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6th Year

Maths

Ordinary Level

Strand 4 of 5 Topics:

Algebra

Complex Numbers

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6th Year Easter Revision CoursesSUBJECT LEVEL DATES TIME

Accounting H Monday 21st March – Friday 25th March 8:00am - 9:30am

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Applied Maths H Monday 28th March – Friday 1st April 8:00am - 9:30am

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Business H Monday 28th March – Friday 1st April 8:00am - 9:30am

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Chemistry Course B* H Monday 28th March – Friday 1st April 2:00pm - 3:30pm

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Maths Paper 1* H Monday 21st March – Friday 25th March 12:00pm - 1:30pm

Maths Paper 1* H Monday 28th March – Friday 1st April 10:00am - 11:30am

Maths Paper 1* H Monday 28th March – Friday 1st April 2:00pm - 3:30pm

Maths Paper 2* H Monday 21st March – Friday 25th March 10:00am - 11:30am

Maths Paper 2* H Monday 21st March – Friday 25th March 2:00pm - 3:30pm



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Spanish H Monday 28th March – Friday 1st April 10:00am - 11:30am

6th Year Oral Preparation CoursesSUBJECT LEVEL DATES TIME

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Business Studies H Monday 28th March – Friday 1st April 8:00am - 9:30am

English H Monday 21st March – Friday 25th March 8:00am - 9:30am

English H Monday 28th March – Friday 1st April 2:00pm - 3:30pm

French H Monday 28th March – Friday 1st April 12:00pm - 1:30pm

Geography H Monday 28th March – Friday 1st April 12:00pm - 1:30pm

German H Monday 21st March – Friday 25th March 8:00am - 9:30am

History H Monday 21st March – Friday 25th March 4:00pm - 5:30pm

Irish H Monday 28th March – Friday 1st April 2:00pm - 3:30pm

Maths H Monday 21st March – Friday 25th March 10:00am - 11:30am

Maths H Monday 21st March – Friday 25th March 12:00pm - 1:30pm

Maths H Monday 28th March – Friday 1st April 10:00am - 11:30am

Maths O Monday 28th March – Friday 1st April 12:00pm - 1:30pm

Science H Monday 28th March – Friday 1st April 2:00pm - 3:30pm

Science H Monday 21st March – Friday 25th March 2:00pm - 3:30pm

Spanish H Monday 21st March – Friday 25th March 12:00pm - 1:30pm

2nd Year Easter Revision CoursesSUBJECT LEVEL DATES TIME

Maths H Monday 21st March – Friday 25th March 2:00pm - 3:30pm

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©The Dublin School of Grinds Page 1

Strand 4 is worth 8% to 16% of The Leaving Cert.

Contents: Algebra

1. Evaluating expressions ........................................................................................................................................................ 2

2. Simplifying algebraic expressions .................................................................................................................................. 3

3. Simplifying algebraic fractions ........................................................................................................................................ 5

4. Solving linear equations ..................................................................................................................................................... 7

5. Solving linear inequalities .................................................................................................................................................. 8

6. Factorising ............................................................................................................................................................................. 11

7. Solving quadratic equations ........................................................................................................................................... 16

8. Forming a quadratic equation from the roots........................................................................................................ 21

9. Finding a letter on its own .............................................................................................................................................. 22

10. Simultaneous equations ................................................................................................................................................... 24

11. Laws of Indices (powers) ................................................................................................................................................ 31

12. Index equations ................................................................................................................................................................... 33

13. Past and probable exam questions ............................................................................................................................. 36

14. Solutions to Algebra ........................................................................................................................................................... 53

Complex Numbers 1. The imaginary number ..................................................................................................................................................... 78

2. Complex numbers ............................................................................................................................................................... 79

3. Adding and subtracting complex numbers ............................................................................................................. 80

4. Multiplying complex numbers ...................................................................................................................................... 81

5. The complex conjugate (the guy with the hat) ...................................................................................................... 84

6. Division of complex numbers ........................................................................................................................................ 85

7. Equality of complex numbers ........................................................................................................................................ 86

8. Quadratic equations with complex numbers ......................................................................................................... 88

9. Argand diagrams ................................................................................................................................................................. 91

10. The modulus of complex numbers (the guy with the jacket).......................................................................... 93

11. Transformations of complex numbers ...................................................................................................................... 96

12. z, z2, z3, z4, z5, z6, etc. ......................................................................................................................................................... 102

13. Past and probable exam questions ........................................................................................................................... 103

14. Solutions to Complex Numbers .................................................................................................................................. 114


Algebra Algebra is worth 4% to 8% but really you use many of the skills learned in algebra throughout the rest of the course. It appears on Paper 1

1. Evaluating expressions

“Evaluating expressions” means “finding the value of something”.

Example 1 Find the value of 3(2x – y) when x = 3 and y = −2. Solution

3(2𝑥 − 𝑦) = 3(2(3) − (−2))

= 3(6 + 2) = 3(8) = 24

Question 1.1 Find the value of 5p + 2(q – 4) when p = 6 and q = –8.

Example 2

Find the value of 𝑥+𝑦−1

𝑥−𝑦+1 when x =

1

4 and y =

5

8

Solution

𝑥 + 𝑦 − 1

𝑥 − 𝑦 + 1=

14

+58

− 1

14

−58

+ 1

=

28

+58

−88

28

−58

+88

=−

18

58

= −1

8×

8

5

= −1

5

Question 1.2

Evaluate 𝑝𝑞−𝑟

2, when p = 3, q =

2

3 and r = 1


2. Simplifying algebraic expressions

To simplify an algebraic expression:

Step 1. Remove the brackets

Step 2. Add the like terms

Some example of like terms:

3x and 5x

-3ab and 7ba

5y2 and 2y2

Example 1

Simplify 3x – 2x(x – 2) Solution

3𝑥 − 2𝑥(𝑥 − 2) = 3𝑥 − 2𝑥2 + 4𝑥 = 7𝑥 − 2𝑥2

Question 2.1

Simplify 4x(x – 3) –x(x + 3)

Example 2 Simplify (2x + 5)(x – 3) Solution

(2𝑥 + 5)(𝑥 − 3) = 2𝑥2 − 6𝑥 + 5𝑥 − 15

= 2𝑥2 − 𝑥 − 15

Question 2.2

Simplify (3x – 2)(4x – 1)

When you have an expression in a bracket squared, always write the two brackets out before multiplying.

Example 3 Simplify (5a – 2b)2 Solution

(5𝑎 − 2𝑏)2 = (5𝑎 − 2𝑏)(5𝑎 − 2𝑏)

= 25𝑎2 − 10𝑎𝑏 − 10𝑎𝑏 + 4𝑏2

= 25𝑎2 − 20𝑎𝑏 + 4𝑏2

Note: The arrows show each step in

the multiplication. You should draw

these arrows for each step to make

sure you complete each step


Question 2.3 Simplify (2x + 3)2

Example 4 Simplify (2x – 3)(x2 – 4x + 1) Solution

(2𝑥 − 3)(𝑥2 − 4𝑥 + 1) = 2𝑥3 − 8𝑥2 + 2𝑥 − 3𝑥2 + 12𝑥 − 3

= 2𝑥3 − 11𝑥2 + 14𝑥 − 3 Question 2.4 Simplify (x + 4)(2x2 – 3x + 5)


3. Simplifying algebraic fractions

The Syllabus requires you to know how to simplify algebraic fractions:

Example 1 Express

𝑥+1

2+

𝑥+5

3 in its simplest form

Solution

(𝑥 + 1)

2 +

(𝑥 + 5)

3

=3(𝑥 + 1) + 2(𝑥 + 5)

(2)(3)

=3𝑥 + 3 + 2𝑥 + 10

6

=5𝑥 + 13

6

Question 3.1

Express 5𝑥−1

4+

2𝑥−3

3 in its simplest form.

Example 2

Express 2

(𝑥+3)+

3

(𝑥+5) in its simplest form

Solution We following the same steps as the previous example even though x is on the bottom.

2

(𝑥 + 3)+

3

(𝑥 + 5)

=2(𝑥 + 5) + 3(𝑥 + 3)

(𝑥 + 3)(𝑥 + 5)

=2𝑥 + 10 + 3𝑥 + 9

(𝑥 + 3)(𝑥 + 5)

=5𝑥 + 19

(𝑥 + 3)(𝑥 + 5)

Note: You don’t have to multiply out the bottom brackets when there are letters involved on the bottom.

Multiply the bottom by the

bottom to get the common

denominator (2x3)

Then 3(x+1) on the left.

Then 2(x+5) on the right.

Then multiply out the brackets

and add the terms.

Multiply the bottom by the bottom to get the common denominator (x + 3)(x + 5)

Then 2(x + 5) on the left

Then 3(x + 3) on the right

Then multiply out the brackets and

add the terms.


Question 3.2

Express 5

(2𝑥+1)+

4

(2𝑥−1) in its simplest form

Solution


4. Solving linear equations

When rearranging equations we must always do the same thing to each side to keep the equation balanced.

Example 1 Find the value of x for the equation 6(2x – 5) = 2(6 + 3x) Solution

6(2𝑥 − 5) = 2(6 + 3𝑥) 12𝑥 − 30 = 12 + 6𝑥

12𝑥 = 12 + 6𝑥 + 30 12𝑥 = 42 + 6𝑥

12𝑥 − 6𝑥 = 42 6𝑥 = 42

𝑥 = 7

Question 4.1

Find the value of x for the equation 10(x + 4) = 3(2x + 5) +1

Note: To remove fractions from linear equations multiple EACH term by the lowest common denominator.

Example 2

Solve the equation 𝑥−7

2=

𝑥+3

6

Solution

𝑥 − 7

2=

𝑥 + 3

6

6(𝑥 − 7)

2=

6(𝑥 + 3)

6

3(𝑥 − 7) = 𝑥 + 3 3𝑥 − 21 = 𝑥 + 3

3𝑥 = 𝑥 + 3 + 21 3𝑥 − 𝑥 = 24

2𝑥 = 24 𝑥 = 12

Question 4.2

Solve the equation 2𝑥−1

3+

𝑥

4=

3

2


5. Solving linear inequalities

There are four symbols the Examiner can use for inequalities:

1. > is greater than.

2. ≥ is greater than or equal to.

3. < is less than.

4. ≤ is less than or equal to.

When solving linear inequalities we solve it just like equations except for one important rule:

Rule: If you change the signs of both sides, it reverses the direction of the inequality:

−𝑥 < −3

𝑥 > 3

You also need to know the different types of numbers:

Type 1. The set of natural numbers, N = {1, 2, 3, 4, … }

Hint: remember N for Normal.

Type 2. The set of integers, Z = { … , −3, −2, −1, 0, 1, 2, 3}

Type 3. The set of real numbers, R contains all the numbers on the number line. A full heavy line is used for R.

Hint: Remember R for Random.

Note: Use a solid dot for ≤ and ≥.

Use a hollow dot for < and >.

Now we will look at some examples:

Example 1 Solve the following inequality, and show the set on the number line. 3𝑥 − 11 ≤ 1 where 𝑥 ∈ 𝑁. Solution

3𝑥 − 11 ≤ 1 3𝑥 ≤ 1 + 11 3𝑥 ≤ 12

𝑥 ≤ 4

Example 2 Solve the following inequality, and show the set on the number line. 2 − 5𝑥 ≤ −13 where 𝑥 ∈ 𝑅. Solution

2 − 5𝑥 ≤ −13 −5𝑥 ≤ −13 − 2

−5𝑥 ≤ −15 5𝑥 ≥ 15

𝑥 ≥ 3

(change the sign and reverse the inequality)


Question 5.1

Solve the following inequality, and show the set on the number line.

𝑥 + 13 > 4 − 2𝑥 where 𝑥 ∈ 𝑍.

Example 3 Solve the following inequality, and show the set on the number line. 3𝑥

5−

𝑥

2<

3

10 where 𝑥 ∈ 𝑅

Solution

3𝑥

5−

𝑥

2<

3

10

10 (3𝑥

5) − 10 (

𝑥

2) < 10 (

3

10)

2(3𝑥) − 5(𝑥) < 3

6𝑥 − 5𝑥 < 3

𝑥 < 3

Question 5.2

Solve the following inequality, and show the set on the number line.

5𝑥−6

2≤ 7 where 𝑥 ∈ 𝑅.


Example 4 Find the solution set A of 11 ≥ 3𝑥 + 2, 𝑥 ∈ 𝑅. Find the solution set B of 3𝑥 + 2 > −7, 𝑥 ∈ 𝑅 Find A∩B and illustrate your answer on a number line. Solution First solve each inequality,

11 ≥ 3𝑥 + 2 11 − 2 ≥ 3𝑥

9 ≥ 3𝑥 3 ≥ 𝑥 𝑥 ≤ 3

3𝑥 + 2 > −7 3𝑥 > −7 − 2

3𝑥 > −9 𝑥 > −3

To find A∩B we combine these results we get −3 < 𝑥 ≤ 3.

Question 5.3

Find the solution set A of 2𝑥 − 3 ≤ 7, 𝑥 ∈ 𝑅.

Find the solution set B of 3𝑥 + 1 ≥ 4, 𝑥 ∈ 𝑅.

Find A∩B and illustrate your answer on a number line.


6. Factorising

We will look at three methods for factorising.

1. Highest Common Factor

Example 1 Factorise the following terms.

i) 3x2 + x ii) 4a2 – 16a

Solution

i) 3𝑥2 + 𝑥 = 𝑥(3𝑥 + 1)

ii) 4𝑎2 − 16𝑎 = 4𝑎(𝑎 − 4)

Question 6.1

Factorise the following terms.

i) 5x2 + 8x

ii) 2p2 – 6p

iii) 6y2 + 30y

2. Difference of two squares

You need to remember the formula:

𝑎2 − 𝑏2 = (𝑎 − 𝑏)(𝑎 + 𝑏)

Example 2 Factorise x2 – y2

Solution

𝑥2 − 𝑦2 = (𝑥 − 𝑦)(𝑥 + 𝑦)

Sometimes you will need to write each term as a square.

Example 3 Factorise 9x2 – 16y2 Solution

9𝑥2 − 16𝑦2 = (3𝑥)2 − (4𝑦)2 = (3𝑥 − 4𝑦)(3𝑥 + 4𝑦)


Question 6.2

Factorise the following expression i) x2 – 16

ii) x2 – 81y4

iii) x2 – 1

3. Quadratics

Within quadratics there are 5 different types:

Type 1: positive and positive

Example 4 Find the factors of the expression x2 + 7x + 10. Solution First, find the factors of the last term: Now ask yourself which of these options sum up to the middle number? The answer is 2 and 5 because 2 + 5 = 7

=> (𝑥 + 2)(𝑥 + 5)

Question 6.3

Factorise x2 + 10x + 21

Factors of 10

1×10 2×5

−1×−10 −2×−5


Type 2: negative and positive

Example 5 Factorise x2 – 9x + 20 Solution First, find the factors of the last term: Now ask yourself which of these options sum up to the middle number? The answer is −4 and −5 because −4 − 5 = −9

=> (𝑥 − 4)(𝑥 − 5)

Question 6.4

Factorise x2 – 5x + 6

Type 3: positive and negative

Example 6 Factorise x2 + 2x – 15 Solution First, find the factors of the last term: Now ask yourself which of these options sum up to the middle number? The answer is −3 and 5 because −3 + 5 = 2

=> (𝑥 − 3)(𝑥 + 5)

Question 6.5

Factorise the x2 + 3x – 18

Factors of 20

1×20 2×10 4×5

−1× −20 −2×−10 −4×−5

Factors of −15

−1×15 −3×5

1×−15 3×−5


Type 4: negative and negative

Example 7 Factorise x2 −3x – 18 Solution First, find the factors of the last term: Now ask yourself which of these options sum up to the middle number? The answer is 3 and −6 because 3 − 6 = −3

=> (𝑥 + 3)(𝑥 − 6)

Question 6.6

Factorise the x2 − 5x – 24.

Type 5: Number in front of the x2

Example 8 Factorise: 5x2 + 8x + 3. Solution First, find the factors of the last term: For this type we use trial and error. The potential answers are. Option 1:

(5𝑥 + 1)(𝑥 + 3)

or

Option 2: (5𝑥 − 1)(𝑥 − 3)

or

Option 3 (5𝑥 + 3)(𝑥 + 1)

or

Option 4: (5𝑥 − 3)(𝑥 − 1)

If you multiply each of these out you will get: Option 1: 5x2 + 16x + 3 Option 2: 5x2 − 16x + 3 Option 3: 5x2 + 8x + 3 Option 4: 5x2 − 8x + 3 => Option 3 is the answer, i.e. (5𝑥 + 3)(𝑥 + 1)

Factors of −18

−1×18 −2×9 −3×6

1 ×−18 2 ×−9 3 ×−6

Factors of 3

1×3 −1 ×−3


Question 6.7

Factorise the following expressions:

i) 2x2 − 11x + 5

ii) 3x2 – x − 2


7. Solving quadratic equations

A quadratic equation has an equals sign in it. There are several methods to solve these, but ignore everything you’ve

heard and just use the formula from the cover of your log tables:

𝑥 =−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎

Note: a is the number in front of the of the x2

b is the number in front of the of the x

c is the number that has no x2 or x beside it (the fancy name for this is the ‘constant’).

Terms and conditions: All terms must be on one side of the equals sign.

Hint: Put all numbers in brackets.

Example 1 Solve the equation x2 – 6x + 4 = 0 correct to two decimal places. Solution

𝑎 = 1, 𝑏 = −6, 𝑐 = 4

𝑥 =−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎

𝑥 =−(−6) ± √(−6)2 − 4(1)(4)

2(1)

𝑥 =6 ± √36 − 16

2

𝑥 =6 ± √20

2

𝑥 =6 + √20

2 𝑜𝑟 𝑥 =

6 − √20

2

𝑥 = 5.24 𝑜𝑟 𝑥 = 0.76

Question 7.1

Solve the following equation correct to two decimal places 3x2 + 7x – 4 = 0


What if the Examiner asks you to give your answer in the form 𝑎 ± √𝑏

Example 2

Find the roots of the equation x2 – 6x + 4 = 0 giving your answer in the form 𝑎 ± √𝑏. Solution

𝑎 = 1, 𝑏 = −6, 𝑐 = 4

𝑥 =−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎

𝑥 =−(−6) ± √(−6)2 − 4(1)(4)

2(1)

𝑥 =6 ± √36 − 16

2

𝑥 =6 ± √20

2

𝑥 =6 ± √4√5

2

𝑥 =6 ± 2√5

2

𝑥 =6

2±

2√5

2

𝑥 = 3 ± √5

Question 7.2

Find the roots of the equation x2 + 6x + 7 = 0 giving your answer in the form 𝑎 ± √𝑏.


Question 7.3

Find the roots of the equation 3x2 – 6x + 2 = 0 giving your answer in the form 𝑎±√𝑏.

𝑐.


The Examiner can also give us equations involving algebraic fractions and we will have to turn them into quadratic

equations:

Example 3

Solve the equation

4

2𝑥 − 3−

2

𝑥=

1

9

Solution

The lowest common denominator is (9)(2𝑥 − 3)(𝑥) we get this by multiplying the bottom three terms.

We then multiply each term by this:

4(9)(2𝑥 − 3)(𝑥)

(2𝑥 − 3)−

2(9)(2𝑥 − 3)(𝑥)

(𝑥)=

1(9)(2𝑥 − 3)(𝑥)

9

36(𝑥) − 18(2𝑥 − 3) = (2𝑥 − 3)(𝑥)

36𝑥 − 36𝑥 + 54 = 2𝑥2 − 3𝑥

54 = 2𝑥2 − 3𝑥

−2𝑥2 + 3𝑥 + 54 = 0

2𝑥2 − 3𝑥 − 54 = 0

Now solve the quadratic equation

𝑎 = 2, 𝑏 = −3, 𝑐 = −54

𝑥 =−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎

𝑥 =−(−3) ± √(−3)2 − 4(2)(−54)

2(2)

𝑥 =3 ± √9 + 432

4

𝑥 =3 ± √441

4

𝑥 =3 ± 21

4

𝑥 =3 + 21

4 𝑜𝑟 𝑥 =

3 − 21

4

𝑥 =24

4 𝑜𝑟 𝑥 =

−18

4

𝑥 = 6 𝑜𝑟 𝑥 = −9

2


Question 7.4

Find the values of x for the following equation correct to 2 decimal places.

2

3𝑥 − 4−

1

7𝑥 + 1=

1

2


8. Forming a quadratic equation from the roots

The Examiner can ask you to form quadratics when given the roots of an equation.

Example 1 Form the quadratic whose two roots −2 and 3. Solution x = −2 and x = 3

Therefore,

x + 2 = 0 and x − 3 = 0

So (x + 2) and ( x − 3) are factors of the quadratic equation.

=> (𝑥 + 2)(𝑥 − 3) = 0

𝑥2 − 3𝑥 + 2𝑥 − 6 = 0

The equation is:

𝑥2 − 𝑥 − 6 = 0

Question 8.1

Form the quadratic whose two roots are 5 and 3


9. Finding a letter on its own

The fancy names for this section are: ‘Manipulation of formula’ or ‘Finding the subject’.

Example 1

If x = 5z +2y, make y the subject of the formula

Solution

𝑥 = 5𝑧 + 2𝑦

𝑥 − 5𝑧 = 2𝑦 (𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡𝑖𝑛𝑔 5𝑧 𝑓𝑟𝑜𝑚 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠)

2𝑦 = 𝑥 − 5𝑧 (𝑟𝑒𝑣𝑒𝑟𝑠𝑖𝑛𝑔 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛)

=> 𝑦 =𝑥−5𝑧

2 (𝑑𝑖𝑣𝑖𝑑𝑖𝑛𝑔 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 𝑏𝑦 2)

Question 9.1

i) If a = 8b -6 make b the subject of the formula.

ii) If 3x – b = 4c make x the subject of the formula.

