Let a, b be non-zero real numbers. Which of the following statements about the quadratic equation
`ax^2 + (a + b)x + b =0`
is neccesarily true ?
(I) It has at least one negative root
(II) It has at least one positive root.
(III) Both its roots are real.

Options

A. (I) and (II) only
B. (I) and (III) only
C. (II) and (III) only
D. All of them

Answer

To solve the problem, we need to analyze the quadratic equation given by:

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

where \( a \) and \( b \) are non-zero real numbers. We will examine the three statements provided:

1. It has at least one negative root.
2. It has at least one positive root.
3. Both its roots are real.

### Step 1: Calculate the Discriminant

The discriminant \( D \) of a quadratic equation \( Ax^2 + Bx + C = 0 \) is given by:

\[ D = B^2 - 4AC \]

For our equation:

- \( A = a \)
- \( B = a + b \)
- \( C = b \)

Thus, the discriminant \( D \) becomes:

\[ D = (a + b)^2 - 4ab \]

### Step 2: Simplify the Discriminant

Now, we simplify \( D \):

\[ D = (a^2 + 2ab + b^2) - 4ab = a^2 - 2ab + b^2 = (a - b)^2 \]

### Step 3: Analyze the Discriminant

Since \( (a - b)^2 \) is always greater than or equal to zero (because the square of any real number is non-negative), we conclude that:

\[ D \geq 0 \]

This means that the roots of the quadratic equation are real. Thus, statement (III) is necessarily true.

### Step 4: Finding the Roots

Next, we find the roots using the quadratic formula:

\[ x = \frac{-B \pm \sqrt{D}}{2A} \]

Substituting our values:

\[ x = \frac{-(a + b) \pm \sqrt{(a - b)^2}}{2a} \]

This simplifies to:

\[ x = \frac{-(a + b) \pm |a - b|}{2a} \]

### Step 5: Analyzing the Roots

Now we consider two cases based on the value of \( a \) and \( b \):

1. **Case 1**: If \( a \geq b \)

- Then \( |a - b| = a - b \)
- The roots become:
\[ x_1 = \frac{-(a + b) + (a - b)}{2a} = \frac{-2b}{2a} = -\frac{b}{a} \]
\[ x_2 = \frac{-(a + b) - (a - b)}{2a} = \frac{-2a}{2a} = -1 \]

Both roots \( x_1 \) and \( x_2 \) are negative if \( b > 0 \), and one root is negative if \( b < 0 \).

2. **Case 2**: If \( a < b \)

- Then \( |a - b| = b - a \)
- The roots become:
\[ x_1 = \frac{-(a + b) - (b - a)}{2a} = \frac{-2b}{2a} = -\frac{b}{a} \]
\[ x_2 = \frac{-(a + b) + (b - a)}{2a} = \frac{-2a}{2a} = -1 \]

Again, both roots are negative if \( a > 0 \), and one root is negative if \( a < 0 \).

### Conclusion

From our analysis:

- **Statement (I)**: It has at least one negative root is true.
- **Statement (II)**: It has at least one positive root is not necessarily true, as both roots can be negative.
- **Statement (III)**: Both its roots are real is true.

Thus, the correct statements are (I) and (III).

### Final Answer

The statements that are necessarily true are:
- (I) It has at least one negative root.
- (III) Both its roots are real.