The roots of the equation `2a^(2)x^(2) - 2abx + b^(2) = 0 ` when a `lt`0 and b`gt`0 are :

Options

A. Sometimes complex
B. Always irrational
C. Always complex
D. Always real

Explanation

To find the roots of the quadratic equation \(2a^2x^2 - 2abx + b^2 = 0\) under the conditions \(a < 0\) and \(b > 0\), 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:
- \(A = 2a^2\)
- \(B = -2ab\)
- \(C = b^2\)

### 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 \(C\):
\[ D = (-2ab)^2 - 4(2a^2)(b^2) \]

### Step 3: Simplify the discriminant
Calculating \(D\):
\[ D = 4a^2b^2 - 4(2a^2)(b^2) \]
\[ D = 4a^2b^2 - 8a^2b^2 \]
\[ D = -4a^2b^2 \]

### Step 4: Analyze the discriminant
Since \(a < 0\) and \(b > 0\), we know that \(a^2 > 0\) and \(b^2 > 0\). Therefore, the product \(4a^2b^2\) is positive, which means:
\[ D = -4a^2b^2 < 0 \]

### Step 5: Conclusion about the roots
Since the discriminant \(D\) is negative, it indicates that the roots of the quadratic equation are complex.

Thus, the roots of the equation \(2a^2x^2 - 2abx + b^2 = 0\) when \(a < 0\) and \(b > 0\) are **always complex**.

### Final Answer:
The roots are always complex.

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