If `ax^(2)+bx+c=0` and `cx^(2)+bx+a=0 (a, b, c in R)` have a common non - real root, then

Options

A. `-2|a|lt b lt 2 |a|`
B. `-2|c|lt|b|lt2|c|`
C. `a=pm c`
D. a = c

Answer

To solve the problem, we need to analyze the given quadratic equations and the conditions for them to have a common non-real root.

### Step 1: Write down the given equations
We have two quadratic equations:
1. \( ax^2 + bx + c = 0 \)
2. \( cx^2 + bx + a = 0 \)

### Step 2: Understand the condition for common roots
For these two equations to have a common non-real root, they must share at least one root. Let's denote the common root as \( r \).

### Step 3: Substitute the common root into both equations
Since \( r \) is a root of both equations, we can substitute \( r \) into both equations:
1. \( ar^2 + br + c = 0 \) (from the first equation)
2. \( cr^2 + br + a = 0 \) (from the second equation)

### Step 4: Set the two equations equal to each other
Since both equations equal zero, we can set them equal to each other:
\[ ar^2 + br + c = cr^2 + br + a \]

### Step 5: Simplify the equation
By simplifying, we can eliminate \( br \) from both sides:
\[ ar^2 + c = cr^2 + a \]
Rearranging gives:
\[ ar^2 - cr^2 = a - c \]
Factoring out \( r^2 \) from the left side:
\[ (a - c)r^2 = a - c \]

### Step 6: Analyze the equation
If \( a \neq c \), we can divide both sides by \( a - c \):
\[ r^2 = 1 \]
This implies that \( r = 1 \) or \( r = -1 \), which are real roots. However, we need a common non-real root, so we must have:
\[ a - c = 0 \]
Thus, we conclude:
\[ a = c \]

### Step 7: Use the condition for non-real roots
For the quadratic equations to have non-real roots, the discriminant must be less than zero. The discriminant \( D \) for the first equation is:
\[ D_1 = b^2 - 4ac \]
For the second equation, it is:
\[ D_2 = b^2 - 4ca \]
Since \( a = c \), we can write:
\[ D_1 = b^2 - 4a^2 \]
\[ D_2 = b^2 - 4a^2 \]
Both discriminants must be less than zero:
\[ b^2 - 4a^2 < 0 \]

### Step 8: Solve the inequality
This implies:
\[ b^2 < 4a^2 \]
Taking square roots gives:
\[ |b| < 2|a| \]

### Conclusion
Thus, we have established that:
1. \( a = c \)
2. \( |b| < 2|a| \)