The condition that one root of `px^(2)+qx+r=0` may be double of the other is

Options

A. `q^(2)=9rp`
B. `2q^(2)=9rp`
C. `q^(2)=rp`
D. None of these

Answer

To find the condition that one root of the quadratic equation \( px^2 + qx + r = 0 \) is double the other, we can follow these steps:

### Step 1: Define the Roots
Let one root be \( \alpha \). Then, the other root will be \( 2\alpha \).

### Step 2: Use the Sum of Roots
According to Vieta's formulas, the sum of the roots of the quadratic equation \( ax^2 + bx + c = 0 \) is given by:
\[ \text{Sum of roots} = -\frac{b}{a} \]
For our equation, the sum of the roots is:
\[ \alpha + 2\alpha = 3\alpha \]
Thus, we have:
\[ 3\alpha = -\frac{q}{p} \]
From this, we can express \( \alpha \):
\[ \alpha = -\frac{q}{3p} \quad \text{(Equation 1)} \]

### Step 3: Use the Product of Roots
Again, according to Vieta's formulas, the product of the roots is given by:
\[ \text{Product of roots} = \frac{c}{a} \]
For our equation, the product of the roots is:
\[ \alpha \cdot (2\alpha) = 2\alpha^2 \]
Thus, we have:
\[ 2\alpha^2 = \frac{r}{p} \]
Substituting \( \alpha \) from Equation 1 into this equation gives:
\[ 2\left(-\frac{q}{3p}\right)^2 = \frac{r}{p} \]

### Step 4: Simplify the Equation
Calculating \( \left(-\frac{q}{3p}\right)^2 \):
\[ \left(-\frac{q}{3p}\right)^2 = \frac{q^2}{9p^2} \]
Substituting this back into the product equation:
\[ 2 \cdot \frac{q^2}{9p^2} = \frac{r}{p} \]
This simplifies to:
\[ \frac{2q^2}{9p^2} = \frac{r}{p} \]

### Step 5: Clear the Denominator
Multiplying both sides by \( 9p^2 \) gives:
\[ 2q^2 = 9rp \]

### Conclusion
Thus, the condition that one root of \( px^2 + qx + r = 0 \) is double the other is:
\[ 2q^2 = 9rp \]