Consider the following the next two items that follow : Let `alpha and beta` be the roots of the equation `x^(2) - (1 - 2a^(2)) x + (1 - 2a^(2)) = 0`
Under what condition equation have real roots?

Options

A. `a^(2) lt (1)/(2)`
B. `a^(2) gt (1)/(2)`
C. `a^(2) gt 1`
D. `a^(2) epsilon ((1)/(3),(1)/(2))`

Answer

To determine the condition under which the quadratic equation \( x^2 - (1 - 2a^2)x + (1 - 2a^2) = 0 \) has real roots, we need to analyze its discriminant.

### Step-by-Step Solution:

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

2. **Write the discriminant formula**:
The discriminant \( D \) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by:
\[ D = b^2 - 4ac \]
For the roots to be real, we need \( D \geq 0 \).

3. **Calculate the discriminant**:
Substitute the values of \( a \), \( b \), and \( c \) into the discriminant formula:
\[ D = [-(1 - 2a^2)]^2 - 4(1)(1 - 2a^2) \]
Simplifying this gives:
\[ D = (1 - 2a^2)^2 - 4(1 - 2a^2) \]

4. **Expand the expression**:
Now, expand \( (1 - 2a^2)^2 \):
\[ D = (1 - 4a^2 + 4a^4) - (4 - 8a^2) \]
This simplifies to:
\[ D = 1 - 4a^2 + 4a^4 - 4 + 8a^2 \]
Combining like terms results in:
\[ D = 4a^4 + 4a^2 - 3 \]

5. **Set the discriminant greater than or equal to zero**:
For the roots to be real, we need:
\[ 4a^4 + 4a^2 - 3 \geq 0 \]

6. **Let \( t = a^2 \)**:
Substitute \( t \) for \( a^2 \):
\[ 4t^2 + 4t - 3 \geq 0 \]

7. **Factor the quadratic**:
To solve \( 4t^2 + 4t - 3 = 0 \), we can use the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 4 \), \( b = 4 \), and \( c = -3 \):
\[ t = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 4 \cdot (-3)}}{2 \cdot 4} \]
\[ t = \frac{-4 \pm \sqrt{16 + 48}}{8} \]
\[ t = \frac{-4 \pm \sqrt{64}}{8} \]
\[ t = \frac{-4 \pm 8}{8} \]
This gives us:
\[ t = \frac{4}{8} = \frac{1}{2} \quad \text{and} \quad t = \frac{-12}{8} = -\frac{3}{2} \]
Since \( t = a^2 \) must be non-negative, we discard \( t = -\frac{3}{2} \).

8. **Determine the intervals**:
The quadratic \( 4t^2 + 4t - 3 \) opens upwards (since the coefficient of \( t^2 \) is positive). The roots are \( t = \frac{1}{2} \). The quadratic will be non-negative outside the interval defined by the roots:
\[ t \leq -\frac{3}{2} \quad \text{or} \quad t \geq \frac{1}{2} \]
Since \( t = a^2 \geq 0 \), we conclude:
\[ a^2 \geq \frac{1}{2} \]

### Final Condition:
Thus, the condition for the quadratic equation to have real roots is:
\[ a^2 \geq \frac{1}{2} \]