Quadratic with misread coefficients: For the true equation x^2 + p x + q = 0, one solver (wrong p) found roots 2 and 6; another solver (wrong q) found roots 2 and −9. Determine the correct roots and the correct quadratic.

Introduction / Context:The question leverages Vieta’s formulas: for a monic quadratic x^2 + px + q = 0, the sum of roots is −p and the product is q. Using the information about the two incorrect attempts, we can deduce the true p and q, and hence the correct roots and equation.

Given Data / Assumptions:

• True equation: x^2 + p x + q = 0.
• Wrong-p attempt yielded roots 2 and 6, so that attempt used the same q but incorrect p.
• Wrong-q attempt yielded roots 2 and −9, so that attempt used the same p but incorrect q.

Concept / Approach:From the wrong-p attempt: product of roots is still q, thus q = 2*6 = 12. From the wrong-q attempt: sum of roots is still −p, thus −p = 2 + (−9) = −7 ⇒ p = 7. With p and q known, solve the correct quadratic for actual roots.

Step-by-Step Solution:

Verification / Alternative check:

Why Other Options Are Wrong:

• Other choices list roots/equations that do not match p = 7 and q = 12 determined from the two attempts.

Common Pitfalls:

• Mixing up which parameter (p or q) remains correct in each incorrect attempt.
• Forgetting signs in the sum −p and product q relations.

Final Answer: