If d,e,f are G.P. and the two quadratic equations
`ax^(2)+2bx+c=0anddx^(2)+2ex+f=0` have a common root, then

Options

A. `(d)/(a),(e)/(b),(f)/(c)` are in H.P.
B. `(d)/(a),(e)/(b),(f)/(c)` are in G.P.
C. dbf=aef+cde
D. `b^(2)df=ace^(2)`

Answer

To solve the problem step by step, we need to analyze the given quadratic equations and the conditions provided.

### Step 1: Understand the Given Information
We have two quadratic equations:
1. \( ax^2 + 2bx + c = 0 \)
2. \( dx^2 + 2ex + f = 0 \)

It is given that \( d, e, f \) are in a Geometric Progression (G.P.), which means:
\[ e^2 = df \]
Also, these two equations have a common root.

### Step 2: Find the Discriminant of the Second Equation
The discriminant \( D \) of the second quadratic equation \( dx^2 + 2ex + f = 0 \) is given by:
\[ D = (2e)^2 - 4df = 4e^2 - 4df \]
Substituting \( e^2 = df \) into the discriminant, we get:
\[ D = 4e^2 - 4df = 4df - 4df = 0 \]
This indicates that the second equation has equal roots.

### Step 3: Find the Roots of the Second Equation
Since the discriminant is zero, the roots of the second equation can be found using the formula:
\[ x = \frac{-b}{2a} \]
For our equation, this becomes:
\[ x = \frac{-2e}{2d} = \frac{-e}{d} \]
Thus, the common root is:
\[ r = \frac{-e}{d} \]

### Step 4: Substitute the Common Root into the First Equation
Now, we substitute \( r = \frac{-e}{d} \) into the first equation \( ax^2 + 2bx + c = 0 \):
\[ a\left(\frac{-e}{d}\right)^2 + 2b\left(\frac{-e}{d}\right) + c = 0 \]
This simplifies to:
\[ \frac{ae^2}{d^2} - \frac{2be}{d} + c = 0 \]

### Step 5: Substitute \( e^2 = df \)
Using the relation \( e^2 = df \):
\[ \frac{a(df)}{d^2} - \frac{2be}{d} + c = 0 \]
This simplifies to:
\[ \frac{af}{d} - \frac{2be}{d} + c = 0 \]
Multiplying through by \( d \) gives:
\[ af - 2be + cd = 0 \]

### Step 6: Rearranging the Equation
Rearranging the equation gives:
\[ af + cd = 2be \]

### Step 7: Express in Terms of Ratios
Dividing through by \( f \):
\[ \frac{a}{d} + \frac{c}{f} = \frac{2b}{e} \]
This shows that the ratios \( \frac{a}{d}, \frac{c}{f}, \frac{2b}{e} \) are equal.

### Step 8: Conclusion
From the above steps, we conclude that if \( \frac{a}{d} + \frac{c}{f} = \frac{2b}{e} \), then the ratios \( \frac{d}{a}, \frac{e}{b}, \frac{f}{c} \) are in Harmonic Progression (H.P.).

### Final Answer
Thus, the answer to the question is that \( \frac{d}{a}, \frac{e}{b}, \frac{f}{c} \) are in H.P.

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