Mathematics Grade 12 Questions and AnswersExponents and Surds Questions and Answers Mathematics Grade 12 Pdf Download

Exponents and Surds Questions and Answers Mathematics Grade 12 Pdf Download

Exponents and Surds Questions and Answers Mathematics Grade 12 Pdf. The number of times the number appears or is multiplied is the exponent. In 4 3 , 4 is the base and is the exponent. The values in the square root or cube root or any other roots, which cannot be further simplified into whole numbers or integers, are known as a surd.

EXPONENTS AND SURDS QUESTIONS AND ANSWERS GRADE 12

Activity 1
Write in simplest form without using a calculator (show all working).

  1. √8  × √2
  2. 3√4 × 3√2
  3.  9 + √45
    3
  4. (2  + √5 ) (2  −  √5 )

[10]

Solutions

  1. √8 × √8 = √8×2 = √16 = 4 (1)
  2. 3√4 × 3√2  =  3√4×2 = 3√8  = 2 (2)
  3. 9+ √45 = 9+3√5 = 3( 3+ √5 ) = 3 + √5 (3)
    3             3               3
  4. ( 2 + √5 ) (2  −  √5 )
    = 2 × 2 – √5 × √5 = 4 – 5 = –1 (2)
    Or multiply out the brackets:
    ( 2 + √5 ) (2  −  √5 ) = 4 + 2√5 – 2 √5 – √5 .√5 = 4 – 5 = –1 (2)
    [10] 

Activity 2 Interpret a graph

1. Complete the table for each number by marking the correct columns.
Nonreal numberReal number
ℝ		Rational number
ℚ		Irrational number
ℚ′		Integer
ℤ		Whole number
ℕ0		Natural number
ℕ
a) 13
b) 5,121212…
c) √–6
d) 3π
e) 0 = 0
9
f) √17
g)3√64 = 4
h) 22
7

(23)
2. Which of the following numbers are rational and which are irrational?

  1. √16
  2. √8
  3. √ 9
    4
  4. √6¼
  5. √47
  6. 22
    7
  7. 0,347347…
  8. π − (− 2)
  9. 2 + √2
  10. 1,121221222… (10)

[33]

Solutions

1. Complete the table for each number by marking the correct columns.
Nonreal numberReal number
ℝ		Rational number
ℚ		Irrational number
ℚ′		Integer
ℤ		Whole number
ℕ0		Natural number
ℕ
a) 13 ✓ ✓ ✓ ✓ ✓(5)
b) 5,121212… ✓ ✓ (2)
c) √–6 ✓ (1)
d) 3π ✓ ✓ (2)
e) 0 = 0
9		 ✓ ✓ ✓ ✓ (4)
f) √17 ✓ ✓ (2)
g)3√64 = 4 ✓ ✓ ✓ ✓ ✓ (5)
h) 22
7		 ✓ ✓ (2)

2.

  1. √16 (rational)
  2. √8 (irrational)
  3. √ 9  =  3  (rational)
    4      2
  4. √6¼ =  √25  =  5  (rational)
    4        2
  5. √47 (irrational)
  6. 22 (rational)
    7
  7. 0,347347…(rational, because it is a recurring decimal) 3 (1)
  8. π − (− 2) (irrational, because π is irrational) 3 (1)
  9. 2 + √2 (irrational, because √2 is irrational) 3 (1)
  10. 1,121221222…(irrational, because it is a non-recurring and non-terminating decimal) 3 (1)
    [33]

Activity 3
Calculate

  1. −3 (( −2a3)2 + √9a12)     √9a12  = (32a12)½
  2.       5(2a4)3
    (5a3)2 − 5a6 [5]
Solutions

  1. −3 (( −2a3)2 + √9a12)  simplify exponents inside the brackets and the square root
    = −3(4a6+ 3a6) add like terms inside the bracket
    = –3(7a6 ) = –21a6 simplify (3)
  2.   5(2a4)3      simplify brackets at the top and the bottom first
    (5a3)2 − 5a6
    =     5(8a12)     =  40a 12 = 2a6(2)
    +25a6 – 5a6      20a6
    [5] 

Activity 4
Simplify the following. Write answers with positive exponents where necessary.

  1.   a -3
    b-2
  2.  4a7b–4c–1
    d–2e5
  3.  x–1+ y-1
    [5]
Solutions

  1.  a -3 = b2
    b-2     a3
  2. 4a7b−4c−1 =  4a7d2
    d-2e5          b4c1e5
  3. x–1+ y–1 = 1  +  1 = y + x
    x      y      xy
    [5] 

1.3.5 Working with surd (root) signs
The exponential rule 18 can be used to simplify certain expressions.

Activity 5
1. Rewrite these expressions without surd signs and simplify if possible.

