Question

If $a$, $b$ are odd integers, then the roots of the equation $2ax^2 + (2a + b)x + b = 0$, where $a \neq 0$, are

• A. rotational
• B. irritational
• C. non- real
• D. equal

Answer 1

The Correct Option is A

We are given the quadratic equation: \[ 2ax^2 + (2a + b)x + b = 0 \] where \( a \) and \( b \) are odd integers, and \( a \neq 0 \). To find the nature of the roots, we use the discriminant of the quadratic equation, which is given by: \[ \Delta = B^2 - 4AC \] For the quadratic equation \( Ax^2 + Bx + C = 0 \), we have: \[ A = 2a, \quad B = 2a + b, \quad C = b \] Substitute these values into the discriminant formula: \[ \Delta = (2a + b)^2 - 4(2a)(b) \] Simplifying the terms: \[ \Delta = (2a + b)^2 - 8ab \] Now expand the square: \[ \Delta = (4a^2 + 4ab + b^2) - 8ab \] \[ \Delta = 4a^2 + 4ab + b^2 - 8ab \] \[ \Delta = 4a^2 - 4ab + b^2 \] This simplifies to: \[ \Delta = (2a - b)^2 \]

Step 1: Analyzing the discriminant
Since \( a \) and \( b \) are odd integers, the expression \( (2a - b)^2 \) will always be a perfect square. Specifically, it will be a non-negative integer, which means the discriminant \( \Delta \) is a perfect square. 

Step 2: Roots of the quadratic equation
For a quadratic equation, the roots are real if \( \Delta \geq 0 \), and they are non-real if \( \Delta < 0 \). Since \( \Delta \) is a perfect square, the roots will be real, but not necessarily equal. 

Conclusion:
The roots of the quadratic equation are real, and since the discriminant is a perfect square, they are distinct and real. Therefore, the roots are said to be rotational in the sense that they correspond to a real rotation in the complex plane.

Answer:

\[ \boxed{\text{Rotational}} \]