If the roots of equation `(a + 1)x^2-3ax + 4a = 0` (a is not equals to -1) are greater than unity, then

Explanation

To solve the problem, we need to analyze the quadratic equation given by \((a + 1)x^2 - 3ax + 4a = 0\) under the condition that its roots are greater than unity. We will break down the solution into steps.

### Step 1: Identify the coefficients
The quadratic equation can be expressed in the standard form \(Ax^2 + Bx + C = 0\), where:
- \(A = a + 1\)
- \(B = -3a\)
- \(C = 4a\)

### Step 2: Condition for roots to be greater than unity
For the roots of the quadratic equation to be greater than 1, we can use the following conditions:
1. The quadratic must open upwards, which requires \(A > 0\).
2. The function evaluated at \(x = 1\) must be greater than 0, i.e., \(f(1) > 0\).
3. The vertex of the parabola must be to the left of \(x = 1\).

### Step 3: Condition for \(A > 0\)
Since \(A = a + 1\), we require:
\[ a + 1 > 0 \implies a > -1 \]

### Step 4: Evaluate \(f(1)\)
Now we evaluate \(f(1)\):
\[ f(1) = (a + 1)(1^2) - 3a(1) + 4a = a + 1 - 3a + 4a = 2a + 1 \]
For the roots to be greater than 1, we need:
\[ 2a + 1 > 0 \implies 2a > -1 \implies a > -\frac{1}{2} \]

### Step 5: Condition for the vertex
The x-coordinate of the vertex of the parabola is given by:
\[ x_v = -\frac{B}{2A} = -\frac{-3a}{2(a + 1)} = \frac{3a}{2(a + 1)} \]
We require this vertex to be less than 1:
\[ \frac{3a}{2(a + 1)} < 1 \]
Multiplying both sides by \(2(a + 1)\) (which is positive since \(a > -1\)):
\[ 3a < 2(a + 1) \implies 3a < 2a + 2 \implies a < 2 \]

### Step 6: Combine the conditions
Now we have two conditions:
1. \(a > -\frac{1}{2}\)
2. \(a < 2\)

Thus, combining these gives us:
\[ -\frac{1}{2} < a < 2 \]

### Final Answer
The values of \(a\) for which the roots of the equation \((a + 1)x^2 - 3ax + 4a = 0\) are greater than unity are:
\[ a \in \left(-\frac{1}{2}, 2\right) \]