Consecutive Integer Quadratics A focused 10th-grade algebra lesson on modeling real-world and algebraic relationships using quadratic equations with consecutive odd and even integers. Features structured reference materials and an in-depth teacher solutions key.
AR Angie Rice
Consecutive Quadratics Practice Guide
A comprehensive two-page reference and worked-example guide for 10th-grade algebra covering quadratic applications with consecutive odd and even integers. It presents five detailed, step-by-step examples spanning product relations, sum of squares, geometric area modeling, multi-solution cases, and Pythagorean theorem setups.
Consecutive Quadratics Answer Key
A comprehensive two-page teacher answer key and scoring guide for the five consecutive integer quadratic applications. Includes detailed algebraic solutions, standard forms, factorizations, extraneous root justifications, a 4-point scoring rubric per problem, and a teacher misconception diagnostic chart.
Consecutive Quadratics Worksheet
A five-page student worksheet containing ten comprehensive quadratic application word problems involving consecutive odd and even integers. Features expanded, extra-large partitioned blank work areas for defining variables, setting up standard-form equations, factoring, and contextual checks.
Consecutive Quadratics Teacher Guide
A five-page comprehensive teacher guide and solutions manual matching the ten-problem student worksheet. Provides detailed step-by-step algebraic solutions, 4-point grading rubrics for all 10 problems, pedagogical notes, error diagnostics, and differentiation strategies.
Consecutive Integer Quadratics A focused 10th-grade algebra lesson on modeling real-world and algebraic relationships using quadratic equations with consecutive odd and even integers. Features structured reference materials and an in-depth teacher solutions key.
AR Angie Rice
Consecutive Quadratics Practice Guide Algebra 1 • Grade 10 Quadratic Word Problems
Consecutive Integer Quadratic Applications
Reference & Worked Examples
Page 1 of 2
Core Setup Reference: Defining Consecutive Integer Variables
Standard Consecutive
\(n, \; n+1, \; n+2, \dots\)
Increments by 1 (any integer)
Consecutive Even Integers
\(n, \; n+2, \; n+4, \dots\)
Increments by 2 (where \(n\) is even)
Consecutive Odd Integers
\(n, \; n+2, \; n+4, \dots\)
Increments by 2 (where \(n\) is odd!)
Key Rule: Both consecutive odd and consecutive even integers increase by +2 because odd numbers are separated by an even number (and vice versa).
1
Product of Two Consecutive Even Integers
Consecutive Even
Problem: The product of two consecutive positive even integers is 168. Find the two integers.
Step 1: Define Variables & Form Equation
Let the first positive even integer \(= n\).
\(n(n + 2) = 168\)
Step 2: Rewrite in Standard Form
\(n^2 + 2n - 168 = 0\)
Step 3: Factor & Solve
Find factors of \(-168\) that add up to \(+2\): \((+14) \times (-12) = -168\).
\((n + 14)(n - 12) = 0\)
Step 4: Interpret & Check Context
Since the integers must be positive , reject \(n = -14\).
Check: \(12 \times 14 = 168\) ✓ Answer: 12 and 14
2
Sum of Squares of Consecutive Odd Integers
Consecutive Odd
Problem: Find two consecutive positive odd integers such that the sum of their squares is 202.
Step 1: Set Up Algebraic Equation
Let first odd integer \(= n\), second \(= n + 2\).
Step 2: Expand & Simplify
\(n^2 + (n^2 + 4n + 4) = 202\)
Step 3: Divide by GCF & Factor
Divide the entire equation by \(2\):
\(n^2 + 2n - 99 = 0\)
Step 4: Conclude & Verify
Reject \(-11\) because problem specifies positive .
Check: \(9^2 + 11^2 = 81 + 121 = 202\) ✓ Answer: 9 and 11
Watch the sign: Always verify whether problems require positive integers or allow negative integers! Continued on Page 2 →
Applied Consecutive Integer Modeling
Geometry, Multi-term Relations, and Pythagorean Applications
Part II
Page 2 of 2
3
Geometric Area Problem (Consecutive Even)
Geometry Connection
Problem: A rectangular community garden plot has a width and length that are consecutive even integers in meters. If the area is \(224\text{ m}^2\), find the dimensions and the perimeter.