  1. 3√5
  2. 4√16
  3. 3√–32
    [3]

10

Activity 6
Simplify the following and leave answers with positive exponents where necessary:
(a4)n–1. ( a2b)–3n
(ab)–2n. b–n
[4]

Solution
(a4)n–1. ( a2b)–3n = a4n−4 . a– 6n. b−3n
 (ab)–2n. b–n           a−2n. b−2n . b–n
=  a4n–4– 6n +2 n. b −3n + 2n + n
=  a−4. b0
= 1  . 1 =  1
a4         a4
[4]

Activity 7
Simplify the following and leave answers with positive exponents where necessary:

  1.  273 – 2x.9x-1
    812-x
  2.  6.5x +1 – 2.5x +2
    5x+3
  3. 22009 − 22012
    22010[13]

Solutions

  1. . 27 3−2x. 9x−1= (33)3−2x. (32)x−1  = 39−6x. 32x−2
    812-x                (34)2−x                  38-4x
    = 3 9−6x+2x−2−8+4x
    = 3 −1  = 1
    3 (4)
  2.  6.5x +1 – 2.5x +2 =  6.5x .51  − 2.5x . 52
    5x+3                       5x. 53
    =  30 − 50 = − 20 = − 4
    125        125     25 (4)
  3. 22009 − 22012 = 22009 (1 − 23) = (22009 1 − 8)
    22010             22010                  22010
    = 22009 (− 7)
    22010
    =  22009−22010  ×− 7
    =  2−1 × − 7 = ½ × − 7 = −7/2 (5)
    [13]

Activity 8
Solve for x:

  1. 3 ( 9x+3 ) = 272x–1
  2. 32x–12 = 1
  3. 2x = 0,125
  4. 10x ( x+1 ) = 100
  5. 5x + 5x+1  = 30
  6. 5 2+x – 5x = 5x. 23 + 1
  7. 5x + 15.5 −x  =  2
  8. 19
    {31]
Solutions
Remember: When adding or subtracting terms, you need to factorise first.

  1. 3(9x+3 ) = 272x–1
    31(32)x + 3 = (33)2x – 1 prime bases
    31+2 x+6  = 36 x–3 same bases
    ∴ 7 + 2x = 6x – 3  equate exponents
    –4x = – 3 – 7
    x = −10 =  5
    − 4      2
    =(3)
  2. 32 x −12  = 1
    32 x –12 = 30 make same bases by putting 1 = 30
    ∴ 2x – 12 = 0 3 equate exponents
    2x = 12
    x = 6 (3)
  3. 2x  = 0,125 convert to a common fraction
    2x  = 125  =  1  =  1  simplify
    1 000    8      23
    2x = 2−3 same bases
    ∴ x = –3 equate exponents (3)
  4. 10 x(x+1) = 100
    10 x(x+ 1) = 102 same bases
    ∴  x (x + 1) = 2 equate exponents
    x 2 + x – 2 = 0 set quadratic equation = 0
    (x + 2)(x – 1) = 0 factorise the trinomial
    x + 2 = 0 or x – 1 = 0 make each factor = 0
    x = –2 3 x = 1 (4)
  5. 52 + x – 5x  =  5x · 23 + 1
    52 + x − 5x – 5x · 23 = 1 like terms
    52 + x – 24·5x  = 1
    52. 5x – 24· 5 x  = 1 factorise (Common Factor)
    5x (52 – 24 )  =  1 33
    5x (1) =  1
    5x  = 50 ∴ x = 03 (4)
  6. 5x +  5x+1   = 30
    5x + 5x. 51 = 30 factorise
    5x (1 + 51 ) = 30 common factor 5x
    5x ( 6 ) = 30 3 divide 30 by 6
    5x = 5 same bases
    ∴ x = 1 3 equate exponents (4)
  7. 5x + 15.5 –x  =  2
    ∴ 5x + 15  =  2
    5x
    × 5 x ∴  5x. 5x + 5 x.15 =  2.5 x
    ∴  5x.5x + 15 = 2.5 x
    ∴  5x. 5x − 2.5x  + 15 = 0
    ∴ ( 5x − 5 ) (5x + 3 )  = 0
    ∴  5x  = 5 or 5x  =  − 3 (no solution)
    ∴ x = 1 (5)
  8. 20
    [31] 

Activity 9
Solve for x:
15

Activity 10
Solve these equations and check your solutions.
1. √3x + 4 − 5 = 0 (3)
2. √3x − 5 − x = 5 (5)
[8]

Solutions

  1. √3 x + 4  − 5 = 0
    √3 x + 4  = 5 (isolate the radical )
    ( √3 x + 4) 2  =  52  (square both sides of the equation)
    3x + 4 = 25
    3x = 21
    x = 7
    Check:
    LHS: √3(7) + 4   − 5
    =  √21 + 4   − 5
    =  √25   − 5
    = 0
    = RHS
    ∴ x = 7 is a solution (3)
  2. √3x − 5  − x = 5
    √3x − 5 = x − 5 (always isolate the radical first)
    ( √3x − 5 ) 2 =  ( x − 5)2 (square both sides)
    3x –5 = x2 –10x + 25  Remember: (x– 5 ) 2  ≠  x 2  + 25
    0 = x2 – 13x + 30  (quadratic equation, set  =  0)
    0  = (x – 10)(x – 3 ) (factorise the trinomial and make each factor  =  0)
    x  =  10 or x =  3
    Check your answer:
    If x = 10
    LHS:
    √3(10) − 5 − 10
    =  √25   − 10
    =  −5 = RHS
    If x = 3
    LHS
    √3(3) − 5 − 3
    =  √4  − 3
    =  −1 ≠  RHS (5)
    ∴ x ≠ 3 and only x = 10 is a solution.
    [8] 

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