Back out the correct quadratic from two different mistakes: One student mis-copies the coefficient of x and gets roots −9 and −1. Another mis-copies the constant term and gets roots 8 and 2. Find the correct quadratic equation.

Introduction / Context:This problem describes two different erroneous versions of the same intended quadratic. One student changed only the x-coefficient, the other only the constant term. Using the sums and products of the resulting roots, we can reconstruct the correct coefficients that are consistent with both stories.

Given Data / Assumptions:

• Correct equation is monic: x^2 + Bx + C = 0
• Wrong-x-coefficient student obtains roots −9 and −1 ⇒ sum −10, product 9
• Wrong-constant student obtains roots 8 and 2 ⇒ sum 10, product 16
• Each student changes only one coefficient (B or C) relative to the correct equation

Concept / Approach:From the first student: product equals the correct C, because only B changed. So C = 9. From the second student: sum equals the negative of the correct B, because only C changed. Since 8 + 2 = 10, we get −B = 10 ⇒ B = −10. Thus the correct equation is x^2 − 10x + 9 = 0.

Step-by-Step Solution:

Verification / Alternative check:Check each wrong case: With C fixed at 9, choose B′ = +10 to get roots −9 and −1. With B fixed at −10, choose C′ = 16 to get roots 8 and 2. Both match the described mistakes.

Why Other Options Are Wrong:

• x^2 + 10x + 9 and x^2 − 10x + 16 conflict with the deduced B and C.
• None of the above: Not applicable since x^2 − 10x + 9 works.

Common Pitfalls:Mixing up which coefficient stays the same in each erroneous equation. Remember: changing the x-coefficient alters the sum; changing the constant alters the product.

Final Answer:x^2 − 10x + 9 = 0