Model & Equation:
Let width \(= w\text{ m}\), length \(= w + 2\text{ m}\).
\(w(w + 2) = 224 \implies w^2 + 2w - 224 = 0\)
Factor pairs of \(-224\) with difference of \(2\): \(16 \times 14 = 224\).
Solve & Calculate Perimeter:
\((w + 16)(w - 14) = 0 \implies w = 14 \text{ (reject } -16\text{)}\)
Dimensions: Width \(= \mathbf{14\text{ m}}\), Length \(= 14 + 2 = \mathbf{16\text{ m}}\).
Check: \(14 \times 16 = 224\text{ m}^2\) ✓
4
Three Consecutive Odd Integers (Dual Solution Sets)
Both Signs Valid
Problem: Find three consecutive odd integers such that the product of the first two is \(19\) more than \(4\) times the third integer.
Setup & Linear-Quadratic Balance:
Let the three integers be \(n, \; n+2, \; n+4\).
\(n^2 + 2n = 4n + 16 + 19\)
Factoring & Dual Interpretation:
\((n - 7)(n + 5) = 0 \implies n = 7 \quad \text{or} \quad n = -5\)
Set 1 (\(n=7\)): \(7, 9, 11\) (\(7 \times 9 = 63\); \(4(11)+19 = 63\) ✓)
Set 2 (\(n=-5\)): \(-5, -3, -1\) (\((-5)(-3) = 15\); \(4(-1)+19 = 15\) ✓)
5
Right Triangle Sides (Consecutive Even)
Pythagorean Theorem
Problem: The side lengths of a right triangle are three consecutive positive even integers in centimeters. Find the lengths of the two legs and the hypotenuse.
Geometric Setup (\(a^2 + b^2 = c^2\)):
Let legs \(= n\) and \(n + 2\); Hypotenuse (longest) \(= n + 4\).
\(n^2 + (n + 2)^2 = (n + 4)^2\)
Solve & State Dimensions:
\((n - 6)(n + 2) = 0 \implies n = 6 \text{ (reject } -2\text{)}\)
Leg 1: \(\mathbf{6\text{ cm}}\) • Leg 2: \(6 + 2 = \mathbf{8\text{ cm}}\)
Hypotenuse: \(6 + 4 = \mathbf{10\text{ cm}}\)
Check: \(6^2 + 8^2 = 36 + 64 = 100 = 10^2\) ✓
10th Grade Pitfall Checklist (Avoid Common Mistakes!)
1. Odd Does NOT Mean +1 or +3 Odd integers increase by 2: if \(n=5\), the next odd is \(5+2=7\). Never write \(n+1\) for consecutive odds!
2. Don't Forget FOIL \((n+2)^2 = n^2 + 4n + 4\), not \(n^2 + 4\). Forgetting the middle term is the #1 algebraic error.
3. Check Problem Constraints Geometric lengths must be \(> 0\). But general integer problems often have negative solution sets!
Consecutive Quadratics Answer Key Teacher Resource Complete Solutions & Scoring
Consecutive Quadratics Answer Key
Algebra 1 / Grade 10
Page 1 of 2
Standard 4-Point Analytic Rubric (Per Problem) Total: 20 Points (5 Problems × 4 Pts)
1 Pt: Let Statement Correct variables with parity (e.g., \(n, n+2\)).
1 Pt: Standard Form Accurate expansion to \(ax^2 + bx + c = 0\).
1 Pt: Solving Process Valid factoring or quadratic formula work.
1 Pt: Context & Check Both integers stated; extraneous root rejected/kept.
1
Product of Consecutive Positive Even Integers
Answer: 12 and 14
Problem: The product of two consecutive positive even integers is 168. Find the integers.
Model & Equation
Let \(n\) = first even integer (\(n > 0\)).
Second even integer \(= n + 2\).
