Let `alpha and beta` be the roots of the quadratic equation `ax^(2)+bx+c=0, c ne 0,` then form the quadratic equation whose roots are `(1-alpha)/(alpha) and (1-beta)/(beta)`.

Answer

To find the quadratic equation whose roots are \(\frac{1 - \alpha}{\alpha}\) and \(\frac{1 - \beta}{\beta}\), we will follow these steps:

### Step 1: Identify the roots and their relationships
Given the quadratic equation \(ax^2 + bx + c = 0\), the roots \(\alpha\) and \(\beta\) satisfy:
- Sum of the roots: \(\alpha + \beta = -\frac{b}{a}\)
- Product of the roots: \(\alpha \beta = \frac{c}{a}\)

### Step 2: Find the new roots
The new roots are:
\[ r_1 = \frac{1 - \alpha}{\alpha} \quad \text{and} \quad r_2 = \frac{1 - \beta}{\beta} \]

### Step 3: Calculate the sum of the new roots
The sum of the new roots \(r_1 + r_2\) can be calculated as follows:
\[ r_1 + r_2 = \frac{1 - \alpha}{\alpha} + \frac{1 - \beta}{\beta} \]
Combining the fractions:
\[ = \left(\frac{1}{\alpha} - 1\right) + \left(\frac{1}{\beta} - 1\right) = \left(\frac{1}{\alpha} + \frac{1}{\beta}\right) - 2 \]
Using the relationship for the sum of roots:
\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} = \frac{-\frac{b}{a}}{\frac{c}{a}} = -\frac{b}{c} \]
Thus, we have:
\[ r_1 + r_2 = -\frac{b}{c} - 2 = -\frac{b + 2c}{c} \]

### Step 4: Calculate the product of the new roots
The product of the new roots \(r_1 \cdot r_2\) is given by:
\[ r_1 \cdot r_2 = \left(\frac{1 - \alpha}{\alpha}\right) \cdot \left(\frac{1 - \beta}{\beta}\right) = \frac{(1 - \alpha)(1 - \beta)}{\alpha \beta} \]
Expanding the numerator:
\[ = \frac{1 - (\alpha + \beta) + \alpha \beta}{\alpha \beta} = \frac{1 + \frac{b}{a} + \frac{c}{a}}{\frac{c}{a}} = \frac{a + b + c}{c} \]

### Step 5: Form the new quadratic equation
Using the sum and product of the new roots, we can form the quadratic equation:
\[ x^2 - (r_1 + r_2)x + (r_1 \cdot r_2) = 0 \]
Substituting the values:
\[ x^2 + \frac{b + 2c}{c}x + \frac{a + b + c}{c} = 0 \]
Multiplying through by \(c\) to eliminate the fraction:
\[ c x^2 + (b + 2c)x + (a + b + c) = 0 \]

### Final Answer
The required quadratic equation is:
\[ c x^2 + (b + 2c)x + (a + b + c) = 0 \]