Given that `alpha and beta` are the roots of the equation `2x^(2)-3x+4=0`, find an equation whose roots are `alpha+(1)/(alpha) and beta+(1)/(beta)`.

Answer

To find an equation whose roots are \( \alpha + \frac{1}{\alpha} \) and \( \beta + \frac{1}{\beta} \), where \( \alpha \) and \( \beta \) are the roots of the equation \( 2x^2 - 3x + 4 = 0 \), we can follow these steps:

### Step 1: Identify the coefficients and roots of the original quadratic equation
The given quadratic equation is:
\[ 2x^2 - 3x + 4 = 0 \]
From this, we can identify:
- \( a = 2 \)
- \( b = -3 \)
- \( c = 4 \)

### Step 2: Calculate the sum and product of the roots
Using Vieta's formulas, we know:
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} = -\frac{-3}{2} = \frac{3}{2} \) (Equation 1)
- The product of the roots \( \alpha \beta = \frac{c}{a} = \frac{4}{2} = 2 \) (Equation 2)

### Step 3: Find the new roots
We need to find the new roots:
\[ \alpha + \frac{1}{\alpha} \quad \text{and} \quad \beta + \frac{1}{\beta} \]

#### Step 3a: Calculate the sum of the new roots
The sum of the new roots can be calculated as follows:
\[ \left( \alpha + \frac{1}{\alpha} \right) + \left( \beta + \frac{1}{\beta} \right) = (\alpha + \beta) + \left( \frac{1}{\alpha} + \frac{1}{\beta} \right) \]
Using the identity \( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} \):
\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\frac{3}{2}}{2} = \frac{3}{4} \]
Thus, the sum of the new roots becomes:
\[ \frac{3}{2} + \frac{3}{4} = \frac{6}{4} + \frac{3}{4} = \frac{9}{4} \]

#### Step 3b: Calculate the product of the new roots
The product of the new roots is:
\[ \left( \alpha + \frac{1}{\alpha} \right) \left( \beta + \frac{1}{\beta} \right) = \alpha \beta + \frac{1}{\alpha \beta} + \frac{\alpha}{\beta} + \frac{\beta}{\alpha} \]
Using the identity \( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha \beta} \):
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta = \left( \frac{3}{2} \right)^2 - 2 \cdot 2 = \frac{9}{4} - 4 = \frac{9}{4} - \frac{16}{4} = -\frac{7}{4} \]
Now substituting back:
\[ \alpha \beta + \frac{1}{\alpha \beta} + \frac{\alpha^2 + \beta^2}{\alpha \beta} = 2 + \frac{1}{2} + \frac{-\frac{7}{4}}{2} = 2 + \frac{1}{2} - \frac{7}{8} \]
Converting to a common denominator:
\[ 2 = \frac{16}{8}, \quad \frac{1}{2} = \frac{4}{8} \quad \Rightarrow \quad \frac{16}{8} + \frac{4}{8} - \frac{7}{8} = \frac{13}{8} \]

### Step 4: Form the new quadratic equation
Now we can form the new quadratic equation using the sum and product of the new roots:
\[ x^2 - \left( \text{sum of roots} \right)x + \text{product of roots} = 0 \]
Substituting the values:
\[ x^2 - \frac{9}{4}x + \frac{13}{8} = 0 \]
To eliminate the fractions, multiply the entire equation by 8:
\[ 8x^2 - 18x + 13 = 0 \]

### Final Answer
The equation whose roots are \( \alpha + \frac{1}{\alpha} \) and \( \beta + \frac{1}{\beta} \) is:
\[ 8x^2 - 18x + 13 = 0 \]