When an equation involves a fraction the first step is to remove it:

Example 2

Given that 𝑐 =𝑎+2𝑏

3, express b in terms of a and c.

Solution

𝑐 =𝑎 + 2𝑏

3

3𝑐 = 𝑎 + 2𝑏 (𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑦𝑖𝑛𝑔 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 𝑏𝑦 3)

3𝑐 − 𝑎 = 2𝑏 (𝑆𝑢𝑏𝑡𝑟𝑎𝑐𝑡𝑖𝑛𝑔 𝑎 𝑓𝑟𝑜𝑚 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠)

3𝑐 − 𝑎

2= 𝑏 (𝐷𝑖𝑣𝑖𝑑𝑖𝑛𝑔 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 𝑏𝑦 3)

=> 𝑏 =3𝑐 − 𝑎

2


Question 9.2

Given that 𝑥 =𝑦−4𝑧

6, express y in terms of x and z.

A more difficult and likely Leaving Cert type question the Examiner likes is as follows:

Example 3

If 𝑎 =7𝑏+5

𝑏−2 make b the subject of the formula.

Solution

𝑎 =7𝑏 + 5

𝑏 − 2

𝑎 =7𝑏 + 5

(𝑏 − 2) (𝑝𝑢𝑡 𝑏 − 2 𝑖𝑛 𝑏𝑟𝑎𝑐𝑘𝑒𝑡𝑠)

𝑎(𝑏 − 2) = 7𝑏 − 5 (𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 𝑏𝑦 (𝑏 − 2))

𝑎𝑏 − 2𝑎 = 7𝑏 − 5 (𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑜𝑢𝑡 𝑡ℎ𝑒 𝑏𝑟𝑎𝑐𝑘𝑒𝑡𝑠)

𝑎𝑏 = 7𝑏 − 5 + 2𝑎 (𝑎𝑑𝑑 2𝑎 𝑡𝑜 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠)

𝑎𝑏 − 7𝑏 = 2𝑎 − 5 (𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡 7𝑏 𝑓𝑟𝑜𝑚 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 𝑡𝑜 𝑔𝑒𝑡 𝑡ℎ𝑒 𝑏′𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑙𝑒𝑓𝑡)

𝑏(𝑎 − 7) = 2𝑎 − 5 (𝑓𝑎𝑐𝑡𝑜𝑟𝑖𝑠𝑒 𝑜𝑢𝑡 𝑡ℎ𝑒 𝑏 𝑜𝑛 𝑡ℎ𝑒 𝑙𝑒𝑓𝑡)

=> 𝑏 =2𝑎 − 5

𝑎 − 7 (𝑑𝑖𝑣𝑖𝑑𝑒 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 𝑏𝑦 (𝑎 − 7))

Question 9.3

If 2𝑎 =3𝑘

𝑘−3 make k the subject of the formula.


10. Simultaneous equations

The Leaving Cert syllabus requires you to know how to do 2 types of simultaneous equations. This section also comes

up in coordinate geometry of the line and the circle.

Type 1: Two linear equations

What you need to do is eliminate one of the unknowns. This method is used in coordinate geometry to find the point of

intersection of two lines.

Example 1 Solve the following simultaneous equations.

2𝑥 + 𝑦 = 9 3𝑥 + 5𝑦 = 17

Solution

First, name your equations

(1) 2𝑥 + 𝑦 = 9

(2) 3𝑥 + 5𝑦 = 17

Next, multiply equation (1) by –5 to make the y values the same size but opposite signs.

(1) × (−5) − 10𝑥 − 5𝑦 = −45

(2) 3𝑥 + 5𝑦 = 17

−7𝑥 = −28

7𝑥 = 28

𝑥 = 4

Now substitute x = 4 into equation (1)

(1) 2(4) + 𝑦 = 9

8 + 𝑦 = 9

𝑦 = 9 − 8

𝑦 = 1

=> 𝑥 = 4 𝑎𝑛𝑑 𝑦 = 1

Note: If the Examiner asks you to “verify your answers”, you simply sub your answers back into both equations and

show they are correct.

(add equations 1 & 2)


Question 10.1

Solve the following simultaneous equations and verify your answers.

𝑝 − 𝑞 = 4

𝑝 + 2𝑞 = 1


Example 2 Solve the following simultaneous equations.

3𝑥 + 4𝑦 = 18 4𝑥 − 3𝑦 − 1

Solution First, name your equations

(1) 3𝑥 + 4𝑦 = 18

(2) 4𝑥 − 3𝑦 = −1

Next, multiply equation (1) by 3 and (2) by 4 to make the y values the same size but opposite signs.

(1) × 3 9𝑥 + 12𝑦 = 54

(2) × 4 16𝑥 − 12𝑦 = −4

25𝑥 = 50

𝑥 = 2

Now substitute x = 2 back into equation (1)

(1) 3(2) + 4𝑦 = 18

6 + 4𝑦 = 18

4𝑦 = 18 − 6

4𝑦 = 12

𝑦 = 3

=> 𝑥 = 2 𝑎𝑛𝑑 𝑦 = 3

Question 10.2 Solve the simultaneous equations

3𝑟 + 2𝑠 = 13 4𝑟 − 3𝑠 = 6



In the Leaving Cert exam you can be given simultaneous equations with fractions in them.

Let’s look at an example:

Example 3 Solve the following simultaneous equations

2𝑥 − 3𝑦 = 24 5𝑥

3−

𝑦

2= 12

Solution First, number each of the equations.

1) 2𝑥 − 3𝑦 = 24

2) 5𝑥

3−

𝑦

2= 12

Next remove the fractions from equation two by multiplying by the lowest common denominator. 1) 2𝑥 − 3𝑦 = 24

2) 6 (5𝑥

3) − 6 (

𝑦

2) = (6)12

=> 10𝑥 − 3𝑦 = 72

Proceed like the previous example to find one of the unknowns.

1) × (–1) −2𝑥 + 3𝑦 = −24

2) 10𝑥 − 3𝑦 = 72

8𝑥 = 48

𝑥 =48

8

𝑥 = 6 Substituting x = 6 into equation 1

1) 2𝑥 − 3𝑦 = 24

2(6) − 3𝑦 = 24 12 − 3𝑦 = 24 −3𝑦 = 24 − 12 −3𝑦 = 12

−𝑦 =12

3

−𝑦 = 4 𝑦 = −4


(Multiplying across by 6 to

remove fractions)


Question 10.3

Solve the following simultaneous equations

2𝑎 − 5𝑏 = 19

4𝑎

2+

5𝑏

3= −1


Type two: One linear & one quadratic

You need to know this method of simultaneous equations to find the point of intersection between a circle and a line.

Example 4 𝑥 − 𝑦 + 1 = 0

𝑥2 + 𝑦2 = 25 Solution

First, number the equations

1) 𝑥 − 𝑦 + 1 = 0

2) 𝑥2 + 𝑦2 = 25

Rearrange 1) to find y on its own.

𝑥 − 𝑦 + 1 = 0

−𝑦 = −𝑥 − 1

𝑦 = 𝑥 + 1

Next, substitute x + 1 into equation 2) for y.

𝑥2 + 𝑦2 = 25

𝑥2 + (𝑥 + 1)2 = 25

𝑥2 + (𝑥 + 1)(𝑥 + 1) = 25

𝑥2 + 𝑥2 + 1𝑥 + 1𝑥 + 1 = 25

2𝑥2 + 2𝑥 + 1 = 25

Rearrange the equation to have everything on the left.

2𝑥2 + 2𝑥 + 1 − 25 = 0

2𝑥2 + 2𝑥 − 24 = 0


𝑎 = 2, 𝑏 = 2, 𝑐 = −24

𝑥 =−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎

𝑥 =−(2) ± √(2)2 − 4(2)(−24)

2(2)

𝑥 =−2 ± √4 + 192

4

𝑥 =−2 ± √196

4

𝑥 =−2 ± 14

4

𝑥 =−2 + 14

4 𝑜𝑟 𝑥 =

−2 − 14

4

𝑥 =12

4 𝑜𝑟 𝑥 =

−16

4

𝑥 = 3 𝑜𝑟 𝑥 = −4

Finally we substitute the x values back into equation 1. x = 3

1) 𝑥 − 𝑦 + 1 = 0 (3) − 𝑦 + 1 = 0 −𝑦 = −3 − 1

−𝑦 = −4 𝑦 = 4

=> 𝑥 = 3 & 𝑦 = 4

x = −4 1) 𝑥 − 𝑦 + 1 = 0

(−4) − 𝑦 + 1 = 0 −𝑦 = 4 − 1 −𝑦 = 3

𝑦 = −3 => 𝑥 = −4 & 𝑦 = −3


Question 10.4 Solve the following simultaneous equations

𝑥 − 𝑦 − 4 = 0

𝑦2 + 3𝑥 = 16 Solution


11. Laws of Indices (powers)

On page 21 of the log tables you will see the following nine laws:

Question 11.1

i) Write 16 in the form 2n.

ii) Write 27 in the form 3n.

iii) Write 125 in the form 5n.

iv) Write 49 in the form 7n.

Solution


Question 11.2

Simplify each of the following

i) 𝑥2 × 𝑥3

ii) 𝑥8

𝑥5

iii) (𝑦4)2

Example 1

Simplify 125

√5 writing it in the form 𝑎𝑏 .

Solution

125

√5=

53

512

= 53 × 5−12

= 5(3−12

)

= 𝟓𝟓𝟐

Question 11.3

Simplify 81

√3 writing it in the form 𝑎𝑏 .


12. Index equations

The Examiner can ask you questions where we must use the indices (powers) to find the unknown.

Example 1 Solve the equation 53𝑥−1 = 58. Solution

53𝑥−1 = 58

If both sides have the same number at the base we can say the powers are equal (both have 5 as the base).

3𝑥 − 1 = 8

3𝑥 = 8 + 1

3𝑥 = 9

3𝑥

3=

9

3

𝑥 = 3

Question 12.1

Solve the equations

i) 35𝑥 = 320

ii) 45𝑥−1 = 44

Example 2 Solve the equation 32𝑥+1 = 27 Solution This time the we must write the number on the right with the same base as the left.

32𝑥+1 = 33

Now they have the same base we can put the indices equal to each other

=> 2𝑥 + 1 = 3

2𝑥 = 3 − 1

2𝑥 = 2

𝑥 = 1


Question 12.2 Solve the equation 2𝑥−2 = 16

Example 3 Solve the equation 272𝑥 = 9𝑥+6 Solution Write each side to the same base.

(33)2𝑥 = (32)𝑥+6

36𝑥 = 32(𝑥+6)

Now we can write the indices equal to each other.

