1.1 Easier algebra with exponents

• Discuss the following two problems, and make two more rules to cover these cases.

1.1 $\frac{a^3}{a^3}$ 1.2 $\frac{a^3}{a^5}$

2 WHEN THE EXPONENT IS ZERO

• The answer to 1.1 is a ⁰ when we apply the rule for division.
• But we know that the answer must be 1, because the numerator and denominator are the same.
• So, we can say that anything with a zero as exponent must be equal to 1.
• The rule is now: a ⁰ = 1 also 1 = a ⁰ . A few examples:

3 ⁰ = 1; k ⁰ = 1; (ab ² ) ⁰ = 1 ; (n+1) ⁰ = 1; ${\left(\frac{a^{3}b}{{\left({\text{ab}}^2\right)}^2}\right)}^0=1$ and

1 = (anything) ⁰ , in other words, we can change a 1 to anything that suits us, if necessary!

3 WHEN THE EXPONENT IS NEGATIVE

• Look again at 1.2. According to the rule, the answer is a ^–2 . But what does it mean?
• $\frac{a^3}{a^5}=\frac{a\times a\times a}{a\times a\times a\times a\times a}=\frac{1}{a\times a}=\frac{1}{a^2}$ .
• So the rule is: $a^{-x}=\frac{1}{a^x}$ and vice versa.
• From now on we always try to write answers with positive exponents, where possible.
• The rule also means: $\frac{1}{a^{-x}}=a^x$ and vice versa. These examples are important:

${\text{ab}}^2{c}^{-3}=\frac{{\text{ab}}^2}{c^3}$

${\mathrm{2x}}^{-m}y=\frac{\mathrm{2y}}{x^m}$

end of CLASS WORK

HOMEWORK ASSIGNMENT

• Simplify without a calculator and leave answers without negative exponents.

1. $x^3{y}^2\times 3^2{x}^{2}y\times 2{\text{xy}}^4$

2. $\frac{x^4}{3{\text{xy}}^2}\times \frac{{\mathrm{6x}}^2}{x^{3}y}\times {\mathrm{2x}}^7{y}^3\times \frac{{\mathrm{4x}}^2{y}^4}{\mathrm{2y}}$

3. ${\left(5^x\right)}^3-{\left(5^3\right)}^x$

4. ${\left({\mathrm{2a}}^2{b}^5{c}^{3}d\right)}^2\times \mathrm{2a}{\left({\text{bc}}^{2}d\right)}^3\times 4\text{ab}{\left({\text{cd}}^3\right)}^2$

5. $6{\left(\frac{x^2}{y}\right)}^2\times {\left(\frac{\mathrm{2x}}{y^3}\right)}^3\times \frac{x^4}{3\text{xy}}$

6. ${\left({\mathrm{2a}}^2\right)}^3+{\left(\text{12}{a}^3\right)}^0-{\mathrm{8a}}^6$

7. $x^3{y}^{-4}\times {\left(3^{-1}{x}^2{y}^{-1}\right)}^{-3}\times {\left(2{\text{xy}}^3\right)}^2$

end of HOMEWORK ASSIGNMENT

CLASS WORK

• Let us make sure that we can replace variables with numerical values properly.

1 To calculate the perimeter of a rectangle with side lengths 17cm and 13,5 cm, we use normal formula:

• Perimeter = 2 [ length + breadth ]

• Put brackets in place of the variables: = 2 [ ( ) + ( ) ]
• Fill in the values: = 2 [ (17) + (13,5) ]
• Remove brackets and simplify = 2 [ 17 + 13,5 ]according to the rules: = 2 × 20,5
• Remember the units (if any): = 41 cm

2 What is the value of $\frac{x^3\times y^4\times x^2{y}^5}{x^4{y}^8}$ if x = 3 and y = 2 ?

• There are two possibilities: first substitute and then simplify or simplify first and then substitute. Here are both methods:

$\frac{x^3\times y^4\times x^2{y}^5}{x^4{y}^8}$ = $\frac{{\left(3\right)}^3\times {\left(2\right)}^4\times {\left(3\right)}^2\times {\left(2\right)}^5}{{\left(3\right)}^4\times {\left(2\right)}^8}$ = $\frac{\text{27}\times \text{16}\times 9\times \text{32}}{\text{81}\times \text{128}}$ = 3 × 2 = 6

$\frac{x^3\times y^4\times x^2{y}^5}{x^4{y}^8}$ = $\frac{x^3\times x^2\times y^4\times y^5}{x^4{y}^8}$ = $\frac{x^5\times y^9}{x^4\times y^8}$ = $x^{5-4}\times y^{9-8}$ = x × y = (3) × (2) = 6

• Without errors, the answers will be the same.

3.1 Calculate the perimeter of the square with side length 6,5 cm

3.2 Calculate the area of the rectangle with side lengths 17 cm and 13,5 cm.

3.3 If a = 5 and b = 1 and c = 2 and d = 3, calculate the value of: ${\left({\mathrm{2a}}^2{b}^5{c}^{3}d\right)}^2\times \mathrm{2a}{\left({\text{bc}}^{2}d\right)}^3\times 4\text{ab}{\left({\text{cd}}^3\right)}^2$ .

end of CLASS WORK

Assessment

Exponents ω

 good  average  not so good

Assessment

Memorandum

TEST 2

1. Scientific Notation

1.1 Write the following values as ordinary numbers:

1.1.1 2,405 × 10 ¹⁷

1.1.2 6,55 × 10 ^–9

1.2 Write the following numbers in scientific notation:

1.2.1 5 330 110 000 000 000 000

1.2.2 0,000 000 000 000 000 013 104

1.3 Do the following calculations and give your answer in scientific notation:

1.3.1 (6,148 × 10 ¹¹ ) × (9 230 220 000 000 000)

1.3.2 (1,767 × 10 ^–6 )  (6,553 × 10 ^–4 )

2. Exponents

Simplify and leave answers without negative exponents. (Do not use a calculator.)

