The expression `ax^2+2bx+c`, where 'a' is non-zero real number, has same sign as that of 'a' for every real value of x,then roots of quadratic equation `ax^2+ (b-c) x-2b-c -a-0` are: (a) real and equal (b) real and unequal (c) non-real having positive real part(d) non-real having negative real part

Options

A. ral and equal
B. real and unequal
C. non-real having positive ral part
D. non-real having negative real part

Answer

To solve the problem, we need to analyze the given quadratic expression and the quadratic equation to determine the nature of its roots.

### Step-by-Step Solution:

1. **Understanding the Given Expression**:
We have the expression \( ax^2 + 2bx + c \) which has the same sign as \( a \) for every real value of \( x \). This implies that the quadratic does not cross the x-axis, meaning it has no real roots.

2. **Discriminant Analysis**:
For a quadratic equation \( ax^2 + bx + c \), the discriminant \( D \) is given by:
\[ D = b^2 - 4ac \]
Since the expression \( ax^2 + 2bx + c \) has the same sign as \( a \), it means:
\[ D < 0 \implies (2b)^2 - 4ac < 0 \implies 4b^2 < 4ac \implies b^2 < ac \]

3. **Analyzing the Given Quadratic Equation**:
We need to analyze the quadratic equation:
\[ ax^2 + (b - c)x - (2b + c + a) = 0 \]
Let's denote the coefficients as follows:
- \( A = a \)
- \( B = b - c \)
- \( C = -(2b + c + a) \)

4. **Calculating the Discriminant of the New Equation**:
The discriminant \( D' \) of the new quadratic equation is:
\[ D' = (b - c)^2 - 4a(-2b - c - a) \]
Simplifying this:
\[ D' = (b - c)^2 + 4a(2b + c + a) \]

5. **Expanding and Simplifying**:
Expanding \( D' \):
\[ D' = (b - c)^2 + 8ab + 4ac + 4a^2 \]
We know from earlier that \( 4ac > 4b^2 \), which implies \( 4ac - 4b^2 > 0 \). Thus:
\[ D' > 0 \text{ (since all terms are positive)} \]

6. **Conclusion about the Roots**:
Since \( D' > 0 \), the quadratic equation \( ax^2 + (b - c)x - (2b + c + a) = 0 \) has real and unequal roots.

### Final Answer:
The roots of the quadratic equation are **(b) real and unequal**.