If the equations `x^(2) + 2x + 3 lambda =0 and 2x^(2) + 3x + 5 lambda ` = 0 have a non- zero common roots. then `lambda = ` (a)1 (b)-1 (c)3 (d)None of these

Options

A. 1
B. `-1`
C. 3
D. None of these

Answer

To find the value of \( \lambda \) such that the equations \( x^2 + 2x + 3\lambda = 0 \) and \( 2x^2 + 3x + 5\lambda = 0 \) have a non-zero common root, we can follow these steps:

### Step 1: Let the common root be \( c \).
Assume that \( c \) is the common root of both equations.

### Step 2: Substitute \( c \) into the first equation.
Substituting \( c \) into the first equation:
\[ c^2 + 2c + 3\lambda = 0 \]
From this, we can express \( 3\lambda \):
\[ 3\lambda = -c^2 - 2c \]

### Step 3: Substitute \( c \) into the second equation.
Now substitute \( c \) into the second equation:
\[ 2c^2 + 3c + 5\lambda = 0 \]
We can express \( 5\lambda \) in terms of \( c \):
\[ 5\lambda = -2c^2 - 3c \]

### Step 4: Relate the two expressions for \( \lambda \).
From the two equations for \( \lambda \):
1. \( \lambda = \frac{-c^2 - 2c}{3} \)
2. \( \lambda = \frac{-2c^2 - 3c}{5} \)

Set them equal to each other:
\[ \frac{-c^2 - 2c}{3} = \frac{-2c^2 - 3c}{5} \]

### Step 5: Cross-multiply to eliminate the fractions.
Cross-multiplying gives:
\[ 5(-c^2 - 2c) = 3(-2c^2 - 3c) \]
Expanding both sides:
\[ -5c^2 - 10c = -6c^2 - 9c \]

### Step 6: Rearrange the equation.
Rearranging gives:
\[ -5c^2 + 6c^2 - 10c + 9c = 0 \]
This simplifies to:
\[ c^2 - c = 0 \]

### Step 7: Factor the equation.
Factoring gives:
\[ c(c - 1) = 0 \]
Thus, \( c = 0 \) or \( c = 1 \).

### Step 8: Exclude the zero root.
Since we need a non-zero common root, we take \( c = 1 \).

### Step 9: Substitute \( c = 1 \) back to find \( \lambda \).
Using \( c = 1 \) in the expression for \( \lambda \):
\[ 3\lambda = -1^2 - 2(1) = -1 - 2 = -3 \]
Thus,
\[ \lambda = \frac{-3}{3} = -1 \]

### Final Answer:
The value of \( \lambda \) is \( -1 \).