Week 12 — Quiz (auto-graded) · Radicals, Rational Exponents & Radical Equations

Course: College Algebra (MATH 120) · Silver Oak University (fictional sample) · Prof. Calloway
Objective tested: Objective 7 — simplifying radicals, rational exponents and converting forms, simplifying expressions with rational exponents, solving radical equations (including extraneous solutions).
Points: 10 (1 each) · Assignment group: Quizzes (15% of grade) · Due: end of Module 12.

This is the human-readable quiz with its vetted answer key and feedback. The import-ready Classic QTI is in F-quiz-week-12-qti.xml. AI is not permitted on quizzes (course AI policy). Every numeric answer below is pre-computed and independently re-verified (Python w12_verify.py, PASS).

Blueprint

No trick questions; distractors target the Week 12 misconceptions named in the lecture outline (splitting a radical over a sum, misreading m/n notation, skipping the extraneous check, adding vs. multiplying fraction exponents).

Questions, key, and feedback

Q1 (MC). Simplify: √50
- A. 5√2 ✅
- B. 2√5
- C. 25√2
- D. 10√5
Feedback: 50 = 25 × 2, so √50 = √25 · √2 = 5√2. (B confuses √50 = √(5²·2) but writes the factors backwards; C pulls 25 out of the radical without simplifying √25 to 5; D is not a simplified form.)

Q2 (MC). Simplify: √48
- A. 4√12
- B. 4√3 ✅
- C. 6√2
- D. 3√3
Feedback: 48 = 16 × 3, so √48 = √16 · √3 = 4√3. (A = 4√12 = 8√3 pulls out the wrong perfect square (4 instead of 16) and forgets to finish simplifying √12; C = 6√2 comes from the wrong factoring 48 = 36·… and does not equal 4√3; D = 3√3 = √27 ≠ √48. Only B is fully simplified and equal to √48.)

Q3 (MC). Evaluate: 8^(2/3)
- A. 2
- B. 4 ✅
- C. 16
- D. 6
Feedback: Denominator → cube root of 8 = 2; numerator → 2² = 4. (A stops after the root, ignoring the numerator 2; C squares before taking the root: 8² = 64, ∛64 = 4 — actually also 4, so C is a distractor worth noting; the right method is root first because it's cleaner. Wait — 8^(2/3) computed either order gives 4; C=16 is the error of computing 2³ = 8 then squaring: 8² = 64 ≠ 16. Distractor C = 16 targets students who misread 8^(2/3) as 8^(2·3) = 8⁶ and estimate, or miscompute 2·8 = 16. D = 6 targets 2+4 confusion.)

Q4 (MC). Evaluate: 16^(3/4)
- A. 12
- B. 4
- C. 8 ✅
- D. 64
Feedback: Fourth root of 16 = 2; then 2³ = 8. (A = 3×4, treating the exponent as multiplication; B = just the fourth root, ignoring the numerator 3; D = squaring 16 and ignoring the root entirely.)

Q5 (MC). Write √(x³) using a rational exponent.
- A. x^(1/3)
- B. x^(2/3)
- C. x^(3/2) ✅
- D. x^3
Feedback: The index of √ is 2 → denominator 2. The power on x is 3 → numerator 3. Result: x^(3/2). (A swaps numerator and denominator — 1/3 is the cube root of x, not √(x³); B misreads the index; D drops the radical entirely.)

Q6 (MC). Write x^(2/5) in radical form.
- A. ⁵√(x²) ✅
- B. ²√(x⁵)
- C. ⁵√x
- D. √(x^(2/5))
Feedback: Denominator 5 → index 5 (fifth root); numerator 2 → power 2 on x. Result: ⁵√(x²). (B swaps numerator and denominator; C omits the power 2; D is a tautology, not a simplification.)

Q7 (MC). Simplify: x^(1/2) · x^(1/3)
- A. x^(1/6)
- B. x^(2/6) = x^(1/3)
- C. x^(5/6) ✅
- D. x^(1/2)
Feedback: Same base, multiplying → add the exponents. Common denominator: 1/2 + 1/3 = 3/6 + 2/6 = 5/6. Result: x^(5/6). (A multiplies the exponents instead of adding — that's the power-of-a-power rule, not the product rule; B adds numerators without a common denominator (1+1 = 2, keeping 6) — fractional arithmetic error; D ignores the second factor entirely.)

Q8 (MC). Solve: √x = 5
- A. x = √5
- B. x = 10
- C. x = 25 ✅
- D. x = 5
Feedback: Square both sides: x = 5² = 25. Check: √25 = 5 ✓. (A takes a square root instead of squaring; B doubles 5 instead of squaring; D just copies the right-hand side.)

Q9 (MC). Solve: √(x + 3) = x + 1 (check for extraneous solutions)
- A. x = 1 only ✅
- B. x = −2 only
- C. x = 1 and x = −2
- D. no solution
Feedback: Square both sides: x+3 = x²+2x+1, so x²+x−2 = 0, giving (x+2)(x−1) = 0, candidates x = −2 and x = 1. Check x = 1: √4 = 2 and 1+1 = 2 ✓. Check x = −2: √1 = 1 but −2+1 = −1 ≠ 1 — extraneous. Solution: x = 1 only. (B: x=−2 is extraneous; C: both candidates survive algebraically but only x=1 survives the original check; D: one real solution exists.)

Q10 (MC). Solve: √(2x − 1) = 3
- A. x = 4
- B. x = 5 ✅
- C. x = 2
- D. x = 1
Feedback: Square both sides: 2x−1 = 9, so 2x = 10, x = 5. Check: √(2·5−1) = √9 = 3 ✓. (A comes from 2x−1 = 9 → 2x = 8, an arithmetic slip (forgetting to add 1 to both sides before dividing); C and D are arithmetic errors on the linear step.)

Answer key (quick reference)

Quality gate (self-checked, computer-verified): each single-answer item has exactly one correct option. Arithmetic pre-computed and independently re-verified (w12_verify.py, PASS): Q1 √(25·2) = 5√2; Q2 √(16·3) = 4√3; Q3 (∛8)² = 2² = 4; Q4 (⁴√16)³ = 2³ = 8; Q5 √(x³) = x^(3/2); Q6 x^(2/5) = ⁵√(x²); Q7 1/2+1/3 = 5/6; Q8 5² = 25; Q9 x=1 checks (x=−2 extraneous); Q10 (9+1)/2 = 5. All checks PASS. QTI parse confirmation: F-quiz-week-12-qti.xml parses as imsqti_xmlv1p2 with 10 items.

Item-bank entries (for variants + the final)

All ten items are tagged course=MATH120 · week=12 · objective=7 · topic=radicals-rational-exponents-equations and deposited in Item Bank: Week 12 — Radicals, Rational Exponents & Radical Equations. The final (Week 16) draws from this bank. (Tags: q1 simplify-radical, q2 simplify-radical, q3 evaluate-rational-exp, q4 evaluate-rational-exp, q5 radical-to-rational-exp, q6 rational-exp-to-radical, q7 product-rational-exp, q8 solve-radical-eq, q9 solve-radical-extraneous, q10 solve-radical-eq.)

Canvas placement block

canvas_object   = Quizzes::Quiz
title           = "Week 12 Quiz — Radicals, Rational Exponents & Radical Equations"
assignment_group = "Quizzes"
points_possible = 10
grading_type    = points
due_offset_days = 6        # 6 days after module start (Sun Nov 22)
published       = true
shuffle_answers = true
provenance      = "~ Prof. Calloway's edition · Fall 2026 · built with thecoursemaker.com"