Question:
If \( \alpha, \beta \) are the roots of \( x^2 - 3x + 1 = 0 \), then the equation whose roots are \( \frac{1}{\alpha - 2} \) and \( \frac{1}{\beta - 2} \) is:
If \( \alpha, \beta \) are the roots of \( x^2 - 3x + 1 = 0 \), then the equation whose roots are \( \frac{1}{\alpha - 2} \) and \( \frac{1}{\beta - 2} \) is:
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For quadratic equations with roots, use Vieta's formulas to find the sum and product of the roots, and then apply the transformations as needed.
Updated On: Apr 22, 2026
- \( x^2 + y^2 + 4x - 10y + 25 = 0 \)
- \( x^2 - x - 1 = 0 \)
- \( x^2 + x - 1 = 0 \)
- None of these
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The Correct Option is B
Solution and Explanation
Step 1: Find the sum and product of the roots of the original equation.
We are given the quadratic equation: \[ x^2 - 3x + 1 = 0 \] From Vietaâs formulas, the sum and product of the roots \( \alpha \) and \( \beta \) are: \[ \alpha + \beta = 3, \quad \alpha \beta = 1 \]
Step 2: Apply the transformation to find the new equation.
The new roots are \( \frac{1}{\alpha - 2} \) and \( \frac{1}{\beta - 2} \). The sum and product of the new roots are given by: \[ \frac{1}{\alpha - 2} + \frac{1}{\beta - 2} = \frac{(\beta - 2) + (\alpha - 2)}{(\alpha - 2)(\beta - 2)} = \frac{\alpha + \beta - 4}{(\alpha - 2)(\beta - 2)} \] Substitute \( \alpha + \beta = 3 \): \[ \frac{3 - 4}{(\alpha - 2)(\beta - 2)} = \frac{-1}{(\alpha - 2)(\beta - 2)} \] Now, expand \( (\alpha - 2)(\beta - 2) \): \[ (\alpha - 2)(\beta - 2) = \alpha \beta - 2(\alpha + \beta) + 4 = 1 - 6 + 4 = -1 \] Thus, the sum of the new roots is: \[ \frac{-1}{-1} = 1 \]
Step 3: Find the product of the new roots.
The product of the new roots is: \[ \frac{1}{(\alpha - 2)(\beta - 2)} = \frac{1}{-1} = -1 \]
Step 4: Form the quadratic equation with the new roots.
The quadratic equation whose roots are the new roots is: \[ x^2 - (\text{sum of roots}) x + \text{product of roots} = x^2 - 1x - 1 = 0 \]
Step 5: Conclusion.
Thus, the required equation is \( x^2 - x - 1 = 0 \), which corresponds to option (B).
We are given the quadratic equation: \[ x^2 - 3x + 1 = 0 \] From Vietaâs formulas, the sum and product of the roots \( \alpha \) and \( \beta \) are: \[ \alpha + \beta = 3, \quad \alpha \beta = 1 \]
Step 2: Apply the transformation to find the new equation.
The new roots are \( \frac{1}{\alpha - 2} \) and \( \frac{1}{\beta - 2} \). The sum and product of the new roots are given by: \[ \frac{1}{\alpha - 2} + \frac{1}{\beta - 2} = \frac{(\beta - 2) + (\alpha - 2)}{(\alpha - 2)(\beta - 2)} = \frac{\alpha + \beta - 4}{(\alpha - 2)(\beta - 2)} \] Substitute \( \alpha + \beta = 3 \): \[ \frac{3 - 4}{(\alpha - 2)(\beta - 2)} = \frac{-1}{(\alpha - 2)(\beta - 2)} \] Now, expand \( (\alpha - 2)(\beta - 2) \): \[ (\alpha - 2)(\beta - 2) = \alpha \beta - 2(\alpha + \beta) + 4 = 1 - 6 + 4 = -1 \] Thus, the sum of the new roots is: \[ \frac{-1}{-1} = 1 \]
Step 3: Find the product of the new roots.
The product of the new roots is: \[ \frac{1}{(\alpha - 2)(\beta - 2)} = \frac{1}{-1} = -1 \]
Step 4: Form the quadratic equation with the new roots.
The quadratic equation whose roots are the new roots is: \[ x^2 - (\text{sum of roots}) x + \text{product of roots} = x^2 - 1x - 1 = 0 \]
Step 5: Conclusion.
Thus, the required equation is \( x^2 - x - 1 = 0 \), which corresponds to option (B).
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