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 \( x^2 + (a-1)x + a = 0 \) and determine the conditions under which both roots are positive.

### Step 1: Identify the conditions for positive roots
For a quadratic equation \( ax^2 + bx + c = 0 \), the roots are given by:
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} \)
- The product of the roots \( \alpha \beta = \frac{c}{a} \)

For the equation \( x^2 + (a-1)x + a = 0 \):
- \( a = 1 \)
- \( b = a - 1 \)
- \( c = a \)

### Step 2: Conditions for the roots to be positive
1. **Product of the roots**: \( \alpha \beta = a > 0 \) (Equation 1)
2. **Sum of the roots**: \( \alpha + \beta = -(a - 1) > 0 \) which simplifies to \( 1 - a > 0 \) or \( a < 1 \) (Equation 2)

### Step 3: Discriminant condition
For the roots to be real and distinct, the discriminant must be positive:
\[ D = b^2 - 4ac = (a-1)^2 - 4(1)(a) > 0 \]
Expanding this:
\[ D = (a^2 - 2a + 1) - 4a = a^2 - 6a + 1 > 0 \]
This is a quadratic inequality.

### Step 4: Solve the discriminant inequality
To find the roots of the quadratic \( a^2 - 6a + 1 = 0 \):
Using the quadratic formula:
\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{6 \pm \sqrt{36 - 4}}{2} = \frac{6 \pm \sqrt{32}}{2} = \frac{6 \pm 4\sqrt{2}}{2} = 3 \pm 2\sqrt{2} \]
The roots are \( 3 + 2\sqrt{2} \) and \( 3 - 2\sqrt{2} \).

### Step 5: Analyze the intervals
The quadratic \( a^2 - 6a + 1 \) opens upwards (since the coefficient of \( a^2 \) is positive), so it is positive outside the interval defined by its roots:
- \( a < 3 - 2\sqrt{2} \) or \( a > 3 + 2\sqrt{2} \)

### Step 6: Combine conditions
Now we need to combine the conditions from Equations 1, 2, and the discriminant:
1. From Equation 1: \( a > 0 \)
2. From Equation 2: \( a < 1 \)
3. From the discriminant: \( a < 3 - 2\sqrt{2} \) or \( a > 3 + 2\sqrt{2} \)

### Step 7: Determine the intersection of intervals
Since \( 3 - 2\sqrt{2} \approx 0.172 \) (approximately), we can conclude:
- The interval \( a > 0 \) and \( a < 1 \) gives \( 0 < a < 1 \).
- The condition \( a < 3 - 2\sqrt{2} \) is satisfied within this interval.

Thus, the complete solution set of real values of \( a \) for which both roots are positive is:
\[ \boxed{(0, 3 - 2\sqrt{2})} \]