If the ratio of the roots of `a_(1)x^(2) +b_(1)x+c_(1)=0` be equal to the ratio of the roots of `a_(2)x^(2)+b_(2)+c_(2)=0," then "(a_(1))/(a_(2)), (b_(1))/(b_(2)), (c_(1))/(c_(2))` are in

Options

A. A.P.
B. G.P.
C. H.P.
D. None

Answer

To solve the problem, we need to show that if the ratio of the roots of the quadratic equations \( a_1x^2 + b_1x + c_1 = 0 \) and \( a_2x^2 + b_2x + c_2 = 0 \) are equal, then the ratios \( \frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2} \) are in a geometric progression (GP).

### Step 1: Define the roots of the first equation
Let the roots of the first equation \( a_1x^2 + b_1x + c_1 = 0 \) be \( \alpha \) and \( m\alpha \), where \( m \) is the ratio of the roots.

### Step 2: Use the sum and product of roots
According to Vieta's formulas:
- The sum of the roots \( \alpha + m\alpha = -\frac{b_1}{a_1} \)
- The product of the roots \( \alpha \cdot m\alpha = \frac{c_1}{a_1} \)

From the sum of roots:
\[ \alpha(1 + m) = -\frac{b_1}{a_1} \quad \text{(1)} \]

From the product of roots:
\[ \alpha^2 m = \frac{c_1}{a_1} \quad \text{(2)} \]

### Step 3: Define the roots of the second equation
Let the roots of the second equation \( a_2x^2 + b_2x + c_2 = 0 \) be \( \beta \) and \( m\beta \).

### Step 4: Use the sum and product of roots for the second equation
Using Vieta's formulas again:
- The sum of the roots \( \beta + m\beta = -\frac{b_2}{a_2} \)
- The product of the roots \( \beta \cdot m\beta = \frac{c_2}{a_2} \)

From the sum of roots:
\[ \beta(1 + m) = -\frac{b_2}{a_2} \quad \text{(3)} \]

From the product of roots:
\[ \beta^2 m = \frac{c_2}{a_2} \quad \text{(4)} \]

### Step 5: Divide equations (1) and (3)
Dividing equation (1) by equation (3):
\[ \frac{\alpha(1 + m)}{\beta(1 + m)} = \frac{-\frac{b_1}{a_1}}{-\frac{b_2}{a_2}} \]
This simplifies to:
\[ \frac{\alpha}{\beta} = \frac{b_1 a_2}{b_2 a_1} \quad \text{(5)} \]

### Step 6: Divide equations (2) and (4)
Dividing equation (2) by equation (4):
\[ \frac{\alpha^2 m}{\beta^2 m} = \frac{\frac{c_1}{a_1}}{\frac{c_2}{a_2}} \]
This simplifies to:
\[ \frac{\alpha^2}{\beta^2} = \frac{c_1 a_2}{c_2 a_1} \quad \text{(6)} \]

### Step 7: Relate equations (5) and (6)
From equation (5), we have:
\[ \frac{\alpha}{\beta} = \frac{b_1 a_2}{b_2 a_1} \]
Squaring both sides gives:
\[ \left(\frac{\alpha}{\beta}\right)^2 = \left(\frac{b_1 a_2}{b_2 a_1}\right)^2 \]
Substituting into equation (6):
\[ \frac{b_1^2 a_2^2}{b_2^2 a_1^2} = \frac{c_1 a_2}{c_2 a_1} \]
Cross-multiplying gives:
\[ b_1^2 a_2 c_2 = c_1 a_1 b_2^2 \]

### Step 8: Conclusion
This implies that:
\[ b_1^2 = \frac{c_1 a_1}{a_2 c_2} \Rightarrow a_1 : a_2 = b_1 : b_2 = c_1 : c_2 \]
Thus, \( \frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2} \) are in geometric progression.