If both roots of the equation `x^2+x+a=0` exceeds 'a' then

Options

A. ` 2 lt a lt 3`
B. `a gt 3`
C. `-3 lt a lt 3`
D. ` a lt -2`

Answer

To solve the problem, we need to find the range of 'a' for which both roots of the equation \(x^2 + x + a = 0\) exceed 'a'.

### Step 1: Ensure the roots exist
For the quadratic equation to have real roots, the discriminant must be greater than zero. The discriminant \(D\) is given by:

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

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

\[ D = 1^2 - 4 \cdot 1 \cdot a = 1 - 4a \]

Setting the discriminant greater than zero gives:

\[ 1 - 4a > 0 \]

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

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

This means:

\[ a \in (-\infty, \frac{1}{4}) \]

### Step 3: Ensure both roots exceed 'a'
Next, we need to ensure that both roots of the equation exceed 'a'. The roots of the quadratic equation can be found using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{D}}{2a} \]

Substituting the values we have:

\[ x = \frac{-1 \pm \sqrt{1 - 4a}}{2} \]

Let the roots be \(r_1\) and \(r_2\). For both roots to exceed 'a', we need:

\[ r_1 > a \quad \text{and} \quad r_2 > a \]

Since \(r_1\) and \(r_2\) are given by:

\[ r_1 = \frac{-1 + \sqrt{1 - 4a}}{2} \quad \text{and} \quad r_2 = \frac{-1 - \sqrt{1 - 4a}}{2} \]

We will focus on the root \(r_1\) since it is the larger root. Thus, we need:

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

### Step 4: Solve the inequality for 'a'
Multiplying through by 2 (and reversing the inequality if necessary):

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

Rearranging gives:

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

Squaring both sides (and ensuring both sides are non-negative):

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

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 roots of the equation \(a(a + 2) = 0\) are \(a = 0\) and \(a = -2\). The intervals to test are:

1. \(a < -2\)
2. \(-2 < a < 0\)
3. \(a > 0\)

Testing these intervals, we find that the inequality holds in:

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

### Step 7: Combining the conditions
Now, we combine the two conditions we found:

1. From the discriminant: \(a < \frac{1}{4}\)
2. From the roots exceeding 'a': \(a \in (-2, 0)\)

The intersection of these two conditions is:

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

### Final Result
Thus, the range of 'a' for which both roots of the equation \(x^2 + x + a = 0\) exceed 'a' is:

\[ \boxed{(-2, 0)} \]