The values of 'a' for which the roots of the equation `x^(2) + x + a = 0` are real and exceed 'a' are

Options

A. `0 lt a lt 1//4`
B. `a lt 1//4`
C. `a lt - 2`
D. `-2 lt a lt 0`

Answer

To solve the problem, we need to find the values of \( a \) for which the roots of the equation \( x^2 + x + a = 0 \) are real and exceed \( a \). We will break this down into steps.

### Step 1: Determine the condition for real roots
For the quadratic equation \( x^2 + x + a = 0 \) to have real roots, the discriminant must be greater than zero. The discriminant \( D \) is given by:

\[ D = b^2 - 4ac \]

In our case, \( a = 1 \), \( b = 1 \), and \( c = a \). Thus, we have:

\[ D = 1^2 - 4(1)(a) = 1 - 4a \]

Setting the discriminant greater than zero for real roots:

\[ 1 - 4a > 0 \]

### Step 2: Solve the inequality
Rearranging the inequality gives:

\[ 1 > 4a \implies a < \frac{1}{4} \]

### Step 3: Determine the condition for roots exceeding \( a \)
Let the roots of the equation be \( \alpha \) and \( \beta \). We need both roots to be greater than \( a \). The roots can be expressed using the quadratic formula:

\[ \alpha, \beta = \frac{-b \pm \sqrt{D}}{2a} = \frac{-1 \pm \sqrt{1 - 4a}}{2} \]

For both roots to be greater than \( a \), we need:

\[ \frac{-1 + \sqrt{1 - 4a}}{2} > a \quad \text{and} \quad \frac{-1 - \sqrt{1 - 4a}}{2} > a \]

### Step 4: Solve the first inequality
Starting with the first inequality:

\[ \frac{-1 + \sqrt{1 - 4a}}{2} > a \]

Multiplying through by 2 (and reversing the inequality if necessary) gives:

\[ -1 + \sqrt{1 - 4a} > 2a \]

Rearranging yields:

\[ \sqrt{1 - 4a} > 2a + 1 \]

Squaring both sides (valid since both sides are positive for the range we are considering):

\[ 1 - 4a > (2a + 1)^2 \]

Expanding the right side:

\[ 1 - 4a > 4a^2 + 4a + 1 \]

### Step 5: Rearranging the inequality
Rearranging gives:

\[ 0 > 4a^2 + 8a \]

Factoring out \( 4a \):

\[ 0 > 4a(a + 2) \]

This implies:

\[ a(a + 2) < 0 \]

### Step 6: Finding the intervals
The product \( a(a + 2) < 0 \) is satisfied when \( a \) is between the roots of the equation \( a(a + 2) = 0 \), which are \( a = 0 \) and \( a = -2 \). Thus, the solution is:

\[ -2 < a < 0 \]

### Step 7: Combine the conditions
From the first condition, we found \( a < \frac{1}{4} \). The second condition gives us \( -2 < a < 0 \). Therefore, the combined range of \( a \) is:

\[ -2 < a < 0 \]

### Final Answer
The values of \( a \) for which the roots of the equation \( x^2 + x + a = 0 \) are real and exceed \( a \) are:

\[ a \in (-2, 0) \]