\(n(n + 2) = 168\)
\(n^2 + 2n - 168 = 0\)
Factoring & Roots
Factors of \(-168\) summing to \(+2\):
\((n + 14)(n - 12) = 0\)
\(n = -14 \quad \text{or} \quad n = 12\)
Reject \(-14\) (condition: positive ).
Grading Breakdown
[1 pt] Sets \(n\) and \(n+2\)
[1 pt] Reaches \(n^2+2n-168=0\)
[1 pt] Factors to \((n+14)(n-12)\)
[1 pt] Explains rejection of \(-14\)
2
Sum of Squares of Consecutive Positive Odd Integers
Answer: 9 and 11
Problem: Find two consecutive positive odd integers such that the sum of their squares is 202.
Expansion & GCF
\(n^2 + (n + 2)^2 = 202\)
\(n^2 + n^2 + 4n + 4 = 202\)
\(2n^2 + 4n - 198 = 0\)
Divide by 2: \(n^2 + 2n - 99 = 0\)
Factoring & Selection
\((n + 11)(n - 9) = 0\)
\(n = -11 \quad \text{or} \quad n = 9\)
Reject \(-11\) (must be positive odd).
\(n = 9 \implies n + 2 = 11\)
Grading Breakdown
[1 pt] Sets \(n^2 + (n+2)^2 = 202\)
[1 pt] FOILs correctly (\(4n+4\))
[1 pt] Divides by 2 & factors
[1 pt] Verifies \(81 + 121 = 202\)
3
Garden Area & Perimeter (Consecutive Even)
Consecutive Quadratics Worksheet Algebra 1 • Practice Worksheet
Consecutive Integer Quadratic Applications
10 Practice Problems
Page 1 of 5
Name:
Date:
Period:
1
Consecutive Even Integers: Product Relation
4 Points
The product of two consecutive positive even integers is \(224\). Find the two integers.
Define Variables & Write Standard Equation (\(ax^2+bx+c=0\)):
Solve (Factoring / Quadratic Formula) & Verify:
Final Answer:
The two consecutive even integers are:
2
Consecutive Odd Integers: Product Relation
4 Points
The product of two consecutive positive odd integers is \(195\). Find the two integers.
Define Variables & Write Standard Equation (\(ax^2+bx+c=0\)):
Solve (Factoring / Quadratic Formula) & Verify:
Final Answer:
The two consecutive odd integers are:
Consecutive Quadratics Worksheet • Questions 3–4 on Page 2 →
Sum of Squares Word Problems
Expand \((n+2)^2\) completely using FOIL before simplifying.
Consecutive Quadratics
Page 2 of 5
3
Consecutive Even Integers: Sum of Squares
4 Points
Find two consecutive positive even integers such that the sum of their squares is \(340\).
Setup Sum of Squares & Simplify to Standard Form:
Divide by GCF, Factor & Conclude:
Final Answer:
The two consecutive even integers are:
4
Consecutive Odd Integers: Sum of Squares
4 Points
Find two consecutive positive odd integers such that the sum of their squares is \(130\).
Setup Sum of Squares & Simplify to Standard Form:
Divide by GCF, Factor & Conclude:
Final Answer:
The two consecutive odd integers are:
Consecutive Quadratics Worksheet • Questions 5–6 on Page 3 →
Geometric Area Word Problems
Remember to label dimensions and perimeter with proper units.
Consecutive Quadratics
Page 3 of 5
5
Rectangular Patio: Dimensions & Perimeter
4 Points
A landscape contractor is paving a rectangular patio whose length and width are consecutive positive even integers in feet. If the area is \(120\text{ ft}^2\), determine both dimensions and the perimeter of fencing needed.
Let Statements & Area Model Equation:
Consecutive Quadratics Teacher Guide Teacher Facilitation & Answer Guide
Consecutive Quadratics Teacher Guide
10 Full Solutions & Rubrics
Page 1 of 5 (Questions 1–2)
Recommended Lesson Pacing (60–75 Min or 2 Class Periods): Warm-up & Variable Definition (10m) → Direct Modeling & Q1–Q4 (25m) → Paired Problem Solving Q5–Q8 (25m) → Complex Applications Q9–Q10 (15m).
