The roots of the equation `2(a ^2 +b^2)x^2+2(a+b)x+1=0` are

Options

A. real and equal
B. real and unequal
C. complex numbers
D. cannot be determind

Answer

To find the roots of the equation \(2(a^2 + b^2)x^2 + 2(a + b)x + 1 = 0\), we will analyze the nature of the roots using the discriminant.

### Step-by-Step Solution

1. **Identify the coefficients**:
The given quadratic equation can be compared to the standard form \(ax^2 + bx + c = 0\).
- Here, \(a = 2(a^2 + b^2)\)
- \(b = 2(a + b)\)
- \(c = 1\)

2. **Calculate the discriminant**:
The discriminant \(D\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by:
\[ D = b^2 - 4ac \]
Substituting the values of \(a\), \(b\), and \(c\):
\[ D = [2(a + b)]^2 - 4[2(a^2 + b^2)][1] \]

3. **Expand the discriminant**:
\[ D = 4(a + b)^2 - 8(a^2 + b^2) \]

4. **Further simplify the expression**:
Expanding \(4(a + b)^2\):
\[ D = 4(a^2 + 2ab + b^2) - 8(a^2 + b^2) \]
\[ D = 4a^2 + 8ab + 4b^2 - 8a^2 - 8b^2 \]
\[ D = (4a^2 - 8a^2) + (4b^2 - 8b^2) + 8ab \]
\[ D = -4a^2 - 4b^2 + 8ab \]
\[ D = -4(a^2 - 2ab + b^2) \]
\[ D = -4(a - b)^2 \]

5. **Determine the nature of the roots**:
Since \((a - b)^2\) is always non-negative, \(-4(a - b)^2\) is always non-positive. Therefore:
- If \(D < 0\), the roots are imaginary (complex).
- If \(D = 0\), the roots are real and equal.
- If \(D > 0\), the roots are real and unequal.

Here, since \(D = -4(a - b)^2\) is less than 0 for all \(a\) and \(b\) (unless \(a = b\) which gives \(D = 0\)), we conclude that the roots are imaginary (complex).

### Conclusion:
The roots of the equation \(2(a^2 + b^2)x^2 + 2(a + b)x + 1 = 0\) are **imaginary (complex)**.

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