The ratio of roots of the equations `ax^(2) + bx + c = 0 and px^(2) + qx + r = 0` are equal If `D_(1) and D_(2)` are respectively discriminants, then what is `(D_(1))/(D_(2))` equal to ?

Options

A. `(a^(2))/(p^(2))`
B. `(b^(2))/(q^(2))`
C. `(c^(2))/(r^(2))`
D. None of these

Answer

To solve the problem, we need to find the ratio of the discriminants of two quadratic equations given that the ratio of their roots is equal.

### Step-by-Step Solution:

1. **Understanding the Quadratic Equations**:
We have two quadratic equations:
- \( ax^2 + bx + c = 0 \) (Equation 1)
- \( px^2 + qx + r = 0 \) (Equation 2)

2. **Roots of the Quadratic Equations**:
Let the roots of Equation 1 be \( \alpha \) and \( \beta \), and the roots of Equation 2 be \( \gamma \) and \( \delta \). Given that the ratio of the roots of both equations is equal, we can express this as:
\[ \frac{\alpha}{\beta} = \frac{\gamma}{\delta} \]

3. **Using the Relationship of Roots**:
For a quadratic equation \( ax^2 + bx + c = 0 \), the sum and product of the roots can be given by:
- Sum of roots \( \alpha + \beta = -\frac{b}{a} \)
- Product of roots \( \alpha \beta = \frac{c}{a} \)

Similarly, for the second equation \( px^2 + qx + r = 0 \):
- Sum of roots \( \gamma + \delta = -\frac{q}{p} \)
- Product of roots \( \gamma \delta = \frac{r}{p} \)

4. **Finding the Discriminants**:
The discriminant \( D \) for a quadratic equation \( ax^2 + bx + c = 0 \) is given by:
\[ D = b^2 - 4ac \]
Therefore, the discriminants for our equations are:
- \( D_1 = b^2 - 4ac \)
- \( D_2 = q^2 - 4pr \)

5. **Establishing the Ratio of Discriminants**:
From the property of the ratio of the roots, we know that:
\[ \frac{\alpha}{\beta} = \frac{\gamma}{\delta} \implies \frac{\alpha}{\beta} = k \text{ (some constant)} \]
This leads to the conclusion that:
\[ \frac{b^2}{q^2} = \frac{D_1}{D_2} \]
Thus, we can express the ratio of the discriminants as:
\[ \frac{D_1}{D_2} = \frac{b^2}{q^2} \]

6. **Final Result**:
Therefore, the ratio of the discriminants \( \frac{D_1}{D_2} \) is equal to \( \frac{b^2}{q^2} \).

### Conclusion:
The final answer is:
\[ \frac{D_1}{D_2} = \frac{b^2}{q^2} \]