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

### 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:

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

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

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

Thus, the roots are:

\[ \alpha = \frac{3 + \sqrt{5}}{2}, \quad \beta = \frac{3 - \sqrt{5}}{2} \]

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

From Vieta's formulas, we know:

\[ \alpha + \beta = 3, \quad \alpha \beta = 1 \]

### Step 3: Find the sum of the new roots \( \frac{1}{\alpha - 2} + \frac{1}{\beta - 2} \)

We can express the sum of the new roots as:

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

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

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

### Step 4: Calculate \( (\alpha - 2)(\beta - 2) \)

Expanding the product:

\[ (\alpha - 2)(\beta - 2) = \alpha \beta - 2(\alpha + \beta) + 4 \]

Substituting \( \alpha \beta = 1 \) and \( \alpha + \beta = 3 \):

\[ = 1 - 2 \cdot 3 + 4 = 1 - 6 + 4 = -1 \]

### Step 5: Find the sum of the new roots

Thus, we have:

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

### Step 6: Find the product of the new roots \( \frac{1}{\alpha - 2} \cdot \frac{1}{\beta - 2} \)

Calculating the product:

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

### Step 7: Form the new quadratic equation

The new quadratic equation can be formed using the sum and product of the roots:

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

Substituting the values we found:

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

Thus, the required equation is:

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