6𝑥 = 2(𝑥 + 6)

6𝑥 = 2𝑥 + 12

6𝑥 − 2𝑥 = 12

4𝑥 = 12

4𝑥

4=

12

4

𝑥 = 3

Question 12.3

Solve the equation 42𝑥 = 8𝑥+1

Law 3: (𝑎𝑚)𝑛 = 𝑎𝑚𝑛


Example 4

Solve the equation 25𝑥 =1

125.

Solution This one takes a little more work to write each base to the same power.

(52)𝑥 =1

53

52𝑥 = 5−3 Now we have each side with a base of 5

2𝑥 = −3

𝑥 = −3

2

Question 12.4

Solve the equation 8𝑥 =1

32.

Law 3: (𝑎𝑚)𝑛 = 𝑎𝑚𝑛 (left)

Law 5: 𝑎−𝑚 =1

𝑎𝑚 (right)


13. Past and probable exam questions

Question 1


Question 2


Question 3


Question 4

Let 𝑥 ∈ ℝ.


Question 5


Question 6


Question 7


Question 8


Question 9


Question 10


14. Solutions to Algebra

Question 1.1

p = 6 q = 8 5𝑝 + 2(𝑞 − 4) = 5(6) + 2(8 − 4)

= 30 + 2(4) = 30 + 8

= 38

Question 1.2

p = 3 q = 2

3 r = 1

𝑝𝑞 − 𝑟

2=

3 (23

) − 1

2

=2 − 1

2

=1

2

Question 2.1

4𝑥(𝑥 – 3)– 𝑥(𝑥 + 3) = 4𝑥2 − 12𝑥 − 𝑥2 − 3𝑥 = 3 − 15𝑥

Question 2.2

(3𝑥 – 2)(4𝑥 – 1) = 12𝑥2 − 3𝑥 − 8𝑥 + 2 = 12𝑥2 − 11𝑥 + 2

Question 2.3

(2𝑥 + 3)2 = (2𝑥 + 3)(2𝑥 + 3) = 4𝑥2 + 6𝑥 + 6𝑥 + 9

= 4𝑥2 + 12𝑥 + 9

Question 2.4

(𝑥 + 4)(2𝑥2 – 3𝑥 + 5) = 2𝑥3 − 3𝑥2 + 5𝑥 + 8𝑥2 − 12𝑥 + 20 = 2𝑥3 + 5𝑥2 − 7𝑥 + 20

Question 3.1

(5𝑥 − 1)

4+

(2𝑥 − 3)

3=

3(5𝑥 − 1) + 4(2𝑥 − 3)

(4)(3)

=15𝑥 − 3 + 8𝑥 − 12

12

=23𝑥 − 15

12


Question 3.2

5

(2𝑥 + 1)+

4

(2𝑥 − 1)=

5(2𝑥 − 1) + 4(2𝑥 + 1)

(2𝑥 + 1)(2𝑥 − 1)

=10𝑥 − 5 + 8𝑥 + 4

(2𝑥 + 1)(2𝑥 − 1)

=18𝑥 − 1

(2𝑥 + 1)(2𝑥 − 1)

Question 4.1

10(𝑥 + 4) = 3(2𝑥 + 5) + 1 10𝑥 + 40 = 6𝑥 + 15 + 1 10𝑥 − 6𝑥 = 15 + 1 − 40

4𝑥 = −24 𝑥 = −6

Question 4.2

Lowest common denominator = 12 2𝑥 − 1

3+

𝑥

4=

3

2

12(2𝑥 − 1)

3+

12(𝑥)

4=

12(3)

2

4(2𝑥 − 1) + 3𝑥 = 6(3) 8𝑥 − 4 + 3𝑥 = 18

8𝑥 + 3𝑥 = 18 + 4 11𝑥 = 22

𝑥 = 2

Question 5.1

𝑥 + 13 > 4 − 2𝑥 𝑥 + 2𝑥 > 4 − 13

3𝑥 > −9 𝑥 > −3

Question 5.2

5𝑥 − 6

2≤ 7

2(5𝑥 − 6)

2≤ (7)2

5𝑥 − 6 ≤ 14 5𝑥 ≤ 14 + 6

5𝑥 ≤ 20 𝑥 ≤ 4


Question 5.3

2𝑥 − 3 ≤ 7 2𝑥 ≤ 7 + 3

2𝑥 ≤ 10 𝑥 ≤ 5

3𝑥 + 1 ≥ 4 3𝑥 ≥ 4 − 1

3𝑥 ≥ 3 𝑥 ≥ 1

To find A∩B we combine these results we get 1 ≤ 𝑥 ≤ 5.

Question 6. 1

i) 5𝑥2 + 8𝑥 = 𝑥(5𝑥 + 8)

ii) 2𝑝2 − 6𝑝 = 2𝑝(𝑝 − 3)

iii) 6𝑦2 + 30𝑦 = 6𝑦(𝑦 + 5)

Question 6.2

i) 𝑥2 − 16 = 𝑥2 − 42

= (𝑥 − 4)(𝑥 + 4) ii)

𝑥2 − 81𝑦4 = 𝑥2 − (9𝑦2)2 = (𝑥 − 9𝑦2)(𝑥 + 9𝑦2)

iii) 𝑥2 − 1 = 𝑥2 − 12

= (𝑥 − 1)(𝑥 + 1)

Question 6.3

𝑥2 + 10𝑥 + 21 First, find the factors of the last term. Now ask yourself which of these options sum up to the middle number? The answer is 3 and 7 because 3 + 7 = 10

=> (𝑥 + 3)(𝑥 + 7)

Question 6.4

𝑥2 − 5𝑥 + 6 First, find the factors of the last term. Now ask yourself which of these options sum up to the middle number? The answer is −2 and −3 because −2 − 3 = −5

=> (𝑥 − 2)(𝑥 − 3)

Factors of 21

1×21 3×7

−1×−21 −3×−7

Factors of 21

1×6 2×3

−1 ×−6 −2×−3


Question 6.5 𝑥2 + 3𝑥 − 18

Now ask yourself which of these options sum up to the middle number? The answer is −3 and 6 because −3 + 6 = 3

=> (𝑥 − 3)(𝑥 + 6)

Question 6.6

𝑥2 − 5𝑥 − 24 Now ask yourself which of these options sum up to the middle number? The answer is 3 and −8 because 3 − 8 = −5

=> (𝑥 + 3)(𝑥 − 8)

Question 6.7

i) 2x2 − 11x + 5

For this type we use trial and error. The potential answers are. Option 1:

(2𝑥 + 1)(𝑥 + 5) Option 2:

(2𝑥 + 5)(𝑥 + 1) Option 3

(2𝑥 − 1)(𝑥 − 5) Option 4:

(2𝑥 − 5)(𝑥 − 1) If you multiply each of these out you will get: Option 1: 2x2 + 11x + 5 Option 2: 2x2 + 7x + 5 Option 3: 2x2 − 11x + 5 Option 4: 2x2 − 7x + 5 => Option 3 is the answer

ii) 3x2 – x − 2 For this type we use trial and error. The potential answers are. Option 1:

(3𝑥 + 1)(𝑥 − 2) Option 2:

(3𝑥 + 2)(𝑥 − 1) Option 3:

(3𝑥 − 1)(𝑥 + 2) Option 4:

(3𝑥 − 2)(𝑥 + 1) If you multiply each of these out you will get: Option 1: 3x2 –5x − 2 Option 2: 3x2 − x – 2 Option 3: 3x2 + 5x – 2 Option 4: 3x2 + x – 2 => Option 2 is the answer

Factors of −18

−1×18 −2×9 −3×6

1×−18 2×−9 3×−6

Factors of −24

−1×24 −2×12 −3×8 −4×6

1 ×−24 2 ×−12 3 ×−8 4 ×−6

Factors of 5

1×5 −1 ×−5

Factors of -2

1×−2 −1 ×2


Question 7.1

3𝑥2 + 7𝑥 − 4 = 0

𝑎 = 3, 𝑏 = 7, = −4

𝑥 =−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎

𝑥 =−(7) ± √(7)2 − 4(3)(−4)

2(3)

𝑥 =−7 ± √49 + 48

6

𝑥 =−7 ± √97

6

𝑥 =−7 + √97

6 𝑜𝑟 𝑥 =

−7 − √97

6

𝑥 = 0.47 𝑜𝑟 𝑥 = 2.81

Question 7.2

𝑥2 + 6𝑥 + 7 = 0 𝑎 = 1, 𝑏 = 6, 𝑐 = 7

𝑥 =−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎

𝑥 =−(6) ± √(6)2 − 4(1)(7)

2(1)

𝑥 =−6 ± √36 − 28

2

𝑥 =−6 ± √8

2

𝑥 =−6 ± √4√2

2

𝑥 =−6 ± 2√2

2

𝑥 =−6

2±

2√2

2

𝑥 = −3 ± √2

Question 7.3

3𝑥2 − 6𝑥 + 2 = 0 𝑎 = 3, 𝑏 = −6, 𝑐 = 2

𝑥 =−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎

𝑥 =−(−6) ± √(−6)2 − 4(3)(2)

2(3)

𝑥 =6 ± √36 − 24

6

𝑥 =6 ± √12

6

𝑥 =6 ± √4√3

6

𝑥 =6 ± 2√3

6

𝑥 =3 ± √3

3


Question 7.4

2

3𝑥 − 4−

1

7𝑥 + 1=

1

2

2(3𝑥 − 4)(7𝑥 + 1)(2)

(3𝑥 − 4)−

1(3𝑥 − 4)(7𝑥 + 1)(2)

(7𝑥 + 1)=

1(3𝑥 − 4)(7𝑥 + 1)(2)

(2)

4(7𝑥 + 1) − 2(3𝑥 − 4) = (3𝑥 − 4)(7𝑥 + 1) 28𝑥 + 4 − 6𝑥 + 8 = 21𝑥2 + 3𝑥 − 28𝑥 − 4

22𝑥 + 12 = 21𝑥2 − 25𝑥 − 4 0 = 21𝑥2 − 25𝑥 − 22𝑥 − 4 − 12

21𝑥2 − 47𝑥 − 16 = 0

𝑎 = 21, 𝑏 = −47, 𝑐 = −16

𝑥 =−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎

𝑥 =−(−47) ± √(−47)2 − 4(21)(−16)

2(21)

𝑥 =47 ± √2209 + 1344

42

𝑥 =47 ± √3553

42

𝑥 =47 ± √3553

42

𝑥 =47 + √3553

42 𝑜𝑟 𝑥 =

47 − √3553

42

𝑥 = −0.30 𝑜𝑟 𝑥 = 2.54

Question 8.1

x = 5 and x = 3 Therefore, x − 5 = 0 and x − 3 = 0 So (x − 5) and ( x − 3) are factors of the quadratic equation.