Total Assessment 40 Points
1
Q1 Solution: Consecutive Even Integers (Product = 224)
Answer: 14 and 16
Step-by-Step Algebraic Model:
• Let \(n\) = 1st positive even integer (\(n \in \{2, 4, 6, \dots\}\))
\(n(n + 2) = 224\)
Factors of \(-224\) that add to \(+2\): \(16 \times (-14) = -224\).
Reject \(n = -16\) (condition: positive ).
Scoring Breakdown (4 Pts):
[1 Pt] Correct let statements \(n\) and \(n+2\).[1 Pt] Standard quadratic form \(n^2 + 2n - 224 = 0\).[1 Pt] Factoring work showing \((n+16)(n-14) = 0\).[1 Pt] Final statement: 14 and 16, rejecting \(-16\).
Teacher Alert: For students struggling to factor \(-224\), suggest finding \(\sqrt{224} \approx 14.9\). Test integer factors immediately adjacent: \(14 \times 16 = 224\).
2
Q2 Solution: Consecutive Odd Integers (Product = 195)
Answer: 13 and 15
Step-by-Step Algebraic Model:
• Let \(n\) = 1st positive odd integer(increments by 2!)
\(n(n + 2) = 195\)
Factors of \(-195\) that add to \(+2\): \(15 \times (-13) = -195\).
Reject \(n = -15\) (must be positive).
Scoring Breakdown (4 Pts):
[1 Pt] Identifies odd integers as \(n\) and \(n+2\).[1 Pt] Translates to \(n^2 + 2n - 195 = 0\).[1 Pt] Solves to find roots \(13\) and \(-15\).
Dim: 14m × 16m • Perim: 60m
Problem: Consecutive even dimensions in meters with area \(224\text{ m}^2\). Find dimensions and perimeter.
Let width \(= w\), length \(= w + 2\).
Factor pairs: \(16 \times 14 = 224\).
\(w = 14 \quad (\text{reject } -16 < 0)\)
Length \(= 14 + 2 = 16\text{ m}\).
\(P = 2(14 + 16) = 60\text{ m}\)
[1 pt] Area setup \(w(w+2)=224\)
[1 pt] Reaches \(w^2+2w-224=0\)
[1 pt] Finds \(w=14\) & length \(=16\)
[1 pt] Computes perimeter \(=60\text{ m}\)
Note: Partial credit can be awarded proportionally for each rubric benchmark. Solutions 4–5 & Misconception Matrix on Page 2 →
Consecutive Quadratics Answer Key (Part II) Advanced Relations, Pythagorean Modeling & Diagnostic Guide
Three Consecutive Odd Integers (Both Signs Valid) Two Valid Sets: {7, 9, 11} & {-5, -3, -1}
Problem: Product of first two consecutive odd integers is 19 more than 4 times the third integer.
Let integers be \(n, \; n+2, \; n+4\).
\(n(n + 2) = 4(n + 4) + 19\)
\(n^2 + 2n = 4n + 16 + 19\)
\(n = 7 \quad \text{or} \quad n = -5\)
Case 1 (\(n=7\)): \(\{7, 9, 11\}\)
Case 2 (\(n=-5\)): \(\{-5, -3, -1\}\)
No "positive" restriction \(\implies\) both valid!
[1 pt] Translates \(4(n+4)+19\) correctly
[1 pt] Reaches \(n^2-2n-35=0\)
[1 pt] Identifies roots \(7\) and \(-5\)
[1 pt] Gives both integer triples (-1 pt if negative set is discarded!)
Right Triangle Sides (Pythagorean Consecutive Even) Legs: 6 cm, 8 cm • Hypotenuse: 10 cm
Problem: Three consecutive positive even integers represent the sides of a right triangle in cm.
Legs: \(n, n+2\); Hypotenuse: \(n+4\).
\(n^2 + (n + 2)^2 = (n + 4)^2\)
\(n^2 + n^2 + 4n + 4 = n^2 + 8n + 16\)
\(n = 6 \quad \text{or} \quad n = -2\)
Reject \(-2\) (length cannot be negative).
