If a, b, c are positive real numbers such that the equations `ax^(2) + bx + c = 0 and bx^(2) + cx + a = 0`, have a common root, then

Options

A. `1: 2: 3`
B. `3: 2: 1`
C. `1: 3 : 2`
D. `3 : 1 : 2 `

Explanation

To solve the problem, we need to find the ratio \( \frac{a}{b} : \frac{b}{c} : \frac{c}{a} \) given that the equations \( ax^2 + bx + c = 0 \) and \( bx^2 + cx + a = 0 \) have a common root.

### Step 1: Let the common root be \( r \).
Since \( r \) is a common root, it satisfies both equations. Therefore, we can write:
1. \( ar^2 + br + c = 0 \) (1)
2. \( br^2 + cr + a = 0 \) (2)

### Step 2: Rearranging the equations.
From equation (1), we can express \( c \):
\[ c = -ar^2 - br \]
Substituting this expression for \( c \) into equation (2):
\[ br^2 + (-ar^2 - br)r + a = 0 \]
This simplifies to:
\[ br^2 - ar^3 - br^2 + a = 0 \]
Thus, we have:
\[ -ar^3 + a = 0 \]
Factoring out \( a \):
\[ a(1 - r^3) = 0 \]

### Step 3: Analyzing the factor.
Since \( a \) is a positive real number, we cannot have \( a = 0 \). Therefore, we must have:
\[ 1 - r^3 = 0 \implies r^3 = 1 \implies r = 1 \]

### Step 4: Substitute \( r = 1 \) back into the equations.
Substituting \( r = 1 \) into equation (1):
\[ a(1)^2 + b(1) + c = 0 \implies a + b + c = 0 \]
This is not possible since \( a, b, c \) are positive real numbers. Thus, we need to find a different approach.

### Step 5: Set up the ratios.
Since both equations have a common root, we can set up the ratios:
\[ \frac{a}{b} = \frac{b}{c} = \frac{c}{a} \]
Let \( k \) be the common ratio. Then we can express:
\[ a = kb, \quad b = kc, \quad c = ka \]

### Step 6: Substitute and solve for \( k \).
Substituting \( b = kc \) into \( a = kb \):
\[ a = k(kc) = k^2c \]
Now substituting \( c = ka \) into \( b = kc \):
\[ b = k(ka) = k^2a \]
This gives us a system of equations:
1. \( a = k^2c \)
2. \( b = k^2a \)
3. \( c = k^2b \)

### Step 7: Express \( a, b, c \) in terms of \( k \).
From \( a = k^2c \), substituting \( c = k^2b \):
\[ a = k^2(k^2b) = k^4b \]
From \( b = k^2a \):
\[ b = k^2(k^4b) = k^6b \]
This implies \( k^6 = 1 \), thus \( k = 1 \).

### Step 8: Conclusion.
Since \( k = 1 \), we have:
\[ a = b = c \]
Thus, the ratio \( a : b : c = 1 : 1 : 1 \).

### Final Answer:
The ratio \( a : b : c = 1 : 1 : 1 \).

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