=> (𝑥 − 5)(𝑥 − 3) = 0 𝑥2 − 3𝑥 − 5𝑥 + 15 = 0

The equation is: 𝑥2 − 8𝑥 + 15 = 0

Question 9.1

i) 𝑎 = 8𝑏 − 6

𝑎 + 6 = 8𝑏 8𝑏 = 𝑎 + 6

𝑏 =𝑎 + 6

8

ii) 3𝑥 − 𝑏 = 4𝑐

3𝑥 = 4𝑐 + 𝑏

𝑥 =4𝑐 + 𝑏

3


Question 9.2

𝑥 =𝑦 − 4𝑧

6

6𝑥 = 𝑦 − 4𝑧 6𝑥 + 4𝑧 = 𝑦

𝑦 = 6𝑥 + 4𝑧

Question 9.3

2𝑎 =3𝑘

𝑘 − 3

2𝑎(𝑘 − 3) = 3𝑘 2𝑎𝑘 − 6𝑎 = 3𝑘 2𝑎𝑘 − 3𝑘 = 6𝑎 𝑘(2𝑎 − 3) = 6𝑎

𝑘 =6𝑎

2𝑎 − 3

Question 10.1


(1) 𝑝 − 𝑞 = 4

(2) 𝑝 + 2𝑞 = 1

Next, multiply equation (1) by 2 to make the y values the same size but opposite signs × (2) 𝑝 − 𝑞 = 4

(1) × (2) 2 𝑝 − 2𝑞 = 8

(2) 𝑝 + 2𝑞 = 1

3𝑝 = 9

𝑝 = 3

Now substitute p = 3 into equation (2)

(2) (3) + 2𝑞 = 1

2𝑞 = 1 − 3

2𝑞 = −2

𝑞 = −1

=> 𝑝 = 3 𝑎𝑛𝑑 𝑞 = −1

Check your answer subbing back into both equations.

(1) 2(3) − 2(−1) = 8

6 + 2 = 9

8 = 8

(2) (3) + 2(−1) = 1

1 = 1



Question 10.2


(1) 3𝑟 + 2𝑠 = 13

(2) 4𝑟 − 3𝑠 = 6

Next, multiply equation (1) by 3 and (2) by 2 to make the y values the same size but opposite signs.

(1) × 3 9𝑟 + 6𝑠 = 39

(2) × 2 8𝑟 − 6𝑠 = 12

17𝑟 = 51

𝑟 = 3

Now substitute r = 3 back into equation (1)

(1) 3(3) + 2𝑠 = 13

9 + 2𝑠 = 13

2𝑠 = 13 − 9

2𝑠 = 4

𝑠 = 2

=> 𝑟 = 3 𝑎𝑛𝑑 𝑠 = 2

Question 10.3

First, number each of the equations. (1) 2𝑎 − 5𝑏 = 19

(2) 4𝑎

2+

5𝑏

3= −1

Next remove the fractions from equation two by multiplying by the lowest common denominator.

(1) 2𝑎 − 5𝑏 = 19

(2) 6 (4𝑎

2) + 6 (

5𝑏

3) = (6)(−1)


(1) × (2) 4𝑎 − 10𝑏 = 38

(2) 12𝑎 + 10𝑏 = −6

16𝑎 = 32

𝑎 =32

16

𝑎 = 2

Substituting a = 2 into equation 1)

(1) 2(2) − 5𝑏 = 19

4 − 5𝑏 = 19

−5𝑏 = 19 − 4

−5𝑏 = 15

−𝑏 =15

5

−𝑏 = 3

𝑏 = −3


(add equations (1) & (2))


Question 10.4

First, number the equations (1) 𝑥 − 𝑦 − 4 = 0

(2) 𝑦2 + 3𝑥 = 16 Rearrange (1) to find y on its own.

𝑥 − 𝑦 − 4 = 0 −𝑦 = −𝑥 + 4

𝑦 = 𝑥 − 4 Next, substitute x − 4 into equation (2) for y.

𝑦2 + 3𝑥 = 16 (𝑥 − 4)2 + 3𝑥 = 16

(𝑥 − 4)(𝑥 − 4) + 3𝑥 = 16 𝑥2 − 4𝑥 − 4𝑥 + 16 + 3𝑥 = 16

𝑥2 − 5𝑥 + 16 = 16 Rearrange the equation to have everything on the left:

𝑥2 − 5𝑥 + 16 − 16 = 0 𝑥2 − 5𝑥 = 0

Now solve the quadratic equation 𝑥2 − 5𝑥 = 0

𝑎 = 1, 𝑏 = −5, 𝑐 = 0

𝑥 =−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎

𝑥 =−(−5) ± √(−5)2 − 4(1)(0)

2(1)

𝑥 =5 ± √25 − 0

2

𝑥 =5 ± √25

5

𝑥 =5 ± 5

2

𝑥 =5 + 5

2 𝑜𝑟 𝑥 =

5 − 5

2

𝑥 =10

2 𝑜𝑟 𝑥 =

0

2

𝑥 = 5 𝑜𝑟 𝑥 = 0 Finally we substitute the x values back into equation (1)

−𝑦 = 4 x = 5 1) 𝑥 − 𝑦 − 4 = 0

(5) − 𝑦 − 4 = 0 −𝑦 = −5 + 4

−𝑦 = −1 𝑦 = 1

=> 𝑥 = 5 & 𝑦 = 1

x = 0 2) 𝑥 − 𝑦 − 4 = 0

(0) − 𝑦 − 4 = 0 −𝑦 = +4

−𝑦 = 4 𝑦 = −4

=> 𝑥 = 0 & 𝑦 = −4

Question 11.1

i) 24 ii) 33 iii) 53 iv) 72

Question 11.2

i) 𝑥2 × 𝑥3 = 𝑥2+3 = 𝑥5

ii) 𝑥8

𝑥5 = 𝑥8−5

= 𝑥3

iii) (𝑦4)2 = 𝑦(4)(2) = 𝑦8


Question 11.3

81

√3=

34

312

= 34 × 3−12

= 3(4−12

)

= 372

Question 12.1

i) 35𝑥 = 320

If both sides have the same number at the base we can say the powers are equal (both have 3 as the base). 5𝑥 = 20 5𝑥

5=

20

5

𝑥 = 4 ii)

43𝑥−1 = 44 If both sides have the same number at the base we can say the powers are equal (both have 4 as the base).

5𝑥 − 1 = 4 5𝑥 = 4 + 1

5𝑥 = 5 5𝑥

5=

5

5

𝑥 = 1

Question 12.2

2𝑥−2 = 16 2𝑥−2 = 24

Now they have the same base we can put them equal to each other:

𝑥 − 2 = 4 𝑥 = 4 + 2

𝑥 = 6

Question 12.3

42𝑥 = 8𝑥+1 (22)2𝑥 = (23)𝑥+1 24𝑥 = 23(𝑥+1)

Now we can write the indices equal to each other: 4𝑥 = 3(𝑥 + 1) 4𝑥 = 3𝑥 + 3 4𝑥 − 3𝑥 = 3

𝑥 = 3

Question 12.4

8𝑥 =1

32

(23)𝑥 =1

25

23𝑥 = 2−5 Now we can write the indices equal to each other:

3𝑥 = −5

𝑥 = −5

3


Solutions to past and probable exam questions

Question 1

𝑥2 − 6𝑥 − 23 = 0

𝑎 = 1, 𝑏 = −6, 𝑐 = −23

a)

𝑥 =−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎

𝑥 =−(−6) ± √(−6)2 − 4(1)(−23)

2(1)

𝑥 =6 ± √36 + 92

2

𝑥 =6 ± √128

2

𝑥 =6 ± √64√25

2

𝑥 =6 ± 8√2

2

𝑥 =6

2±

8√2

2

𝑥 = 3 ± 4√2

b)


(1) 2𝑟 − 𝑠 = 10

(2) 𝑟𝑠 − 𝑠2 = 12

Rearrange (1) to find y on its own.

2𝑟 − 𝑠 = 10

−𝑠 = 10 − 2𝑟

𝑠 = −10 + 2𝑟

𝑠 = 2𝑟 − 10

Next, substitute 2r − 10 into equation (2) for s.

𝑟𝑠 − 𝑠2 = 12

𝑟(2𝑟 − 10) − (2𝑟 − 10)2 = 12

2𝑟2 − 10𝑟 − (2𝑟 − 10)(2𝑟 − 10) = 12

2𝑟2 − 10𝑟 − (4𝑟2 − 20𝑟 − 20𝑟 + 100) = 12

2𝑟2 − 10𝑟 − 4𝑟2 + 20𝑟 + 20𝑟 − 100 = 12

−2𝑟2 + 30𝑟 − 100 = 12

Rearrange the equation to have everything on the left:

−2𝑟2 + 30𝑟 − 100 − 12 = 0

−2𝑟2 + 30𝑟 − 112 = 0

Now change the signs of all the terms:

2𝑟2 − 30𝑟 + 112 = 0

Continued on the next page.


Now solve the quadratic equation using r instead of x. 2𝑟2 − 30𝑟 + 112 = 0

𝑎 = 2, 𝑏 = −30, 𝑐 = 112

𝑟 =−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎

𝑟 =−(−30) ± √(−30)2 − 4(2)(112)

2(2)

𝑟 =30 ± √900 − 896

4

𝑟 =30 ± √4

4

𝑟 =30 ± 2

4

𝑥 =30 + 2

4 𝑜𝑟 𝑥 =

30 − 2

4

𝑥 =32

4 𝑜𝑟 𝑥 =

28

4

𝑥 = 8 𝑜𝑟 𝑥 = 7 Finally we substitute the r values back into equation (1):

r = 8 (1) 2𝑟 − 𝑠 = 10

2(8) − 𝑠 = 10 16 − 𝑠 = 10 −𝑠 = 10 − 16

−𝑠 = −6 𝑠 = 6

=> 𝑟 = 8 & 𝑠 = 6

r = 7 (1) 2𝑟 − 𝑠 = 10

2(7) − 𝑠 = 10 14 − 𝑠 = 10 −𝑠 = 10 − 14

−𝑠 = −4 𝑠 = 4 => 𝑟 = 7 & 𝑠 = 4

c) 272𝑥 = 3𝑥+10

(33)2𝑥 = 3𝑥+10 36𝑥 = 3𝑥+10

Now we can write the indices equal to each other: 6𝑥 = 𝑥 + 10

6𝑥 − 𝑥 = 10 5𝑥 = 10 5𝑥

5=

10

5

𝑥 = 2


Question 2

a) t = 7

(𝑡 − 1)𝑥 = 2 − 5𝑡

((7) − 1)𝑥 = 2 − 5(7)

6𝑥 = 2 − 35

6𝑥 = −33

𝑥 = −33

6

𝑥 = −11

2

b)


(1) 𝑥 − 𝑦 + 5 = 0

(2) 𝑥2 + 𝑦2 = 17


𝑥 − 𝑦 + 5 = 0

−𝑦 = −𝑥 − 5

𝑦 = 𝑥 + 5

Next, substitute x + 1 into equation (2) for y.