Legs: \(6\text{ cm}, 8\text{ cm}\); Hyp: \(10\text{ cm}\)
[1 pt] Designates \(n+4\) as hypotenuse
[1 pt] Expands both sides correctly
[1 pt] Factors \(n^2-4n-12=0\)
[1 pt] Rejects \(-2\); states all 3 sides with cm
Teacher Diagnostic Guide: Common Misconceptions & Recommended Feedback
1. Misconception: \(n + 1\) for Odds
Students intuitively write \(n, n+1\) thinking "odd + 1".
Correction Prompt: "Test with numbers: if \(n=3\), what is \(3+1\)? It's even! You must add 2 to jump to the next odd."
2. Misconception: \((n+2)^2 = n^2 + 4\)
Students frequently distribute the exponent and drop the middle term \(4n\).
Correction Prompt: "Write out the binomial twice: \((n+2)(n+2)\) and draw your FOIL or area model box."
3. Misconception: Premature Rejection
Students reflexively cross out any negative root as 'extraneous'.
Correction Prompt: "Check the question: does it say 'positive integers' or just 'integers'? Negative odd numbers exist!"
Consecutive Integer Quadratic Applications • Teacher Grading Key Verified Algebraic Solutions
Solve for Dimensions & Calculate Perimeter (\(2l+2w\)):
Picture Frame Border: Consecutive Odd Dimensions A custom rectangular picture frame has an outer height and width that are consecutive positive odd integers in inches. If the area enclosed by the frame's outer border is \(63\text{ in}^2\), find its outer dimensions.
Let Statements & Area Equation:
Factoring, Solving & Contextual Rejection:
Consecutive Quadratics Worksheet • Questions 7–8 on Page 4 →
Advanced Relations & Geometry Check whether negative roots are valid or extraneous in each context.
Three Consecutive Odd Integers: Dual Solution Sets Find three consecutive odd integers such that the product of the first two is \(5\) less than \(8\) times the third integer. (Hint: Both positive and negative integer sets exist!)
Setup \(n(n+2)=8(n+4)-5\) & Standard Form:
Factor & State Both Valid Sets of 3 Integers:
Pythagorean Right Triangle: Three Consecutive Even Sides A right-triangular park has three boundary paths whose lengths are consecutive positive even integers in decameters (\(\text{dam}\)). Find the lengths of all three sides and the park's total area.
Pythagorean Setup: \(n^2 + (n+2)^2 = (n+4)^2\):
Solve for Sides & Compute Triangle Area (\(\frac{1}{2}bh\)):
Consecutive Quadratics Worksheet • Questions 9–10 on Page 5 →
Complex Applications & Synthesis (Questions 9 & 10) Multi-step word problems involving sums, products, and uniform borders.
Product & Sum Combination: Consecutive Odd Integers The product of two consecutive positive odd integers is \(1\) less than \(6\) times their sum. Find the two integers.
Set Up \(n(n+2) = 6(n + n + 2) - 1\) & Simplify:
Solve Quadratic, Discard Extraneous Root & Check:
The two consecutive odd integers are:
Uniform Walkway Border: Consecutive Even Pool A rectangular pool has dimensions that are consecutive positive even integers in meters. A concrete walkway of uniform width \(1\text{ meter}\) surrounds the pool. If the total combined area of the pool and walkway is \(168\text{ m}^2\), find the pool's dimensions and the walkway's area alone.
Define Pool Dimensions, Add Uniform 1m Border & Form Equation:
Solve for Dimensions & Calculate Walkway Area (\(168 - \text{Pool Area}\)):
Worksheet Complete: Review all 10 problems before submitting.
[1 Pt] Discards \(-15\) and reports pair: 13 and 15.Common Misconception: Watch for students setting \(n(n+1) = 195\). Point out that consecutive odd integers are separated by an even integer, so the gap is always 2.
