If `a gt 1` , then the roots of the equation `(1-a)x^(2)+3ax-1=0` are

Options

A. one positive and one negative
B. both negative
C. both positive
D. both non real complex

Explanation

To solve the equation \( (1-a)x^2 + 3ax - 1 = 0 \) under the condition \( a > 1 \), we will analyze the roots using the properties of quadratic equations.

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

### Step 2: Calculate the sum of the roots
The sum of the roots \( S \) of a quadratic equation is given by the formula:
\[ S = -\frac{B}{A} = -\frac{3a}{1-a} \]
Since \( a > 1 \), we have \( 1 - a < 0 \). Therefore, the sum of the roots becomes:
\[ S = \frac{3a}{a - 1} \]
Since \( a > 1 \), both \( 3a > 0 \) and \( a - 1 > 0 \), hence \( S > 0 \).

### Step 3: Calculate the product of the roots
The product of the roots \( P \) is given by:
\[ P = \frac{C}{A} = \frac{-1}{1-a} = \frac{1}{a-1} \]
Since \( a > 1 \), we have \( a - 1 > 0 \), thus \( P > 0 \).

### Step 4: Calculate the discriminant
To determine the nature of the roots, we calculate the discriminant \( D \):
\[ D = B^2 - 4AC = (3a)^2 - 4(1-a)(-1) \]
Calculating this gives:
\[ D = 9a^2 + 4(1-a) = 9a^2 + 4 - 4a = 9a^2 - 4a + 4 \]
This is a quadratic in \( a \). The discriminant of this quadratic is:
\[ D' = (-4)^2 - 4(9)(4) = 16 - 144 = -128 \]
Since the discriminant \( D' < 0 \), the quadratic \( 9a^2 - 4a + 4 \) is always positive for all \( a > 1 \). Thus, \( D > 0 \).

### Step 5: Conclusion about the roots
Since the sum of the roots is positive, the product of the roots is positive, and the discriminant is positive, we conclude that both roots are real and positive.

### Final Answer
The roots of the equation \( (1-a)x^2 + 3ax - 1 = 0 \) are **both positive**.

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