𝑥2 + 𝑦2 = 17

𝑥2 + (𝑥 + 5)2 = 17

𝑥2 + (𝑥 + 5)(𝑥 + 5) = 17

𝑥2 + 𝑥2 + 5𝑥 + 5𝑥 + 25 = 17

2𝑥2 + 10𝑥 + 25 = 17


2𝑥2 + 10𝑥 + 25 − 17 = 0

2𝑥2 + 10𝑥 + 8 = 0


𝑎 = 2, 𝑏 = 10, 𝑐 = 8

𝑥 =−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎

𝑥 =−(10) ± √(10)2 − 4(2)(8)

2(2)

𝑥 =−10 ± √100 + 64

4

𝑥 =−10 + 6

4 𝑜𝑟 𝑥 =

−10 − 6

4

𝑥 = −1 𝑜𝑟 𝑥 = −4

Continues on the next page.


Finally we substitute the x values back into equation x = 3

(1) 𝑥 − 𝑦 + 5 = 0 (−1) − 𝑦 + 5 = 0 −𝑦 = −5 + 1

−𝑦 = −4 𝑦 = 4

=> 𝑥 = −1 & 𝑦 = 4

x = −4 (1) 𝑥 − 𝑦 + 5 = 0

(−4) − 𝑦 + 5 = 0 −𝑦 = 4 − 5 −𝑦 = −1

𝑦 = 1

=> 𝑥 = −4 & 𝑦 = 1

ii) Find the value of x – 2y for each set of values:

x = −1 and y = 4 𝑥 − 2𝑦 = (−1) − 2(4)

= −1 − 8 = −9

−9 is the lesser value.

x = −4 and y = 1 𝑥 − 2𝑦 = (−4) − 2(1)

= −4 − 2 = −6

c) i)

(√𝑥 −2

√𝑥) (√𝑥 +

2

√𝑥) = √𝑥√𝑥 +

2√𝑥

√𝑥−

2√𝑥

√𝑥−

4

√𝑥√𝑥

= 𝑥 −4

𝑥

ii)

𝑥 −4

𝑥= 3

𝑥(𝑥) − 𝑥 (4

𝑥) = 𝑥(3)

𝑥2 − 4 = 3𝑥 𝑥2 − 3𝑥 − 4 = 0

Now solve the quadratic equation: 𝑎 = 1, 𝑏 = −3, 𝑐 = −4

𝑥 =−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎

𝑥 =−(−3) ± √(−3)2 − 4(1)(−4)

2(1)

𝑥 =3 ± √25

2

𝑥 =3 + 5

2 𝑜𝑟 𝑥 =

3 − 5

2

𝑥 = 4 𝑜𝑟 𝑥 = −1 X = 4 since x = − 1 is less than zero

iii) Substitute x = 4 into the equation given in part (ii):

(√4 −2

√4) (√4 +

2

√4) = 3

(2 −2

2) (2 +

2

2) = 3

(2 − 1)(2 + 1) = 3 (1)(3) = 3

3 = 3


Question 3

a) 1

2(7𝑥 − 2) + 5 = 2𝑥 + 7

7𝑥 − 2 + 2(5) = 2(2𝑥 + 7)

7𝑥 − 2 + 10 = 4𝑥 + 14

7𝑥 − 4𝑥 = 14 + 2 − 10

3𝑥 = 6

𝑥 = 2

b)

2

3𝑥 − 4−

1

2𝑥 + 1=

1

2

(3𝑥 − 4)(2𝑥 + 1)(2)2

3𝑥 − 4−

(3𝑥 − 4)(2𝑥 + 1)(2)1

2𝑥 + 1=

(3𝑥 − 4)(2𝑥 + 1)(2)1

2

(2𝑥 + 1)(2)2 − (3𝑥 − 4)(2)1 = (3𝑥 − 4)(2𝑥 + 1)

4(2𝑥 + 1) − 2(3𝑥 − 4) = 6𝑥2 + 3𝑥 − 8𝑥 − 4

8𝑥 + 4 − 6𝑥 + 8 = 6𝑥2 + 3𝑥 − 8𝑥 − 4

2𝑥 + 12 = 6𝑥2 − 5𝑥 − 4

0 = 6𝑥2 − 5𝑥 − 2𝑥 − 4 − 12

0 = 6𝑥2 − 7𝑥 − 16

Now solve the quadratic equation:

𝑎 = 6, 𝑏 = −7, 𝑐 = −16

𝑥 =−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎

𝑥 =−(−7) ± √(−7)2 − 4(6)(−16)

2(6)

𝑥 =7 ± √433

12

𝑥 =7 + √433

12 𝑜𝑟 𝑥 =

7 − √433

12

𝑥 = 2.3 𝑜𝑟 𝑥 = −1.2


c)

(𝑎(√𝑎)3

𝑎4=

(𝑎. 𝑎12)

3

𝑎4

=𝑎3𝑎

32

𝑎4

=𝑎

32

𝑎1

= 𝑎32𝑎−1

= 𝑎12

= √𝑎

d)

49𝑥 = 72+𝑥

(72)𝑥 = 72+𝑥

72𝑥 = 72+𝑥

Now that the base is the same, we let the indices equal to each other:

2𝑥 = 2 + 𝑥

2𝑥 − 𝑥 = 2

𝑥 = 2


Question 4

a)

2𝑓 +2

3𝑔 + 1 = 0

𝑓 +1

2𝑔 + 1 = 0

Multiply the first equation by 3 and the second by 2 to get rid of the fractions:

6𝑓 + 2𝑔 + 3 = 0

2𝑓 + 𝑔 + 2 = 0

Rearranging and numbering the equations:

(1) 6𝑓 + 2𝑔 = −3

(2) 2𝑓 + 𝑔 = −2


(1) 6𝑔 + 2𝑔 = −3

(2) × (−3) − 6𝑔 − 3𝑔 = 6

−𝑔 = 3

𝑔 = −3

Substituting g = − 3 into equation 1

(1) 6𝑓 + 2𝑔 = −3

6𝑓 + 2(−3) = −3

6𝑓 − 6 = −3

6𝑓 = −3 + 6

6𝑓 = 3

𝑓 =1

2

b)

5 −3

4𝑥 ≤

19

8

8(5) − 8 (3

4𝑥) ≤ 8 (

19

8)

40 − 2(3𝑥) ≤ 19

−6𝑥 ≤ 19 − 40

−6𝑥 ≤ −21

6𝑥 ≥ 21

𝑥 ≥21

6

𝑥 ≥7

2

c)

𝑥 = 3 − √2

(3 − √2)2 − 6(3 − √2) + 7 = 0

(3 − √2)(3 − √2) − 6(3 − √2) + 7 = 0

9 − 3√2 − 3√2 + 2 − 18 + 6√2 = 0

9 + 2 − 18 + 7 − 6√2 + 6√2 = 0

0 = 0

(add equation (1) & (2)


Question 5

a)

𝑎(𝑥 + 5) = 8

𝑥 + 5 =8

𝑎

𝑥 =8

𝑎− 5

b)


1) 𝑥 − 𝑦 = 1

2) 𝑥2 + 𝑦2 = 25

Rearrange 1) to find y on its own.

𝑥 − 𝑦 = 1

−𝑦 = −𝑥 + 1

𝑦 = 𝑥 − 1

Next, substitute x − 4 into equation 2) for y.

𝑥2 + 𝑦2 = 25

𝑥2 + (𝑥 − 1)2 = 25

𝑥2 + (𝑥 − 1)(𝑥 − 1) = 25

𝑥2 + 𝑥2 − 𝑥 − 𝑥 + 1 = 25

2𝑥2 − 2𝑥 + 1 = 25

Rearrange the equation to have everything on the left:

2𝑥2 − 2𝑥 + 1 − 25 = 0

2𝑥2 − 2𝑥 − 24 = 0


𝑎 = 2, 𝑏 = −2, 𝑐 = −24

𝑥 =−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎

𝑥 =−(−2) ± √(−2)2 − 4(2)(−24)

2(2)

𝑥 =2 ± √4 + 192

2(2)

𝑥 =2 ± √196

4

𝑥 =2 ± 14

4

𝑥 =2 + 14

4 𝑜𝑟 𝑥 =

2 − 14

4

𝑥 =16

4 𝑜𝑟 𝑥 =

−12

4

𝑥 = 4 𝑜𝑟 𝑥 = −3

Continued on the next page.


Finally we substitute the x values back into equation 1 x = 4

1) 𝑥 − 𝑦 = 1 (4) − 𝑦 = 1 −𝑦 = 1 − 4

−𝑦 = −3 𝑦 = 3

=> 𝑥 = 4 & 𝑦 = 3

x = − 3 1) 𝑥 − 𝑦 = 1

(−3) − 𝑦 = 1 −𝑦 = 1 + 3

−𝑦 = 4 𝑦 = −4

=> 𝑥 = −3 & 𝑦 = −4

ii) Substitute both solutions into 𝑥 − 𝑦2

𝑥 = 4 & 𝑦 = 3 𝑥 − 𝑦2 = (4) − (3)2

= 4 − 9 = −5

𝑥 = −3 & 𝑦 = −4 𝑥 − 𝑦2 = (−3) − (−4)2

= −3 − 16 = −19

c) Draw a diagram to represent information given to you in the question:

i) To find the area we need to find x. We can use Pythagoras’s theorem to find x:

(√𝑥)2 + (2√𝑥)2 = (√45)2 𝑥 + 4𝑥 = 45

5𝑥 = 45 𝑥 = 9

𝑙𝑒𝑛𝑔𝑡ℎ = 2√𝑥

= 2√9 = 2(3) = 6𝑐𝑚

𝑤𝑖𝑑𝑡ℎ = √𝑥

= √9 = 3𝑐𝑚

𝐴𝑟𝑒𝑎 = 𝑙𝑒𝑛𝑔𝑡ℎ × 𝑤𝑖𝑑𝑡ℎ

= 6 × 3 = 18𝑐𝑚2

ii) The area of the square is twice the rectangle so it is 36cm2

𝑙𝑒𝑛𝑡ℎ 𝑜𝑓 𝑠𝑖𝑑𝑒 𝑎 𝑠𝑞𝑢𝑎𝑟𝑒 = √𝑎𝑟𝑒𝑎

= √36 = 6𝑐𝑚


Question 6

a) 3(4𝑥 + 5) − 2(6𝑥 + 4) = 12𝑥 + 15 − 12𝑥 − 8

= 12𝑥 − 12𝑥 + 15 − 8

= 7

b) i)

𝑎 = 1, 𝑏 = −4, 𝑐 = 1

𝑥 =−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎

𝑥 =−(−4) ± √(−4)2 − 4(1)(1)

2(1)

𝑥 =4 ± √12

2

𝑥 =4 ± √4√3

2

𝑥 =4 ± 2√3

2

𝑥 = 2 ± √3

ii)

5𝑥

3=

56

75

(3)5𝑥

3= (3)

56

75

5𝑥 =56

25

5𝑥 =56

52

5𝑥 = 565−2

5𝑥 = 5(6−2)

5𝑥 = 54

Now the bases are the same so the indices are equal:

=> 𝑥 = 4


Question 7

a)

2𝑥 = 3(5 − 𝑥)

2𝑥 = 15 − 3𝑥

2𝑥 + 3𝑥 = 15

5𝑥 = 15

𝑥 = 3

b)

𝑥

4−

𝑦

3=

5

6

2𝑥 − 6 = 3𝑦

First, number each of the equations.

