Build a quadratic with reciprocal roots: Given α, β are roots of x^2 − 5x + 6 = 0, construct the quadratic whose roots are 1/α and 1/β.

Introduction / Context:Transforming roots to their reciprocals is a standard maneuver. If r is a root of a monic quadratic, then 1/r is a root of another quadratic whose coefficients can be found via Vieta’s formulas by inverting the sum and product appropriately.

Given Data / Assumptions:

• Original: x^2 − 5x + 6 = 0 ⇒ α + β = 5, αβ = 6
• Target roots: 1/α and 1/β

Concept / Approach:If new roots are r′1 = 1/α and r′2 = 1/β, then their sum is (α + β)/(αβ) and their product is 1/(αβ). Construct the monic quadratic z^2 − (sum) z + (product) = 0, then clear denominators for a clean integer-coefficient equation.

Step-by-Step Solution:

Verification / Alternative check:Factor 6z^2 − 5z + 1 = (3z − 1)(2z − 1) ⇒ roots z = 1/3, 1/2 which are indeed reciprocals of α, β = 3, 2.

Why Other Options Are Wrong:

• Sign errors in the linear or constant term break the sum/product (5/6 and 1/6) required for reciprocal roots.

Common Pitfalls:Mixing up sum/product after inversion. Remember: sum becomes (S/P) and product becomes 1/P for monic originals.

Final Answer:6x^2 − 5x + 1 = 0