The value of `a` for which the equation `2x^(2)-2(2a+1) x+a(a-1)=0` has roots, `alpha and beta` such that `alpha lt a lt beta` is

Options

A. `a ge 0`
B. `a lt 0`
C. `-3 lt a lt 0`
D. none of these

Answer

To find the value of \( a \) for which the equation

\[ 2x^2 - 2(2a + 1)x + a(a - 1) = 0 \]

has roots \( \alpha \) and \( \beta \) such that \( \alpha < a < \beta \), we can follow these steps:

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

### Step 2: Calculate the roots using the quadratic formula
The roots of the quadratic equation can be found using the quadratic formula:

\[ x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \]

Substituting the values of \( A \), \( B \), and \( C \):

\[ x = \frac{2(2a + 1) \pm \sqrt{(-2(2a + 1))^2 - 4 \cdot 2 \cdot a(a - 1)}}{2 \cdot 2} \]

### Step 3: Simplify the expression
Calculating \( B^2 - 4AC \):

\[ B^2 = (-2(2a + 1))^2 = 4(2a + 1)^2 = 4(4a^2 + 4a + 1) = 16a^2 + 16a + 4 \]

\[ 4AC = 4 \cdot 2 \cdot a(a - 1) = 8a^2 - 8a \]

Now, substituting back into the discriminant:

\[ B^2 - 4AC = (16a^2 + 16a + 4) - (8a^2 - 8a) = 16a^2 + 16a + 4 - 8a^2 + 8a = 8a^2 + 24a + 4 \]

### Step 4: Find the roots
Now substituting back into the quadratic formula:

\[ x = \frac{2(2a + 1) \pm \sqrt{8a^2 + 24a + 4}}{4} \]

This simplifies to:

\[ x = \frac{2a + 1 \pm \sqrt{2(2a^2 + 6a + 1)}}{2} \]

### Step 5: Set conditions for roots
Let \( \alpha \) and \( \beta \) be the roots. We need to ensure that \( \alpha < a < \beta \). This implies:

1. \( \alpha < a \)
2. \( a < \beta \)

### Step 6: Analyze the inequalities
Using the roots derived, we can analyze the inequalities:

1. For \( \alpha < a \):
\[ \frac{2a + 1 - \sqrt{2(2a^2 + 6a + 1)}}{2} < a \]
Simplifying gives:
\[ 2a + 1 - \sqrt{2(2a^2 + 6a + 1)} < 2a \]
\[ 1 < \sqrt{2(2a^2 + 6a + 1)} \]
Squaring both sides:
\[ 1 < 2(2a^2 + 6a + 1) \]
\[ 1 < 4a^2 + 12a + 2 \]
\[ 0 < 4a^2 + 12a + 1 \]

2. For \( a < \beta \):
\[ a < \frac{2a + 1 + \sqrt{2(2a^2 + 6a + 1)}}{2} \]
Simplifying gives:
\[ 2a < 2a + 1 + \sqrt{2(2a^2 + 6a + 1)} \]
\[ 0 < 1 + \sqrt{2(2a^2 + 6a + 1)} \]
This is always true.

### Step 7: Final condition
The condition \( 4a^2 + 12a + 1 > 0 \) is a quadratic inequality. The discriminant is:

\[ D = 12^2 - 4 \cdot 4 \cdot 1 = 144 - 16 = 128 > 0 \]

Thus, the quadratic has two real roots, and we can find the intervals where it is positive.

### Step 8: Solve the quadratic inequality
Using the quadratic formula on \( 4a^2 + 12a + 1 = 0 \):

\[ a = \frac{-12 \pm \sqrt{128}}{8} = \frac{-12 \pm 8\sqrt{2}}{8} = -\frac{3}{2} \pm \sqrt{2} \]

The quadratic is positive outside the roots. Therefore, the intervals are:

\[ a < -\frac{3}{2} - \sqrt{2} \quad \text{or} \quad a > -\frac{3}{2} + \sqrt{2} \]

### Conclusion
The value of \( a \) for which the equation has roots \( \alpha \) and \( \beta \) such that \( \alpha < a < \beta \) is:

\[ a \in \left(-\infty, -\frac{3}{2} - \sqrt{2}\right) \cup \left(-\frac{3}{2} + \sqrt{2}, \infty\right) \]