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IEB practice · Paper I

IEB Algebra practice

9 newly generated questions across all 8 algebra topics the IEB examines in Paper I, 131 marks in all. Work each one, then open its answer underneath to check yourself. Nothing here is taken from a past paper.

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Quadratic equations

2 questions · 30 marks

Paper I · Section A · 15 marks

QUESTION 1
  1. (a)
    Solve for x in terms of p: x^2 − 8p^2 = −2px
    (3)
  2. (b)
    If x^2 + 16x + 47 = (x − m)^2 + p, determine the values of m and p.
    (3)
  3. (c)
    Hence write down the minimum value of x^2 + 16x + 47.
    (1)
  4. (d)
    Solve for x: (1 − x)(x + 5) > 0
    (4)
  5. (e)
    Determine the values of p for which x^2 − (p − 2)x + 25 = 0 has equal roots.
    (4)

[15]

Show the answers

(a) Answer: x = −4p or x = 2p

  1. x^2 + 2px − 8p^2 = 0
  2. (x + 4p)(x − 2p) = 0

(b) Answer: m = −8 and p = −17

  1. (x − m)^2 + p = x^2 − 2mx + m^2 + p
  2. −2m = 16, so m = −8
  3. m^2 + p = 47, so p = −17

(c) Answer: −17

  1. (x − m)^2 ≥ 0, so the least value is p = −17

(d) Answer: −5 < x < 1

  1. −x^2 − 4x + 5 > 0
  2. x^2 + 4x − 5 < 0
  3. (x + 5)(x − 1) < 0
  4. critical values: x = −5 and x = 1

(e) Answer: p = −8 or p = 12

  1. Δ = (p − 2)^2 − 4(1)(25) = (p − 2)^2 − 100
  2. Δ = 0 for equal roots: (p − 2)^2 = 100
  3. p − 2 = ±10

Paper I · Section A · 15 marks

QUESTION 2
  1. (a)
    Solve for x: x(x + 6)(5x + 7) = 0
    (3)
  2. (b)
    Use the quadratic formula to solve for x, correct to two decimal places, if 3x^2 − 4x = 6. Show all working.
    (4)
  3. (c)
    Solve for x: (4 − x)(x + 6) < 0
    (4)
  4. (d)
    Determine the values of p for which x^2 − (p + 2)x + 16 = 0 has equal roots.
    (4)

[15]

Show the answers

(a) Answer: x = −6 or x = −7/5 or x = 0

  1. x = 0 or x + 6 = 0 or 5x + 7 = 0

(b) Answer: x = −0.90 or x = 2.23

  1. 3x^2 − 4x − 6 = 0
  2. x = (−(−4) ± √((−4)^2 − 4(3)(−6)))/(2(3))
  3. x = (4 ± √(88))/6

(c) Answer: x < −6 or x > 4

  1. −x^2 − 2x + 24 < 0
  2. x^2 + 2x − 24 > 0
  3. (x + 6)(x − 4) > 0
  4. critical values: x = −6 and x = 4

(d) Answer: p = −10 or p = 6

  1. Δ = (p + 2)^2 − 4(1)(16) = (p + 2)^2 − 64
  2. Δ = 0 for equal roots: (p + 2)^2 = 64
  3. p + 2 = ±8

Exponential equations and exponent laws

3 questions · 43 marks

Paper I · Section A · 13 marks

QUESTION 3
  1. (a)
    Solve for x: 5^(2x) − 26(5^x) + 25 = 0
    (4)
  2. (b)
    Solve for x: log_2 (x^2 + x − 12) = 3
    (4)
  3. (c)
    Solve for x: √(x + 51) = x − 5
    (5)

[13]

Show the answers

(a) Answer: x = 0 or x = 2

  1. let t = 5^x: t^2 − 26t + 25 = 0
  2. (t − 1)(t − 25) = 0
  3. 5^x = 1, so x = 0
  4. 5^x = 25, so x = 2

(b) Answer: x = −5 or x = 4

  1. x^2 + x − 12 = 2^3
  2. x^2 + x − 20 = 0
  3. (x + 5)(x − 4) = 0

(c) Answer: x = 13

  1. x + 51 = x^2 − 10x + 25
  2. x^2 − 11x − 26 = 0
  3. (x − 13)(x + 2) = 0
  4. check x = −2: √(−2 + 51) = 7 but −2 − 5 = −7, so x = −2 is rejected

