Question

If $\alpha, \beta$ are the roots of $x^2-3x+1=0$ then the equation whose roots are $\left(\frac{1}{\alpha-2}, \frac{1}{\beta-2}\right)$ is

• A. $x^2+x-1=0$
• B. $x^2+x+1=0$
• C. $x^2-x-1=0$
• D. None of these

Answer 1

Answer: (C)

x^2-x-1=0

Answer 2

Let's find the quadratic equation whose roots are related to the roots of the given quadratic equation.

Given equation: $x^2 - 3x + 1 = 0$ Let its roots be $\alpha$ and $\beta$.

From Vieta's formulas for the given equation: Sum of roots: $\alpha + \beta = -(\text{coefficient of } x) / (\text{coefficient of } x^2) = -(-3)/1 = 3$ Product of roots: $\alpha \beta = (\text{constant term}) / (\text{coefficient of } x^2) = 1/1 = 1$

The new roots are $y_1 = \frac{1}{\alpha-2}$ and $y_2 = \frac{1}{\beta-2}$. Let the new quadratic equation be $y^2 - (\text{Sum of new roots})y + (\text{Product of new roots}) = 0$.

Method 1: Using Sum and Product of New Roots

1. Calculate the sum of the new roots ($S'$): $S' = y_1 + y_2 = \frac{1}{\alpha-2} + \frac{1}{\beta-2}$ To add these fractions, find a common denominator: $S' = \frac{(\beta-2) + (\alpha-2)}{(\alpha-2)(\beta-2)}$ $S' = \frac{\alpha+\beta-4}{\alpha\beta - 2\alpha - 2\beta + 4}$ $S' = \frac{\alpha+\beta-4}{\alpha\beta - 2(\alpha+\beta) + 4}$

   Now substitute the values $\alpha+\beta=3$ and $\alpha\beta=1$: $S' = \frac{3-4}{1 - 2(3) + 4}$ $S' = \frac{-1}{1 - 6 + 4}$ $S' = \frac{-1}{-1}$ $S' = 1$

2. Calculate the product of the new roots ($P'$): $P' = y_1 y_2 = \left(\frac{1}{\alpha-2}\right) \left(\frac{1}{\beta-2}\right)$ $P' = \frac{1}{(\alpha-2)(\beta-2)}$ $P' = \frac{1}{\alpha\beta - 2\alpha - 2\beta + 4}$ $P' = \frac{1}{\alpha\beta - 2(\alpha+\beta) + 4}$

   Now substitute the values $\alpha+\beta=3$ and $\alpha\beta=1$: $P' = \frac{1}{1 - 2(3) + 4}$ $P' = \frac{1}{1 - 6 + 4}$ $P' = \frac{1}{-1}$ $P' = -1$

3. Form the new quadratic equation: The new equation is $y^2 - S'y + P' = 0$. Substitute $S'=1$ and $P'=-1$: $y^2 - (1)y + (-1) = 0$ $y^2 - y - 1 = 0$

Method 2: Using Transformation of Roots

Let $y$ be a root of the new equation. The relationship between the old root $x$ (which can be $\alpha$ or $\beta$) and the new root $y$ is given by: $y = \frac{1}{x-2}$

We need to express $x$ in terms of $y$: $y(x-2) = 1$ $yx - 2y = 1$ $yx = 1 + 2y$ $x = \frac{1+2y}{y}$

Now substitute this expression for $x$ into the original equation $x^2 - 3x + 1 = 0$: $\left(\frac{1+2y}{y}\right)^2 - 3\left(\frac{1+2y}{y}\right) + 1 = 0$

To eliminate the denominators, multiply the entire equation by $y^2$: $(1+2y)^2 - 3y(1+2y) + 1 \cdot y^2 = 0$

Expand and simplify: $(1 + 4y + 4y^2) - (3y + 6y^2) + y^2 = 0$ $1 + 4y + 4y^2 - 3y - 6y^2 + y^2 = 0$

Combine like terms: $(4y^2 - 6y^2 + y^2) + (4y - 3y) + 1 = 0$ $(5y^2 - 6y^2) + y + 1 = 0$ $-y^2 + y + 1 = 0$

Multiply by $-1$ to make the leading coefficient positive: $y^2 - y - 1 = 0$

Both methods yield the same result.

Therefore, the correct option is (C) $x^2-x-1=0$.