Teacher Guide • Questions 3–4 Solutions on Page 2 →
Sum of Squares Solutions (Questions 3 & 4) Focus on Binomial Expansion & Common Factor Simplification
Q3 Solution: Consecutive Even Sum of Squares (\(= 340\)) • Let \(n\) = 1st even integer; \(n+2\) = 2nd even integer.
\(n^2 + (n^2 + 4n + 4) = 340\)
Factoring: \((n + 14)(n - 12) = 0 \implies n = -14 \text{ or } n = 12\).
Reject \(-14\). Final integers: \(12\) and \(14\).
Scoring Breakdown (4 Pts):
[1 Pt] Translates sum of squares correctly.[1 Pt] Expands \((n+2)^2\) to \(n^2+4n+4\).[1 Pt] Simplifies by dividing by 2 to get \(n^2+2n-168=0\).[1 Pt] Discards \(-14\); states integers 12 and 14. Pedagogical Note: Dividing by the GCF of \(2\) before factoring turns a challenging quadratic into a simple trinomial factoring problem.
Q4 Solution: Consecutive Odd Sum of Squares (\(= 130\)) • Let \(n\) = first positive odd integer; \(n+2\) = next odd integer.
\(n^2 + (n^2 + 4n + 4) = 130\)
Factoring: \((n + 9)(n - 7) = 0 \implies n = -9 \text{ or } n = 7\).
Reject \(n = -9\) (positive required).
Scoring Breakdown (4 Pts):
[1 Pt] Translates sum of squares correctly.[1 Pt] Correct binomial expansion \(n^2+4n+4\).[1 Pt] Solves \(n^2+2n-63=0\) to get \(n=7, -9\).[1 Pt] States final pair: 7 and 9. Key Teaching Tip: Highlight the algebraic symmetry: both consecutive even and odd integers lead to the exact same left-hand expression \(2n^2 + 4n + 4\).
Teacher Guide • Questions 5–6 Solutions on Page 3 →
Geometry Modeling Solutions (Questions 5 & 6) Physical Dimensions, Perimeter Calculations & Unit Checks
Q5 Solution: Rectangular Patio (Area = 120 ft²) 10 ft × 12 ft • Perim: 44 ft
• Width \(= w\text{ ft}\); Length \(= w + 2\text{ ft}\) (consecutive even).
\(w^2 + 2w - 120 = 0\)
Physical length must be positive (\(w > 0\)), reject \(w = -12\).
Perimeter: \(P = 2(w + l) = 2(10 + 12) = 2(22) = \mathbf{44\text{ ft}}\).
Scoring Breakdown (4 Pts):
[1 Pt] Area equation \(w(w+2) = 120\).[1 Pt] Reaches standard form \(w^2+2w-120=0\).[1 Pt] Solves \(w=10\text{ ft}\) & Length \(=12\text{ ft}\).[1 Pt] Computes perimeter \(44\text{ ft}\) with correct units. Common Pitfall: Remind students that perimeter requires doubling the sum: \(2(l+w)\), not just adding \(l+w\).
Q6 Solution: Picture Frame (Consecutive Odd Area = 63 in²) Width: 7 in • Height: 9 in
• Frame dimensions are consecutive positive odd integers.
\(w(w + 2) = 63\)
Reject negative dimension (\(w > 0\)).
Check: Both 7 and 9 are odd integers; \(7 \times 9 = 63\text{ in}^2\) ✓
Scoring Breakdown (4 Pts):
[1 Pt] Defines consecutive odd dimensions \(w, w+2\).[1 Pt] Forms standard equation \(w^2 + 2w - 63 = 0\).[1 Pt] Factors to \((w+9)(w-7) = 0\).[1 Pt] States outer dimensions with inches. Visual Check: Encourage visual learners to sketch the rectangle with dimensions labeled \(w\) and \(w+2\).
Teacher Guide • Questions 7–8 Solutions on Page 4 →
Advanced Applications (Questions 7 & 8) Multi-Solution Handling & Right Triangle Geometry
Q7 Solution: Three Consecutive Odd (Dual Valid Sets) Set 1: {9, 11, 13} & Set 2: {-3, -1, 1}
• Let integers be \(n, \; n + 2, \; n + 4\).
