The roots of the equation `ax^(2)+bx+c=0` are `alpha and beta`. Form the quadratic equation whose roots are `alpha+(1)/(beta) and beta+(1)/(alpha)`.

Answer

To solve the problem, we need to find the quadratic equation whose roots are \( \alpha + \frac{1}{\beta} \) and \( \beta + \frac{1}{\alpha} \), given that the roots \( \alpha \) and \( \beta \) are from the equation \( ax^2 + bx + c = 0 \).

### Step 1: Find the sum of the new roots

The sum of the new roots can be calculated as follows:

\[ \text{Sum} = \left( \alpha + \frac{1}{\beta} \right) + \left( \beta + \frac{1}{\alpha} \right) = (\alpha + \beta) + \left( \frac{1}{\beta} + \frac{1}{\alpha} \right) \]

Using the formulas for the sum and product of the roots of the original equation, we know:

\[ \alpha + \beta = -\frac{b}{a} \]
\[ \frac{1}{\beta} + \frac{1}{\alpha} = \frac{\alpha + \beta}{\alpha \beta} = \frac{-\frac{b}{a}}{\frac{c}{a}} = -\frac{b}{c} \]

Thus, the sum of the new roots becomes:

\[ \text{Sum} = -\frac{b}{a} - \frac{b}{c} \]

To combine these fractions, we find a common denominator:

\[ \text{Sum} = -\frac{bc + ab}{ac} = -\frac{b(a + c)}{ac} \]

### Step 2: Find the product of the new roots

Now, we calculate the product of the new roots:

\[ \text{Product} = \left( \alpha + \frac{1}{\beta} \right) \left( \beta + \frac{1}{\alpha} \right) \]

Expanding this product gives:

\[ \text{Product} = \alpha\beta + \alpha \cdot \frac{1}{\alpha} + \beta \cdot \frac{1}{\beta} + \frac{1}{\alpha\beta} \]

This simplifies to:

\[ \text{Product} = \alpha\beta + 1 + 1 + \frac{1}{\alpha\beta} = \alpha\beta + 2 + \frac{1}{\alpha\beta} \]

Substituting the values of \( \alpha\beta \):

\[ \text{Product} = \frac{c}{a} + 2 + \frac{a}{c} \]

To combine these terms, we find a common denominator:

\[ \text{Product} = \frac{c^2 + 2ac + a^2}{ac} \]

### Step 3: Form the quadratic equation

Now that we have the sum and product of the new roots, we can form the quadratic equation. The general form of a quadratic equation with roots \( p \) and \( q \) is:

\[ x^2 - (p + q)x + pq = 0 \]

Substituting the sum and product we found:

\[ x^2 - \left(-\frac{b(a+c)}{ac}\right)x + \frac{c^2 + 2ac + a^2}{ac} = 0 \]

This simplifies to:

\[ x^2 + \frac{b(a+c)}{ac} x + \frac{c^2 + 2ac + a^2}{ac} = 0 \]

Multiplying through by \( ac \) to eliminate the fractions gives:

\[ acx^2 + b(a+c)x + (c^2 + 2ac + a^2) = 0 \]

### Final Answer

Thus, the quadratic equation whose roots are \( \alpha + \frac{1}{\beta} \) and \( \beta + \frac{1}{\alpha} \) is:

\[ acx^2 + b(a+c)x + (c^2 + 2ac + a^2) = 0 \]