2.1 ${\mathrm{3a}}^2\text{xy}{\left(3{\text{ab}}^2{x}^{2}y\right)}^3$

2.2 $\frac{{\left(a^0{b}^{0}c\right)}^3}{{\mathrm{6c}}^2{\left({\text{ab}}^3{c}^5\right)}^2}\times \frac{2{\left({\mathrm{3a}}^2{c}^3\right)}^0}{4{\text{abc}}^2}\times \text{18}{b}^4{\left({\mathrm{2a}}^3{c}^4\right)}^2$

3. Substitution

3.1 Simplify: 2 x ² y ³ + ( xy ) ² – 4 x

3.2 Calculate the value of 2 x ² y ³ + ( xy ) ² – 4 x as x = 4 and y = –2

4. Formulae

the formula for the area of a circle is: area = π r ² (r is the radius).

4.1 Calculate the areas of the following circles:

4.1.1 A circle with radius = 12 cm ; round answer to 1 decimal place.

4.1.2 A circle with a diameter 8 m ; approximate to the nearest metre.

TEST 2

1.1.1 240 500 000 000 000 000

1.1.2 0,000 000 006 55

1.2.1 5,330 110 × 10 ¹⁸

1.2.2 1,3104 × 10 ^–17

1.3.1 6,148 × 10 ¹¹ × 9,23022 × 10 ¹⁵

= 6,148 × 9,23022 × 10 ¹¹ × 10 ¹⁵

≈ 56,74 × 10 ²⁶

= 5,674 × 10 ²⁷

1.3.2 $\frac{\mathrm{1,}\text{767}\times {\text{10}}^{-6}}{\mathrm{6,}\text{553}\times {\text{10}}^{-4}}$ = $\frac{\mathrm{1,}\text{767}}{\mathrm{6,}\text{553}}\times {\text{10}}^{-6-(-4)}$ ≈ 0,26 × 10 ^–2 = 2,6 × 10 ^–1

2.1 3 ⁴ a ⁵ x ⁷ y ⁴ = 81 a ⁵ x ⁷ y ⁴

2.2 $\frac{c^3\times 2\times \text{18}{a}^6{b}^4{c}^8}{{\mathrm{6a}}^2{b}^6{c}^{\text{12}}\times 4{\text{abc}}^2}$ = $\frac{\text{36}{a}^6{b}^4{c}^{\text{11}}}{\text{24}{a}^3{b}^7{c}^{\text{14}}}$ = $\frac{{\mathrm{3a}}^3}{{\mathrm{2b}}^3{c}^3}$

3.1 2 x ² y ³ + x ² y ² – 4 x

3.2 2(4) ² (–2) ³ + (4) ² (–2) ² – 4(4) = 2(16)(–8) + (16)(4) – 16 = –256 + 64 – 16 = – 208

4.1.1 opp = π × 12 ² = 452,38934… ≈ 452,4 cm ²

4.1.2 opp = π × 4 ² = 50,26548… ≈ 50 m ²

CLASS WORK

The learners are likely to know the work in the first part already. Those who have not mastered the simplest laws of exponents now have an opportunity to catch up. For the rest it serves as revision in preparation for the new work in the second part.

1.1 4 × 4 × 4 ( p +2) × ( p +2) × ( p +2) × ( p +2) × ( p +2) etc.

1.2 7 ⁴ y ⁵ etc.

1.3 (–7) ⁶ = 7 ⁶ , (–7) ⁶ ≠ –7 ⁶ etc.

2.1 7 ¹⁴ (–2) ¹⁷ = –2 ¹⁷ etc.

3.1 a ^6– y 3 ² ( a + b ) ^p–12 a ⁰

4.1 a ⁵ a etc.

TUTORIAL

The tutorial should be done in silence in class in a fixed time. Recommendation: Mark it immediately – maybe the learners can mark one another’s work.

Answers: 1. a ³

2. xy

3. a ⁶ b ⁸ c ⁸

4. a ⁸ b ⁶

5. 4 x ⁸ y ⁹

6. 1

CLASS WORK

New for most learners in grade 9.

HOMEWORK ASSIGNMENT

Answers:

1. 18 x ⁶ y ⁷

2. 24 x ¹¹ y ³

3. 0

4. 32 a ⁶ b ¹⁴ c ¹⁴ d ¹¹

5. $\frac{\text{16}{x}^{\text{10}}}{y^{\text{12}}}$

6. 1

7. $\frac{\text{108}{y}^5}{x}$

CLASS WORK

Substitution gives lots of trouble because it looks so easy. Learners who leave out steps (or don’t write them down) often make careless mistakes. Force them to use brackets.

2. They should conclude that simplification should come first – after all, this is why we simplify!

3.1 26 cm

3.2 229,5 cm ²

3.3 ≈ 1,45 × 10 ¹⁵