{b}{3a}\right)^2 = \frac{c}{a} \]
Calculating \( \left(-\frac{b}{3a}\right)^2 \):
\[ \left(-\frac{b}{3a}\right)^2 = \frac{b^2}{9a^2} \]
So, substituting this back gives:
\[ 2 \cdot \frac{b^2}{9a^2} = \frac{c}{a} \]

### Step 5: Simplify the Equation
Multiply both sides by \( 9a^2 \):
\[ 2b^2 = 9ac \]

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