For all `lambda in R`, The equation `ax^2+ (b - lambda)x + (a-b-lambda)= 0, a != 0` has real roots. Then

Options

A. `a = b`
B. `b le a lt 0`
C. `b gt a gt 0`
D. `a gt b gt 0`

Answer

To solve the problem, we need to determine the conditions under which the quadratic equation

\[ ax^2 + (b - \lambda)x + (a - b - \lambda) = 0 \]

has real roots for all values of \( \lambda \in \mathbb{R} \), given that \( a \neq 0 \).

### Step-by-Step Solution:

1. **Identify the coefficients**:
The quadratic equation can be expressed in the standard form \( ax^2 + bx + c = 0 \). Here:
- \( A = a \)
- \( B = b - \lambda \)
- \( C = a - b - \lambda \)

2. **Calculate the discriminant**:
The condition for the quadratic equation to have real roots is that the discriminant \( D \) must be non-negative:
\[ D = B^2 - 4AC \geq 0 \]
Substituting the coefficients:
\[ D = (b - \lambda)^2 - 4a(a - b - \lambda) \]

3. **Expand the discriminant**:
Expanding \( D \):
\[ D = (b - \lambda)^2 - 4a(a - b - \lambda) \]
\[ = (b^2 - 2b\lambda + \lambda^2) - 4a^2 + 4ab + 4a\lambda \]
\[ = \lambda^2 + (4a - 2b)\lambda + (b^2 + 4ab - 4a^2) \]

4. **Set the discriminant condition**:
For \( D \) to be non-negative for all \( \lambda \), the quadratic in \( \lambda \) must have a non-positive discriminant:
\[ (4a - 2b)^2 - 4 \cdot 1 \cdot (b^2 + 4ab - 4a^2) \leq 0 \]

5. **Simplify the inequality**:
Expanding and simplifying:
\[ (4a - 2b)^2 - 4(b^2 + 4ab - 4a^2) \leq 0 \]
\[ 16a^2 - 16ab + 4b^2 - 4b^2 - 16ab + 16a^2 \leq 0 \]
\[ 32a^2 - 32ab \leq 0 \]
\[ 32a(a - b) \leq 0 \]

6. **Determine the conditions**:
From \( 32a(a - b) \leq 0 \), we can conclude:
- If \( a > 0 \), then \( a - b \leq 0 \) which implies \( b \geq a \).
- If \( a < 0 \), then \( a - b \geq 0 \) which implies \( b \leq a \).

### Conclusion:
Thus, the conditions for the equation to have real roots for all \( \lambda \in \mathbb{R} \) are:
- If \( a > 0 \), then \( b \geq a \).
- If \( a < 0 \), then \( b \leq a \).