If one root of the equations ` A x ^(2) + b x + c = 0 ` is two and a half times the others, then which of the following is TRUE ?

Options

A. ` 7 B ^(2) = 3 CA `
B. ` 7 B^(2) = 4 CA`
C. ` 7 B^(2) = 36 CA`
D. ` 10 B^(2) = 49 CA`

Answer

To solve the problem step by step, we start with the information given:

1. **Understanding the Roots**:
Let the first root be \( r_1 = A \) and the second root be \( r_2 = \frac{5}{2}A \).

2. **Using the Relationship of Roots**:
According to Vieta's formulas, for a quadratic equation of the form \( Ax^2 + Bx + C = 0 \):
- The sum of the roots \( r_1 + r_2 = -\frac{B}{A} \)
- The product of the roots \( r_1 \cdot r_2 = \frac{C}{A} \)

3. **Calculating the Sum of the Roots**:
\[ r_1 + r_2 = A + \frac{5}{2}A = \frac{7}{2}A \]
Therefore, we have:
\[ -\frac{B}{A} = \frac{7}{2}A \implies B = -\frac{7}{2}A^2 \]

4. **Calculating the Product of the Roots**:
\[ r_1 \cdot r_2 = A \cdot \frac{5}{2}A = \frac{5}{2}A^2 \]
Therefore, we have:
\[ \frac{C}{A} = \frac{5}{2}A^2 \implies C = \frac{5}{2}A^3 \]

5. **Finding the Relationship Between B² and CA**:
Now, we need to find \( B^2 \) and \( CA \):
\[ B^2 = \left(-\frac{7}{2}A^2\right)^2 = \frac{49}{4}A^4 \]
\[ CA = \left(\frac{5}{2}A^3\right)A = \frac{5}{2}A^4 \]

6. **Establishing the Final Relationship**:
Now we can relate \( B^2 \) and \( CA \):
\[ B^2 = \frac{49}{4}A^4 \quad \text{and} \quad CA = \frac{5}{2}A^4 \]
Thus, we can write:
\[ B^2 = \frac{49}{10}CA \]
or equivalently,
\[ CA = \frac{10}{49}B^2 \]

### Conclusion
From the above calculations, we conclude that the relationship between \( B^2 \) and \( CA \) is:
\[ B^2 = \frac{49}{10}CA \]