Two roots of equation of `2x^(2)-7x+12=0` are `alpha " &" beta` then find `(alpha)/(beta) + (beta)/(alpha)=` ?

Answer

To solve the equation \(2x^2 - 7x + 12 = 0\) and find the value of \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\), where \(\alpha\) and \(\beta\) are the roots of the equation, we can follow these steps:

### Step 1: Identify the coefficients
The given quadratic equation is:
\[ 2x^2 - 7x + 12 = 0 \]
Here, the coefficients are:
- \(a = 2\)
- \(b = -7\)
- \(c = 12\)

### Step 2: Calculate the sum and product of the roots
Using Vieta's formulas:
- The sum of the roots \(\alpha + \beta\) is given by:
\[ \alpha + \beta = -\frac{b}{a} = -\frac{-7}{2} = \frac{7}{2} \]
- The product of the roots \(\alpha \beta\) is given by:
\[ \alpha \beta = \frac{c}{a} = \frac{12}{2} = 6 \]

### Step 3: Express \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\)
We know that:
\[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha \beta} \]
To find \(\alpha^2 + \beta^2\), we can use the identity:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]

### Step 4: Calculate \(\alpha^2 + \beta^2\)
Substituting the values we found:
\[ \alpha^2 + \beta^2 = \left(\frac{7}{2}\right)^2 - 2 \times 6 \]
Calculating \(\left(\frac{7}{2}\right)^2\):
\[ \left(\frac{7}{2}\right)^2 = \frac{49}{4} \]
Calculating \(2 \times 6\):
\[ 2 \times 6 = 12 = \frac{48}{4} \]
Now substituting back:
\[ \alpha^2 + \beta^2 = \frac{49}{4} - \frac{48}{4} = \frac{1}{4} \]

### Step 5: Substitute into the expression
Now we can substitute \(\alpha^2 + \beta^2\) and \(\alpha \beta\) into our expression:
\[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha \beta} = \frac{\frac{1}{4}}{6} = \frac{1}{4} \times \frac{1}{6} = \frac{1}{24} \]

### Final Answer
Thus, the value of \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\) is:
\[ \frac{1}{24} \]