EXPONENTS AND SURDS QUESTIONS AND ANSWERS GRADE 12

EXPONENTS AND SURDS QUESTIONS AND ANSWERS GRADE 12 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.

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

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

[10]

Solutions

√8 × √8 = √8×2 = √16 = 4 (1)
³√4 × ³√2  =  ³√4×2 = ³√8  = 2 (2)
9+ √45 = 9+3√5 = 3( 3+ √5 ) = 3 + √5 (3)
3 3 3
( 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 number	Real 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?

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

[33]

Solutions

1. Complete the table for each number by marking the correct columns.
	Nonreal number	Real 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)

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

Activity 3
Calculate

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

Solutions

−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)
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.

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

Solutions

a ^-3 = b²
b^-2 a³
4a⁷b⁻⁴c⁻¹ =  4a⁷d²
d^-2e5   b⁴c¹e⁵
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.

³√5
⁴√16
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=  a^4n–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:

27^{3 – 2x}.9^x-1
81^2-x
6.5^{x +1} – 2.5^{x +2}
5^x+3
2²⁰⁰⁹ − 2²⁰¹²
2²⁰¹⁰[13]

Solutions

. 27 ^3−2x. 9^x−1= (3³)^3−2x. (3²)^x−1= 3^9−6x. 3^2x−2
81^2-x (3⁴)^2−x3^8-4x
= 3 ^{9−6x+2x−2−8+4x}
= 3 ⁻¹  = 1
3 (4)
6.5^{x +1} – 2.5^{x +2} =  6.5^x .5¹  − 2.5^x . 5²
5^x+3 5^x. 5³
=  30 − 50 = − 20 = − 4
125 125 25 (4)
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:

3 ( 9^x+3 ) = 27^2x–1
3^2x–12 = 1
2^x = 0,125
10^{x ( x+1 )} = 100
5^x + 5^x+1  = 30
5 ^2+x – 5^x = 5^x. 23 + 1
5^x+ 15.5 ^−x  =  2
$19$
{31]

Solutions
Remember: When adding or subtracting terms, you need to factorise first.

3(9^x+3 ) = 27^2x–1
3¹(3²)^{x + 3} = (3³)^{2x – 1} prime bases
3^{1+2 x+6}  = 3^{6 x–3} same bases
∴ 7 + 2x = 6x – 3 equate exponents
–4x = – 3 – 7
x = −10 = 5
− 4 2
=(3)
3^{2 x −12}  = 1
3^{2 x –12} = 3⁰ make same bases by putting 1 = 3⁰
∴ 2x – 12 = 0 3 equate exponents
2x = 12
x = 6 (3)
2^x  = 0,125 convert to a common fraction
2^x  = 125 = 1 = 1 simplify
1 000 8 2³
2^x = 2⁻³ same bases
∴ x = –3 equate exponents (3)
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^{2 + x} – 5^x  =  5^x · 23 + 1
5^{2 + x} − 5^x – 5^x· 23 = 1 like terms
5^{2 + x} – 24·5^x  = 1
5². 5x – 24· 5 x  = 1 factorise (Common Factor)
5^x (5² – 24 )  =  1 33
5^x (1) =  1
5^x  = 5⁰ ∴ x = 03 (4)
5^x+  5^x+1   = 30
5^x + 5^x. 5¹ = 30 factorise
5^x (1 + 5¹ ) = 30 common factor 5x
5^x ( 6 ) = 30 3 divide 30 by 6
5^x = 5 same bases
∴ x = 1 3 equate exponents (4)
5^x + 15.5 ^–^x  =  2
∴ 5^x + 15 =  2
5^x
× 5 ^x ∴  5^x. 5^x + 5 ^x.15 =  2.5 ^x
∴  5^x.5^x + 15 = 2.5 ^x
∴ 5^x. 5^x − 2.5^x  + 15 = 0
∴ ( 5^x − 5 ) (5^x + 3 )  = 0
∴  5^x  = 5 or 5^x  =  − 3 (no solution)
∴ x = 1 (5)
$20$
[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

√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)
√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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