(1) 𝑥

4−

𝑦

3=

5

6

(2) 2𝑥 − 6 = 3𝑦

Next remove the fractions from equation (1) by multiplying by the lowest common denominator and rearrange

(2) to put the x and y on he left

(1) 12 (𝑥

4) − 12 (

𝑦

3) = 12 (

5

6)

(2) 2𝑥 − 3𝑦 = 6

(1) 3𝑥 − 4𝑦 = 10

(2) 2𝑥 − 3𝑦 = 6


(1) × (2) 6𝑥 − 8𝑦 = 20

(2) × (-3) −6𝑥 + 9𝑦 = −18

𝑦 = 2

Substituting x = 6 into equation (1)

(1) 3𝑥 − 4𝑦 = 10

3𝑥 − 4(2) = 10

3𝑥 − 8 = 10

3𝑥 = 10 + 8

3𝑥 = 18

𝑥 =18

3

𝑥 = 6

(add equation (1) & (2)


c) i) The quadratic is positive and positive.

First, find the factors of the last term.

Now ask yourself which of these options sum up to the middle number?

The answer is 2 and 2 because 2 + 2 = 4

=> (𝑥 + 2)(𝑥 + 2)

ii)

𝑥2 + 4𝑥 + 4 = (𝑥 + 2)(𝑥 + 2)

= (𝑥 + 2)2

𝑥2 + 2𝑥 + 1 = (𝑥 + 1)(𝑥 + 1)

= (𝑥 + 1)2

√𝑥2 + 4𝑥 + 4 + √𝑥2 + 2𝑥 + 1 = √(𝑥 + 2)2 + √(𝑥 + 1)2

= (𝑥 + 2) + (𝑥 + 1)

= 2𝑥 + 3

iii)

2𝑥 + 3 = 𝑥2

−𝑥2 + 2𝑥 + 3 = 0

𝑥2 − 2𝑥 − 3 = 0

𝑎 = 1, 𝑏 = −2, 𝑐 = −3

𝑥 =−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎

𝑥 =−(−2) ± √(−2)2 − 4(1)(−3)

2(1)

𝑥 =2 ± √16

2

𝑥 =2 + 4

2 𝑜𝑟 𝑥 =

2 − 4

2

𝑥 = 3 𝑜𝑟 𝑥 = −1

x = 3 only (since x = − 1 is less than 0 so it is not an answer.)

Factors of 4

1×4 2×2

−1×−4 −2×−2


Question 8

a) 4𝑥 − 15 < 14

𝑥 < 1 + 15 4𝑥 < 16

𝑥 < 4 but 𝑥 ∈ 𝑁

So the possible values of x are {1, 2, 3}.

b) i) x = 5

2 y =

1

3

𝑥 + 3𝑦 + 5

2𝑥 + 2𝑦=

52

+ 3 (13

) + 5

2 (52

) + 2 (13

)

=

52

+ 1 + 5

5 +23

=

52

+22

+102

153

+23

=

172

173

=17

2×

3

17

=3

2

ii) 2𝑥+3 = 4𝑥

2𝑥+3 = (22)𝑥 2𝑥+3 = 22𝑥

Now that the bases are equal we can let the indices equal each other:

𝑥 + 3 = 2𝑥 𝑥 − 2𝑥 = −3𝑥

−𝑥 = −3 𝑥 = 3

c)

𝑥 −1

𝑥= 2

𝑥(𝑥) − 𝑥 (1

𝑥) = 𝑥(2)

𝑥2 − 𝑥 = 2𝑥 𝑥2 − 2𝑥 − 1 = 0

𝑎 = 1, 𝑏 = −2, 𝑐 = −1

𝑥 =−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎

𝑥 =−(−2) ± √(−2)2 − 4(1)(−1)

2(1)

𝑥 =2 ± √8

2

𝑥 =2 ± 2√2

2

𝑥 = 1 ± √2

ii) 𝑥2 − 2𝑥 − 1 = 0

(1 + √2)2 − 2(1 + √2) − 1 = 0

(1 + √2)(1 + √2) − 2(1 + √2) − 1 = 0

1 + 2√2 + 2 − 2 − 2√2 − 1 = 0 0 = 0


Question 9

a) (i) 2(4 − 3𝑥) + 12 = 7𝑥 − 5(2𝑥 − 7) 8 − 6𝑥 + 12 = 7𝑥 − 10𝑥 + 35

20 − 6𝑥 = −3𝑥 + 35 −6𝑥 + 3𝑥 = 35 − 20

−3𝑥 = 15 3𝑥 = −15

𝑥 = −5 (ii) To verify your answer substitute x = 5 into the equation given

2(4 − 3(−5)) + 12 = 7(−5) − 5(2(−5) − 7)

2(4 + 15) + 12 = −35 − 5(−10 − 7) 2(19) + 12 = −35 − 5(−17)

38 + 12 = −35 + 85 50 = 50

b)


(1) 𝑥 + 𝑦 = 7

(2) 𝑥2 + 𝑦2 = 25


𝑥 + 𝑦 = 7

𝑦 = 7 − 𝑥

Next, substitute 7 − 𝑥 into equation (2) for y.

𝑥2 + 𝑦2 = 25

𝑥2 + (7 − 𝑥)2 = 25

𝑥2 + (7 − 𝑥)(7 − 𝑥) = 25

𝑥2 + 49 − 7𝑥 − 7𝑥 + 𝑥2 = 25

2𝑥2 − 14𝑥 + 49 = 25


2𝑥2 − 14𝑥 + 49 − 25 = 0

2𝑥2 − 14𝑥 + 24 = 0


𝑎 = 2, 𝑏 = −14, 𝑐 = 24

𝑥 =−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎

𝑥 =−(−14) ± √(−14)2 − 4(2)(24)

2(2)

𝑥 =14 ± √196 − 192

4

𝑥 =14 + 2

4 𝑜𝑟 𝑥 =

14 − 2

4

𝑥 = 4 𝑜𝑟 𝑥 = 3

Finally we substitute the x values back into equation (1):

𝑥 = 4 (1) 𝑥 + 𝑦 = 7

(4) + 𝑦 = 7 𝑦 = 7 − 4 𝑦 = 3 => 𝑥 = 4 & 𝑦 = 3

𝑥 = 3 (1) 𝑥 + 𝑦 = 7

(3) + 𝑦 = 7 𝑦 = 7 − 3 𝑦 = 4 => 𝑥 = 3 & 𝑦 = 4


Question 10

(a) 3(4 − 5𝑥) − 2(5 − 6𝑥) 12 − 15𝑥 − 10 + 12𝑥

2 − 3𝑥 (b)

2 − 3𝑥 ≥ −6 −3𝑥 ≥ −8

𝑥 ≤ 2.6 𝑥 = {1, 2}

(c)

=> 𝑔(𝑥) = −5𝑥 + 4


Complex Numbers Complex numbers is worth 4% to 8% of the Leaving Cert. It appears on Paper 1

1. The imaginary number

Complex numbers contain an imaginary number which is a negative square root. To help us with this mathematicians

invented a new number. You are required to know this for the Leaving Certificate:

𝑖 = √−1 or 𝑖2 = −1

We use it to write the square root of a negative number:

Example 1 Write the following negative square roots in the form ki.

i) √−9

ii) √−64 Solution

i)

√−9 = √9 × −1

= √9√−1 = 3𝑖

ii)

√−64 = √64 × −1

= √64√−1 = 8𝑖

Question 1.1 Write the following negative square roots in the form ki.

i) √−25

ii) √−49

iii) √−100

iv) √−144


2. Complex numbers

Complex numbers contain a real part and an imaginary part.

E.g. z is an example of a complex number:

𝑧 = 4 + 3𝑖

Example 1 Write down the real and imaginary parts for each of these

i) 4+5i ii) 1 – 7i iii) 5 iv) −3i

Solution

Complex number Real Part Imaginary part

i) 4 + 5i 4 5

ii) 1 – 7i 1 −7

iii) 5 5 0

iv) – 3i 0 -3

Question 2.1

Complete the following table:


i) −3+9i

ii) ¼ + 5i

iii) 6i

iv) 4+i

v) −7

vi) a + bi

4 is the real

part

3 is the imaginary part

Notice we don’t

mention the i


3. Adding and subtracting complex numbers

When adding and subtracting complex numbers there is a simple rule:

Add or subtract the real parts then add or subtract the imaginary parts separately.

Example 1 Write the following complex numbers in the form a + bi.

i) (4 + 5i) + (2 + 3i) ii) (8 + 2i) – (7 – 6i)

Solution

i) (4 + 5𝑖) + (2 + 3𝑖) = 4 + 5𝑖 + 2 + 3𝑖

= 6 + 8𝑖 ii)

(8 + 2𝑖) − (7 − 6𝑖) = 8 + 2𝑖 − 7 + 6𝑖 = 1 + 8𝑖

Question 3.1

Write the following complex numbers in the form a + bi.

i) (3 + 9i) + (5 – 3i)

ii) (7 – 4 i) – (2 + 3i)

Example 2 If u = 2 + 3i and w = 5 + 6i write u + w as a complex number. Solution

𝑢 + 𝑤 = (2 + 3𝑖) + (5 + 6𝑖) = 2 + 3𝑖 + 5 + 6𝑖

= 7 + 9𝑖

Question 3.2

If z1 = 7 – 3i and z2 = 2 + 5i find

i) z1+z2

ii) z1−z2


4. Multiplying complex numbers

The Examiner can ask you to multiply complex numbers in the Leaving Cert exam. When multiplying complex numbers

we use the same method as algebra.

However, there is an important rule you must remember:

𝑖2 is replaced with −1

We’ll start by multiplying a complex number by a single number.