Paper I · Section A · 14 marks

QUESTION 4
  1. (a)
    If 3^x = k, write 3^(−x) in terms of k.
    (2)
  2. (b)
    Write 9^x in terms of k.
    (2)
  3. (c)
    If log 3 = m and log 5 = n, determine log 0.6 in terms of m and n.
    (2)
  4. (d)
    Determine log 0.15 in terms of m and n.
    (2)
  5. (e)
    Solve for x: log_x 16 = 2
    (2)
  6. (f)
    Solve for x: log_2 (x − 4) + log_2 (x + 8) = 6
    (4)

[14]

Show the answers

(a) Answer: 3^(−x) = 1/k

  1. 3^(−x) = (3^x)^(−1)
  2. = 1/k

(b) Answer: 9^x = k^2

  1. 9^x = (3^x)^2
  2. = k^2

(c) Answer: log 0.6 = m − n

  1. log 0.6 = log(3/5)
  2. = m − n

(d) Answer: log 0.15 = m + n − 2

  1. log 0.15 = log(15/100)
  2. = m + n − 2

(e) Answer: x = 4

  1. x^2 = 16
  2. x = 4 (a base is positive and not 1)

(f) Answer: x = 8

  1. log_2 [(x − 4)(x + 8)] = 6
  2. (x − 4)(x + 8) = 64
  3. x^2 + 4x − 96 = 0
  4. x = 8 or x = −12
  5. x = −12 gives the log of a negative number, so x = 8

Paper I · Section A · 16 marks

QUESTION 5
  1. (a)
    Solve for x in terms of y: 2^(x + 1) · 16^y = 2^y
    (3)
  2. (b)
    Solve for x: √(x + 41) = x − 1
    (5)
  3. (c)
    Solve for x: log_3 (x + 1) − log_3 x = 1
    (4)
  4. (d)
    Solve for x: 5^(2x) − 4(5^x) − 21 = 0
    (4)

[16]

Show the answers

(a) Answer: x = −3y − 1

  1. 2^(x + 1) · 2^(4y) = 2^(y)
  2. x + 1 + 4y = y

(b) Answer: x = 8

  1. x + 41 = x^2 − 2x + 1
  2. x^2 − 3x − 40 = 0
  3. (x − 8)(x + 5) = 0
  4. check x = −5: √(−5 + 41) = 6 but −5 − 1 = −6, so x = −5 is rejected

(c) Answer: x = 1/2

  1. log_3 ((x + 1)/x) = 1
  2. (x + 1)/x = 3
  3. x + 1 = 3x

(d) Answer: x = 1.21

  1. let t = 5^x: t^2 − 4t − 21 = 0
  2. (t − 7)(t + 3) = 0
  3. 5^x = −3 has no solution
  4. 5^x = 7, so x = log_5 7 = 1.21

The nature of the roots

1 questions · 16 marks

Paper I · Section A · 16 marks

QUESTION 6
  1. (a)
    Consider the equation kx^2 + 6x + k + 8 = 0, where k is a real number. Determine the discriminant of the equation in terms of k. Give your answer in factorised form.
    (3)
  2. (b)
    For which values of k will the equation be a quadratic equation with real roots?
    (3)
  3. (c)
    For which integer values of k will the equation be a quadratic equation with rational roots?
    (2)
  4. (d)
    Solve x^2 + 4x − 1 = 0 by completing the square. Give your answers correct to two decimal places.
    (4)
  5. (e)
    Solve for x: x(2 − x) < 0
    (4)

[16]

Show the answers

(a) Answer: Δ = −4(k + 9)(k − 1)

  1. Δ = 6^2 − 4(k)(k + 8)
  2. = −4k^2 − 32k + 36
  3. = −4(k^2 + 8k − 9)
  4. = −4(k + 9)(k − 1)

(b) Answer: −9 ≤ k ≤ 1, k ≠ 0

  1. the equation is quadratic only if k ≠ 0
  2. real roots: Δ ≥ 0, so −4(k + 9)(k − 1) ≥ 0
  3. (k + 9)(k − 1) ≤ 0

(c) Answer: k = −9; −8; −7; −4; −1; 1

  1. Δ must be a perfect square: try the integers from −9 to 1, k ≠ 0
  2. k = −9: Δ = 0; k = −8: Δ = 36; k = −7: Δ = 64; k = −6: Δ = 84; k = −5: Δ = 96; k = −4: Δ = 100; k = −3: Δ = 96; k = −2: Δ = 84; k = −1: Δ = 64; k = 1: Δ = 0

