If `alpha, beta` are the roots of `x^(2)-3x+1=0`, then the equation whose roots are `(1/(alpha-2),1/(beta-2))` is

Options

A. `x^(2)+x-1=0`
B. `x^(2)+x+1=0`
C. `x^(2)-x-1=0`
D. none of these

Answer

To solve the problem, we need to find the equation whose roots are given as \( \frac{1}{\alpha - 2} \) and \( \frac{1}{\beta - 2} \), where \( \alpha \) and \( \beta \) are the roots of the equation \( x^2 - 3x + 1 = 0 \).

### Step 1: Find the roots \( \alpha \) and \( \beta \)

The roots of the quadratic equation \( x^2 - 3x + 1 = 0 \) can be found using the quadratic formula:

\[ \alpha, \beta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 1, b = -3, c = 1 \).

\[ \alpha, \beta = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2} \]

### Step 2: Calculate \( \alpha + \beta \) and \( \alpha \beta \)

From Vieta's formulas, we know:

- The sum of the roots \( \alpha + \beta = 3 \)
- The product of the roots \( \alpha \beta = 1 \)

### Step 3: Find the sum of the new roots

The new roots are \( \frac{1}{\alpha - 2} \) and \( \frac{1}{\beta - 2} \).

To find the sum of these roots:

\[ \frac{1}{\alpha - 2} + \frac{1}{\beta - 2} = \frac{(\beta - 2) + (\alpha - 2)}{(\alpha - 2)(\beta - 2)} = \frac{\alpha + \beta - 4}{(\alpha - 2)(\beta - 2)} \]

Substituting \( \alpha + \beta = 3 \):

\[ \frac{3 - 4}{(\alpha - 2)(\beta - 2)} = \frac{-1}{(\alpha - 2)(\beta - 2)} \]

Next, we calculate \( (\alpha - 2)(\beta - 2) \):

\[ (\alpha - 2)(\beta - 2) = \alpha \beta - 2(\alpha + \beta) + 4 = 1 - 2 \cdot 3 + 4 = 1 - 6 + 4 = -1 \]

Thus, the sum of the new roots is:

\[ \frac{-1}{-1} = 1 \]

### Step 4: Find the product of the new roots

Now we calculate the product of the new roots:

\[ \frac{1}{(\alpha - 2)(\beta - 2)} = \frac{1}{-1} = -1 \]

### Step 5: Form the new quadratic equation

Using the sum and product of the new roots, we can form the quadratic equation:

\[ x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0 \]

Substituting the values we found:

\[ x^2 - 1x - 1 = 0 \]

Thus, the required equation is:

\[ \boxed{x^2 - x - 1 = 0} \]