\(n(n + 2) = 8(n + 4) - 5\)
Both Sets Valid! (No "positive" restriction):
Scoring Breakdown (4 Pts):
[1 Pt] Sets up \(n(n+2) = 8(n+4) - 5\).[1 Pt] Reaches standard form \(n^2 - 6n - 27 = 0\).[1 Pt] Finds roots \(n=9\) and \(n=-3\).[1 Pt] Lists both integer sets (-1 pt if negative set discarded). Critical Insight: Many students erroneously discard \(n = -3\). Explain that negative odd integers exist and satisfy the prompt.
Q8 Solution: Triangular Park (Consecutive Even Sides) Sides: 6, 8, 10 dam • Area: 24 dam²
Pythagorean Theorem Setup:
• Legs: \(n, n+2\); Hypotenuse: \(n+4\).
\(n^2 + n^2 + 4n + 4 = n^2 + 8n + 16\)
Reject \(-2\). Sides: \(\mathbf{6\text{ dam}, 8\text{ dam}, 10\text{ dam}}\).
Area: \(\frac{1}{2}(6)(8) = \mathbf{24\text{ dam}^2}\). Check: \(36 + 64 = 100\) ✓
Scoring Breakdown (4 Pts):
[1 Pt] Correct right triangle setup \(n^2+(n+2)^2=(n+4)^2\).[1 Pt] Simplifies to \(n^2 - 4n - 12 = 0\).[1 Pt] Solves \(n=6\) (rejects \(-2\)); lists all three sides.[1 Pt] Calculates area \(24\text{ dam}^2\) with units. Extension Connection: This is the famous \(3\text{–}4\text{–}5\) right triangle scaled by a factor of 2.
Teacher Guide • Questions 9–10 Solutions on Page 5 →
Complex Synthesis Solutions (Questions 9 & 10) Product-Sum Relations & Uniform Border Applications
Q9 Solution: Product & Sum Combination (Consecutive Odd) Algebraic Translation & Solving:
• Let odd integers be \(n\) and \(n + 2\).
\(n(n + 2) = 6(n + n + 2) - 1\)
Factoring: \((n - 11)(n + 1) = 0 \implies n = 11 \text{ or } n = -1\).
Reject \(n = -1\) (positive required).
Scoring Breakdown (4 Pts):
[1 Pt] Translates sum accurately: \(6(2n+2)-1\).[1 Pt] Standard form quadratic \(n^2 - 10n - 11 = 0\).[1 Pt] Factors to \((n-11)(n+1)=0\).[1 Pt] States pair: 11 and 13 with verification. Teaching Alert: Watch for distribution errors with the sum: students often write \(6(n+2)-1\) instead of \(6(n + n + 2)-1\).
Q10 Solution: Uniform Pool Border (Area = 168 m²) Pool: 10m × 12m • Walkway: 48 m²
• Pool width \(= w\), length \(= w + 2\).
\((w + 2)(w + 4) = 168\)
Reject \(w = -16\). Pool width \(= \mathbf{10\text{ m}}\), length \(= \mathbf{12\text{ m}}\).
Walkway Area: \(168 - (10 \times 12) = 168 - 120 = \mathbf{48\text{ m}^2}\).
Scoring Breakdown (4 Pts):
[1 Pt] Adds \(2\text{m}\) to each dimension: \((w+2)(w+4)\).[1 Pt] Reaches \(w^2 + 6w - 160 = 0\).[1 Pt] Solves \(w = 10\text{ m}\); pool dimensions \(10\text{m} \times 12\text{m}\).[1 Pt] Subtracts pool area to find walkway \(= 48\text{ m}^2\). Border Trap: Remind students that a border surrounds both sides of each dimension, so width increases by \(+2\), not \(+1\).
Unit Assessment Grading Scale (10 Problems • 40 Total Points) Standard Mastery: ≥ 32/40 (80%)
Exemplary (90-100%) 36 – 40 Pts
Proficient (80-89%) 32 – 35 Pts
Developing (70-79%) 28 – 31 Pts
Novice (< 70%) 0 – 27 Pts