(d) Answer: x = −4.24 or x = 0.24

  1. x^2 + 4x = 1
  2. x^2 + 4x + 4 = 1 + 4
  3. (x + 2)^2 = 5
  4. x + 2 = ±√(5)
  5. x = −2 ± √(5)

(e) Answer: x < 0 or x > 2

  1. −x^2 + 2x < 0
  2. x^2 − 2x > 0
  3. x(x − 2) > 0
  4. critical values: x = 0 and x = 2

Inequalities

1 questions · 13 marks

Paper I · Section A · 13 marks

QUESTION 7
  1. (a)
    Solve for x: (6 − x)/(x + 2) < 0
    (4)
  2. (b)
    For which values of x is √((x − 2)(8 − x)) real?
    (3)
  3. (c)
    Solve for x: x(x + 3) > −2
    (3)
  4. (d)
    How many integers satisfy x(x + 4) < 0?
    (3)

[13]

Show the answers

(a) Answer: x < −2 or x > 6

  1. critical values: x = 6 (the numerator is 0) and x = −2 (undefined, x ≠ −2)
  2. the sign of the left-hand side changes at each of them

(b) Answer: 2 ≤ x ≤ 8

  1. (x − 2)(8 − x) ≥ 0

(c) Answer: x < −2 or x > −1

  1. x^2 + 3x + 2 > 0
  2. (x + 2)(x + 1) > 0
  3. critical values: x = −2 and x = −1

(d) Answer: 3

  1. x(x + 4) < 0
  2. −4 < x < 0
  3. the integers −3 to −1: 3 of them

Simultaneous equations

2 questions · 29 marks

Paper I · Section A · 15 marks

QUESTION 8
  1. (a)
    Solve for x and y if y + 3x = −3 and y = x^2 − 7.
    (6)
  2. (b)
    For which values of k will the line y = x + k never meet the parabola y = x^2 − 2?
    (4)
  3. (c)
    An amount of R11 500 is shared equally among a group of people. If there were 2 more people in the group, each person would receive R40 less. How many people are in the group?
    (5)

[15]

Show the answers

(a) Answer: x = −4, y = 9 or x = 1, y = −6

  1. y = −3x − 3
  2. −3x − 3 = x^2 − 7
  3. x^2 + 3x − 4 = 0
  4. (x − 1)(x + 4) = 0
  5. x = −4 or x = 1
  6. y = 9 or y = −6

(b) Answer: k < −9/4

  1. x^2 − 2 = x + k
  2. x^2 − x − 2 − k = 0
  3. no point of intersection: Δ < 0
  4. (−1)^2 − 4(1)(−2 − k) < 0

(c) Answer: 23 people

  1. let there be n people: 11 500/n − 11 500/(n + 2) = 40
  2. 11 500(n + 2) − 11 500n = 40n(n + 2)
  3. n^2 + 2n − 575 = 0
  4. (n − 23)(n + 25) = 0
  5. n = 23 (n = −25 is rejected: the number of people is positive)

Paper I · Section A · 14 marks

QUESTION 9
  1. (a)
    Solve for x and y if 3^(x + y) = 1 and 2^(x − y) = 16.
    (5)
  2. (b)
    For which values of k will the line y = x + k never meet the parabola y = x^2 + 3x + 7?
    (4)
  3. (c)
    The 1292 chairs in a school hall are set out in equal rows. If there were 2 fewer rows, each row would need 8 more chairs. How many rows are there?
    (5)

[14]

Show the answers

(a) Answer: x = 2, y = −2

  1. 3^(x + y) = 3^(0), so x + y = 0
  2. 2^(x − y) = 2^(4), so x − y = 4
  3. adding: 2x = 4, so x = 2
  4. y = 0 − (2) = −2

(b) Answer: k < 6

  1. x^2 + 3x + 7 = x + k
  2. x^2 + 2x + 7 − k = 0
  3. no point of intersection: Δ < 0
  4. (2)^2 − 4(1)(7 − k) < 0

(c) Answer: 19 rows

  1. let there be r rows: 1292/(r − 2) − 1292/r = 8
  2. 1292r − 1292(r − 2) = 8r(r − 2)
  3. r^2 − 2r − 323 = 0
  4. (r − 19)(r + 17) = 0
  5. r = 19 (r = −17 is rejected: the number of rows is positive)

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