Find the quadratic whose roots are reciprocals of the roots of 3x^2 − 20x + 17 = 0.

Introduction / Context:If α and β are roots of a quadratic, then 1/α and 1/β are roots of another quadratic obtained by the substitution x → 1/x and clearing denominators. This is a standard transformation for reciprocal roots.

Given Data / Assumptions:

• Original quadratic: 3x^2 − 20x + 17 = 0
• We need the equation with roots 1/α and 1/β

Concept / Approach:Replace x by 1/x: 3*(1/x^2) − 20*(1/x) + 17 = 0. Multiply both sides by x^2 to clear denominators and obtain a monic-in-form quadratic in x (up to a constant factor).

Step-by-Step Solution:Start: 3/x^2 − 20/x + 17 = 0Multiply by x^2: 3 − 20x + 17x^2 = 0Reorder: 17x^2 − 20x + 3 = 0

Verification / Alternative check:By Vieta: For original, sum = 20/3 and product = 17/3. For reciprocal equation, sum = (α+β)/(αβ) = (20/3)/(17/3) = 20/17, product = 1/(αβ) = 3/17, matching coefficients of 17x^2 − 20x + 3 = 0 after normalization.

Why Other Options Are Wrong:Signs or constants are incorrect for a reciprocal transformation; only option (a) matches the derived equation.

Common Pitfalls:Forgetting to multiply through by x^2 or mishandling the order/signs when rewriting.

Final Answer:17x^2 − 20x + 3 = 0