If `ax^(2)+bx+c=0andbx^(2)+cx+a=0` have a common root then the relation between a,b,c is

Options

A. `a^(2)+b^(2)+c^(2)=2` abc
B. `b^(2)=ac`
C. `a^(3)+b^(3)+c^(3)=3abc`
D. `a=1`

Answer

To find the relation between \( a, b, c \) given that the equations \( ax^2 + bx + c = 0 \) and \( bx^2 + cx + a = 0 \) have a common root, we can follow these steps:

### Step 1: Let the common root be \( r \).
Assume \( r \) is the common root of both equations. Then, we can substitute \( r \) into both equations.

### Step 2: Substitute \( r \) into the first equation.
Substituting \( r \) into the first equation:
\[ ar^2 + br + c = 0 \tag{1} \]

### Step 3: Substitute \( r \) into the second equation.
Substituting \( r \) into the second equation:
\[ br^2 + cr + a = 0 \tag{2} \]

### Step 4: Rearranging both equations.
From equation (1), we can express \( c \):
\[ c = -ar^2 - br \tag{3} \]
From equation (2), we can express \( a \):
\[ a = -br^2 - cr \tag{4} \]

### Step 5: Substitute \( c \) from equation (3) into equation (4).
Substituting equation (3) into equation (4):
\[ a = -br^2 - \left(-ar^2 - br\right)r \]
This simplifies to:
\[ a = -br^2 + ar^3 + br^2 \]
Thus, we have:
\[ a = ar^3 \tag{5} \]

### Step 6: Rearranging equation (5).
Rearranging gives us:
\[ a(1 - r^3) = 0 \]
This implies either \( a = 0 \) or \( r^3 = 1 \).

### Step 7: Analyze the case \( r^3 = 1 \).
If \( r^3 = 1 \), then \( r = 1 \) (since \( r \) is a real root). Substituting \( r = 1 \) into equation (1):
\[ a(1)^2 + b(1) + c = 0 \Rightarrow a + b + c = 0 \tag{6} \]

### Step 8: Conclusion.
Thus, we have derived the relation between \( a, b, c \):
\[ a + b + c = 0 \]

### Final Relation:
The relation between \( a, b, c \) is:
\[ a + b + c = 0 \]