Total marks: 60 · Duration: 90 minutes
Instructions to students
• Answer all questions. • Show all your work; credit may be given for a correct method even if the final answer is incorrect. • Calculators may be used where appropriate. • This assessment covers Common Core high school Algebra standards: creating and solving equations and inequalities, structure in expressions, arithmetic with polynomials and rational expressions, and reasoning with equations. • The total number of points for this assessment is 60. • All information needed is given within each question.
Assessment
Section A — Short-answer Questions (44 points)
1. Linear equations and inequalities
(a) Solve for x: 3(x − 2) + 5 = 2x + 7. [3]
(b) Solve the inequality and graph the solution set on a number line: 4 − 2x < 10. [3]
(c) The formula for the area of a trapezoid is A = ½(b₁ + b₂)h. Rearrange the formula to make h the subject. [2]
2. Quadratic expressions and equations
(a) Factor completely: x² + 7x + 12. [2]
(b) Hence solve: x² + 7x + 12 = 0. [2]
(c) Solve using the quadratic formula, giving exact answers: 2x² − 3x − 2 = 0. [4]
(d) Write x² + 6x + 5 in the form (x + a)² + b (complete the square). [2]
3. Polynomials
(a) Simplify: (2x² + 3x − 1) + (x² − 4x + 5). [2]
(b) Expand and simplify: (x + 3)(x − 5). [2]
(c) Use the Remainder Theorem to find the remainder when x³ − 2x² + 4x − 5 is divided by (x − 2). [3]
(d) Show that (x − 1) is a factor of x³ + 2x² − 5x + 2, and hence explain what this tells you about x = 1. [3]
4. Systems of equations
(a) Solve the system by elimination: 2x + y = 7 x − y = 2 [4]
(b) Solve the system by substitution: y = 2x − 1 3x + y = 14 [4]
5. Rational expressions
(a) Simplify: (x² − 9) / (x² + 4x + 3). [3]
(b) Write as a single simplified fraction: 2/x + 3/(x + 1). [5]
Section B — Extended Questions (16 points)
6. Creating and using equations
A rectangular garden is 3 meters longer than it is wide. Its area is 40 square meters.
(a) Letting the width be w, write an equation in terms of w to represent this situation. [3]
(b) Solve your equation to find the dimensions of the garden. [5]
7. Sequences
A geometric sequence has first term 3 and common ratio 2.
(a) Write down the first four terms of the sequence. [2]
(b) Write an expression for the nth term of the sequence. [3]
(c) Which term of the sequence is equal to 192? [3]
Answer Key and Worked Solutions
Section A
1. (a) 3x − 6 + 5 = 2x + 7 → 3x − 1 = 2x + 7 (M1) → x = 8 (M1 A1). [3] (b) 4 − 2x < 10 → −2x < 6 (M1) → x > −3 (divide by −2, reverse inequality) (M1); number line with an open circle at −3 and arrow to the right (A1). [3] (c) A = ½(b₁ + b₂)h → 2A = (b₁ + b₂)h (M1) → h = 2A / (b₁ + b₂) (A1). [2]
2. (a) (x + 3)(x + 4) (M1 A1). [2] (b) x = −3 or x = −4 (A1 A1). [2] (c) a = 2, b = −3, c = −2. x = [3 ± √(9 + 16)] / 4 = [3 ± √25] / 4 = (3 ± 5)/4 (M1 M1) → x = 2 or x = −½ (A1 A1). [4] (d) x² + 6x + 5 = (x + 3)² − 9 + 5 = (x + 3)² − 4 (M1 A1). [2]
3. (a) 3x² − x + 4 (M1 A1). [2] (b) x² − 5x + 3x − 15 = x² − 2x − 15 (M1 A1). [2] (c) By the Remainder Theorem, remainder = f(2) = 8 − 8 + 8 − 5 = 3 (M1 M1 A1). [3] (d) f(1) = 1 + 2 − 5 + 2 = 0 (M1 M1), so by the Factor Theorem (x − 1) is a factor; this tells us that x = 1 is a root/zero of the polynomial (the graph crosses the x-axis at x = 1) (A1). [3]
4. (a) Add the equations: 3x = 9 → x = 3 (M1 M1); substitute: 2(3) + y = 7 → y = 1 (A1). Solution (3, 1) (A1). [4] (b) Substitute y = 2x − 1 into 3x + y = 14: 3x + 2x − 1 = 14 (M1) → 5x = 15 → x = 3 (M1); y = 2(3) − 1 = 5 (A1). Solution (3, 5) (A1). [4]
5. (a) (x² − 9)/(x² + 4x + 3) = [(x − 3)(x + 3)] / [(x + 1)(x + 3)] (M1 M1) = (x − 3)/(x + 1) (A1), x ≠ −3, −1. [3] (b) Common denominator x(x + 1): [2(x + 1) + 3x] / [x(x + 1)] (M1 M1) = (2x + 2 + 3x)/[x(x+1)] = (5x + 2) / [x(x + 1)] (M1 A1, with 1 mark for correct expansion). [5]
Section B
6. (a) Width w, length (w + 3). Area: w(w + 3) = 40 (M1 M1 A1). [3] (b) w² + 3w − 40 = 0 (M1) → (w + 8)(w − 5) = 0 (M1) → w = 5 or w = −8 (M1); reject the negative, so w = 5 (M1). Dimensions: width 5 m, length 8 m (A1). [5]
7. (a) 3, 6, 12, 24 (M1 A1). [2] (b) nth term = 3 × 2^(n−1) (M1 M1 A1). [3] (c) 3 × 2^(n−1) = 192 → 2^(n−1) = 64 = 2⁶ (M1 M1) → n − 1 = 6 → n = 7 (the 7th term) (A1). [3]