If the roots of the equation `ax^(2)+bx+c=0` are in the ratio `m:n` then

Options

A. `mnb^(2)=ac(m+n)^(2)`
B. `b^(2)(m+n)=mn`
C. `m+n=b^(2)mn`
D. `mnc^(2)=ab(m+n)^(2)`

Answer

To solve the problem, we need to find the relationship between the coefficients of the quadratic equation \( ax^2 + bx + c = 0 \) and the ratio of its roots \( m:n \).

### Step-by-Step Solution:

1. **Assume the Roots**:
Let the roots of the equation be \( \alpha \) and \( \beta \). Given that the roots are in the ratio \( m:n \), we can express the roots as:
\[ \alpha = km \quad \text{and} \quad \beta = kn \]
where \( k \) is a constant.

**Hint**: Use the ratio of the roots to express them in terms of a common variable.

2. **Sum and Product of Roots**:
According to Vieta's formulas, for the quadratic equation \( ax^2 + bx + c = 0 \):
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} \)
- The product of the roots \( \alpha \beta = \frac{c}{a} \)

3. **Express the Sum and Product**:
Substitute \( \alpha \) and \( \beta \) into the sum and product formulas:
\[ \alpha + \beta = km + kn = k(m+n) \]
Thus, we have:
\[ k(m+n) = -\frac{b}{a} \quad \text{(1)} \]

For the product:
\[ \alpha \beta = (km)(kn) = k^2mn \]
Thus, we have:
\[ k^2mn = \frac{c}{a} \quad \text{(2)} \]

4. **Solve for \( k \)**:
From equation (1):
\[ k = -\frac{b}{a(m+n)} \]

5. **Substitute \( k \) into the Product Equation**:
Substitute \( k \) from equation (1) into equation (2):
\[ \left(-\frac{b}{a(m+n)}\right)^2 mn = \frac{c}{a} \]

6. **Simplify the Equation**:
Squaring \( k \):
\[ \frac{b^2}{a^2(m+n)^2} mn = \frac{c}{a} \]
Multiply both sides by \( a^2(m+n)^2 \):
\[ b^2 mn = ac(m+n)^2 \]

7. **Final Result**:
Rearranging gives us the desired relationship:
\[ mn b^2 = ac(m+n)^2 \]

### Conclusion:
The relationship derived shows that if the roots of the equation \( ax^2 + bx + c = 0 \) are in the ratio \( m:n \), then:
\[ mn b^2 = ac(m+n)^2 \]