If both the roots of the equation `x^(2)+(a-1) x+a=0` are positive, the the complete solution set of real values of a is

Options

A. `(0, oo)`
B. `(0, 1)`
C. `(0, 3 - 2sqrt2)`
D. `(3-2sqrt2, 1)`

Answer

To solve the problem, we need to analyze the quadratic equation given by:

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

We want to find the values of \( a \) such that both roots of this equation are positive.

### Step 1: Identify the coefficients
In the quadratic equation \( ax^2 + bx + c = 0 \), we have:
- \( a = 1 \)
- \( b = a - 1 \)
- \( c = a \)

### Step 2: Use the properties of roots
Let \( x_1 \) and \( x_2 \) be the roots of the equation. According to Vieta's formulas:
- The sum of the roots \( x_1 + x_2 = -\frac{b}{a} = -\frac{a-1}{1} = 1 - a \)
- The product of the roots \( x_1 x_2 = \frac{c}{a} = \frac{a}{1} = a \)

### Step 3: Set conditions for positive roots
For both roots \( x_1 \) and \( x_2 \) to be positive, we need to satisfy the following conditions:
1. The sum of the roots must be positive:
\[ x_1 + x_2 > 0 \]
This gives us:
\[ 1 - a > 0 \]
Simplifying this, we find:
\[ a < 1 \]

2. The product of the roots must also be positive:
\[ x_1 x_2 > 0 \]
This gives us:
\[ a > 0 \]

### Step 4: Combine the conditions
From the two conditions derived:
- \( a < 1 \)
- \( a > 0 \)

We can combine these inequalities to find the complete solution set for \( a \):
\[ 0 < a < 1 \]

### Final Answer
The complete solution set of real values of \( a \) is:
\[ (0, 1) \]