If a and b are the odd integers, then the roots of the equation, `2ax^2 + (2a + b)x + b = 0, a!=0`, will be

Options

A. rational
B. irrational
C. non-real
D. none of these

Answer

To solve the quadratic equation \(2ax^2 + (2a + b)x + b = 0\) where \(a\) and \(b\) are odd integers, we will follow these steps:

### Step 1: Identify the coefficients
The given quadratic equation is in the standard form \(Ax^2 + Bx + C = 0\), where:
- \(A = 2a\)
- \(B = 2a + b\)
- \(C = b\)

### Step 2: Calculate the discriminant
The discriminant \(D\) of a quadratic equation is given by the formula:
\[ D = B^2 - 4AC \]
Substituting the values of \(A\), \(B\), and \(C\):
\[ D = (2a + b)^2 - 4(2a)(b) \]

### Step 3: Expand the discriminant
Now we will expand \(D\):
\[ D = (2a + b)(2a + b) - 8ab \]
\[ D = 4a^2 + 4ab + b^2 - 8ab \]
\[ D = 4a^2 - 4ab + b^2 \]
\[ D = 4a^2 - 4ab + b^2 = (2a - b)^2 \]

### Step 4: Analyze the discriminant
Since \(D = (2a - b)^2\), the discriminant is always non-negative, indicating that the roots are real.

### Step 5: Calculate the roots using the quadratic formula
The roots of the quadratic equation can be calculated using the quadratic formula:
\[ x = \frac{-B \pm \sqrt{D}}{2A} \]
Substituting \(B\) and \(D\):
\[ x = \frac{-(2a + b) \pm (2a - b)}{2(2a)} \]

### Step 6: Simplify the roots
Now we will simplify the expression:
1. For the negative sign:
\[ x_1 = \frac{-(2a + b) - (2a - b)}{4a} = \frac{-2a - b - 2a + b}{4a} = \frac{-4a}{4a} = -1 \]

2. For the positive sign:
\[ x_2 = \frac{-(2a + b) + (2a - b)}{4a} = \frac{-2a - b + 2a - b}{4a} = \frac{-2b}{4a} = \frac{-b}{2a} \]

### Conclusion
The roots of the equation \(2ax^2 + (2a + b)x + b = 0\) are:
\[ x_1 = -1 \quad \text{and} \quad x_2 = -\frac{b}{2a} \]

Since both \(a\) and \(b\) are odd integers, \(-1\) is a rational number, and \(-\frac{b}{2a}\) is also rational because the product of two odd integers is odd, and dividing an odd integer by an even integer (which \(2a\) is) yields a rational number.

Thus, the roots of the equation are rational.