Example 1 Multiply the complex number 5 + 6i by 3 Solution

= 3(5 + 6𝑖) = 15 + 18𝑖

Now let’s look at and example where we must also multiply by an imaginary number:

Example 2 Express 5(1 + 2i)+3i(7 + 4i) in the form of x + yi. Solution 5(1 + 2i) + 3i(7 + 4i) = 5 + 10i + 21i + 12𝑖2

= 5 + 10𝑖 + 21𝑖 + 12(−1) = 5 − 12 + 31𝑖

= −7 + 31𝑖

Question 4.1

Simplify the following and write the answers in the form a + bi

i) 3i(5 – 2i)

ii) 7(2 – i) + 9i(1 – 2i)


Next, we will look at another way the Examiner can ask you to multiply complex numbers in the Leaving Cert exam:

Example 3 If u = 5+3i and w = 3 - 2i , find the following terms in the form of a + bi, where a,b ∈R and 𝑖2 = −1.

i) uw ii) w2 iii) 2u+3iw

Solution When doing questions like this put the complex numbers in brackets

i)

𝑢𝑤 = (5 + 3𝑖)(3 − 2𝑖)

= 15 − 10𝑖 + 9𝑖 − 6𝑖2

= 15 − 1𝑖 − 6(−1)

= 15 − 1𝑖 + 6

= 21 − 𝑖

ii)

𝑤2 = (3 − 2𝑖)2

= (3 − 2𝑖)(3 − 2𝑖)

= 9 − 6𝑖 − 6𝑖 + 4𝑖2

= 9 − 12𝑖 + 4(−1)

= 5 − 12𝑖

iii)

2𝑢 + 3𝑖𝑤 = 2(5 + 3𝑖) + 3𝑖(3 − 2𝑖)

= 10 + 6𝑖 + 9𝑖 − 6𝑖2

= 10 + 15𝑖 − 6(−1)

= 16 + 15𝑖


Question 4.2

If z1 = 4 + 7i and z2 = 2 + 3i find the following terms in the form of a + bi, where a,b ∈R and 𝑖2 = −1.

i) z12

ii) z1z2

iii) 3iz1 + 5z2


5. The complex conjugate (the guy with the hat)

To find the ‘complex conjugate’ we change the sign of the imaginary part. The symbol for the complex conjugate is a line

above the letter (the guy with the hat).

Let’s look at an example:

𝑧 = 5 + 9𝑖

The complex conjugate is:

𝑧̅ = 5 − 9𝑖

Remember: only change the sign of the imaginary part.

Question 5.1

Find the complex conjugate of the following complex numbers.

i) -3 + 7i

ii) 1 – 5i

iii) 8 – 2i


6. Division of complex numbers

We have a rule to remember when dividing by a complex number.

Multiply the top and the bottom by the conjugate of the bottom

Note: The conjugate just means change the sign of the imaginary part.

Example 1

Express 5+4𝑖

3−2𝑖 in the form a + bi

Solution The conjugate of the bottom is 3 + 2𝑖. Now multiply the top and the bottom by the 3 + 2𝑖.

5 + 4𝑖

3 − 2𝑖=

5 + 4𝑖

3 − 2𝑖×

3 + 2𝑖

3 + 2𝑖

=(5 + 4𝑖)(3 + 2𝑖)

(3 − 2𝑖)(3 + 2𝑖)

=15 + 10𝑖 + 12𝑖 + 8𝑖2

9 + 6𝑖 − 6𝑖 − 4𝑖2

=15 + 8(−1) + 10𝑖 + 12𝑖

9 − 4(−1)

=15 − 8 + 22𝑖

13

=7 + 22𝑖

13

=7

13+

22𝑖

13

Question 6.1 Express the following in the form a + bi.

11 − 7𝑖

2 + 𝑖


7. Equality of complex numbers

If two complex numbers are equal we can compare the real parts and the imaginary parts. The real parts must be equal

and the imaginary parts must be equal.

For example

If 2 + 3i = x + yi what is the value of x and y?

The real parts must be equal so x = 2.

The imaginary parts must be equal so y = 3.

Example 1 Find the value of x and y, if x(2 + 3i) – 2y = 4 + 9i. Solution First let’s multiply out the brackets

𝑥(2 + 3𝑖) − 2𝑦 = 4 + 9𝑖 2𝑥 + 3𝑥𝑖 − 2𝑦 = 4 + 9𝑖

Now rearrange it bringing the real parts together. (2𝑥 − 2𝑦) + 3𝑥𝑖 = 4 + 9𝑖

Compare the real and imaginary parts Real parts:

2𝑥 − 2𝑦 = 4 2(3) − 2𝑦 = 4

−2𝑦 = 4 − 6 −2𝑦 = −2

=> 𝑦 = 1

Imaginary parts: 3𝑥 = 9

=> 𝑥 = 3

Question 7.1

If 3x + 4yi = 15 + 16i find the value of x and y.

For more complicated questions you may end up with two equations where you will have to solve simultaneous

equations:

Example 2 If (3x + 2y) + (x + 4y)i = 7 – i find the values of x and y. Solution Compare the real parts and the imaginary parts

Real parts 3𝑥 + 2𝑦 = 7

Imaginary parts 𝑥 + 4𝑦 = −1

We now have two equation to solve simultaneously 1) 3𝑥 + 2𝑦 = 7 2) 𝑥 + 4𝑦 = −1

1) 3𝑥 + 2𝑦 = 7 2) × (−3) −3𝑥 − 12𝑦 = 3

−10𝑦 = 10 𝑦 = −1

Put y = -1 into (1): 3𝑥 + 2𝑦 = 7 3𝑥 + 2(−1) = 7

3𝑥 − 2 = 7 3𝑥 = 7 + 2 3𝑥 = 9 => 𝑥 = 3


Question 7.2

If x(1 + 2i) + y(2+3i) = 8 – 14i find the values of x and y.

Hint: Remove the brackets and rearrange the left hand side first.


8. Quadratic equations with complex numbers

To solve quadratic equations we can use the quadratic formula from the cover of the log tables:

𝑥 =−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎

Example 1 Solve the quadratic equation 𝑧2 + 2𝑧 + 10 = 0, in the form x + yi. Solution So, a = 1 b = 2 c = 10

Using the formula

𝑧 =−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎

=−2 ± √22 − 4(1)(10)

2(1)

=−2 ± √4 − 40

2

=−2 ± √−36

2

Remember what we learned in section 1:

=−2 ± 6𝑖

2

=−2

2±

6𝑖

2

= −1 ± 3𝑖

𝑧 = −1 + 3𝑖 𝑜𝑟 𝑧 = −1 − 3𝑖

Note: The second root is always the conjugate’ of the first root.

i.e. Just change the sign of the imaginary part.

Question 8.1

Solve the following equation 𝑧2 − 8𝑧 + 25 = 0 giving your answer in the form a + bi.


You can also be asked to prove or show that a complex number is the root of a quadratic equation:

Example 2 Show that −3 + i is a root of the equation z2 + 6z + 10 = 0 and write down the other root. Solution

Let z = −3 + i. then substitute it into the equation z2 + 6z + 10 = 0.

(−3 + 𝑖)2 + 6(−3 + 𝑖) + 10 = 0

(−3 + 𝑖)(−3 + 𝑖) + 6(−3 + 𝑖) + 10 = 0

9 − 3𝑖 − 3𝑖 + 𝑖2 − 18 + 6𝑖 + 10 = 0

9 − 6𝑖 − 1 + 6𝑖 + 10 = 0

10 − 10 + 6𝑖 − 6𝑖 = 0

0 = 0

=> Therefore −3 + i is a root and the other root is the conjugate −3 − i

Question 8.2

Show that −2 − 4i is a root of the equation z2 + 4z + 20 = 0 and write down the other root.

Example 3 If −6 − i is a root of z2 + az + b = 0 find the value of a and b. Solution If −6 − i is a root we substitute it into the equation

(−6 − 𝑖)2 + 𝑎(−6 − 𝑖) + 𝑏 = 0: 36 + 6𝑖 + 6𝑖 + 𝑖2 − 6𝑎 − 𝑎𝑖 + 𝑏 = 0

Next bring a and b to the right hand side of the equation and put the real parts in brackets and the imaginary parts in brackets.

36 + 12𝑖 − 1 = (6𝑎 − 𝑏) + 𝑎𝑖 35 + 12𝑖 = (6𝑎 − 𝑏) + 𝑎𝑖

Now compare and the real and imaginary parts.

Real Parts: 35 = 6𝑎 − 𝑏 35 = 6(12) − 𝑏 35 = 72 − 𝑏

35 − 72 = −𝑏 −37 = −𝑏

=> 𝑏 = 37

Imaginary Parts: 𝑎 = 12


Question 8.3

If 5+4i is a root of z2 + kz + l = 0 find the value of k and l.


9. Argand diagrams

We can plot complex numbers on a graph. This graph is known as an Argand diagram. In an Argand diagram we do not

have an x-axis and y-axis.

The horizontal axis is the real axis (Re).

The vertical axis is the imaginary axis (Im)

The complex number 𝑧 = 5 + 2𝑖 can be represented as the coordinates (5, 2). The 5 is the real part and the 2 is the

imaginary part. So remember the coordinates are (real, imaginary)

Complex number Coordinates

z1 = −3 + i (−3, 1)

z2 = −5 – 3i (−5, −3)

z3 = 4 (4, 0)

z4 = -2i (0, −2)

z5 = 4 −4i (4, −4)

We can draw these points on the diagram shown below.

Question 9.1

Represent the following points on an Argand diagram

i) 6 + 4i

ii) −3 + 2i

iii) −6


Question 9.2

If 𝑢 = 5 + 3𝑖 and 𝑤 = −4 + 𝑖, plot the following on an Argand diagram

i) 𝑢

ii) 𝑤

iii) �̅�

iv) �̅�

v) 𝑢 + 𝑤


10. The modulus of complex numbers (the guy with the jacket)

The modulus of a complex number is the distance from the origin (0, 0) to the complex number.

The modulus of 𝑧 is written as |𝑧| (the guy with the jacket).

We find the modulus of 𝑧 using the formula:

|𝑧| = √(𝑎)2 + (𝑏)2

Example 1 If w = 4 – 3i find the modulus of w. Solution

|𝑤| = √(4)2 + (−3)2

= √16 + 9

= √25 = 5

Question 10.1

Find the modulus of the complex number 12 + 5i.

Note that we

don’t put the i in


The Examiner can ask you to find the modulus in different ways. For example he can ask you to add or multiply two

complex numbers within a modulus. You can also be asked to compare the moduli of complex numbers.

Example 2

If u = 5 + 2i and w = 2 + 4i, find

i) |u|

ii) |w|

iii) |u + w|

iv) Prove that |u| + |w| > |u + w|

Solution

i)

|𝑢| = √(5)2 + (2)2

= √25 + 4

= √29

ii)

| 𝑤| = √(2)2 + (4)2

= √4 + 16

= √20

iii)

𝑢 + 𝑤 = 5 + 2𝑖 + 2 + 4𝑖

= 7 + 6𝑖

|𝑢 + 𝑤| = √(7)2 + (6)2

= √49 + 36

= √85

iv)

|𝑢| + |𝑤| = √29 + √20

= 9.86

|𝑢 + 𝑤| = √85

= 9.22

Since 9.86>9.22

=> |𝑢| + |𝑤| > |𝑢 + 𝑤|


Question 10.2

Given that z1 = 3 – i and z2 = −5 + 2i

i) Find |z1|

ii) Find |z2|

iii) Find |z1.z2|

iv) Using your answers from part i) to iii) prove that |z1|.|z2| = |z1.z2|

Note: z1.z2 is the same as z1×z2


11. Transformations of complex numbers

The syllabus requires you to know how different operations can affect complex numbers:

There are 3 types.

Type 1: Addition of complex numbers

When one complex number is added to another number it causes a translation.

Example 1 𝑧1 = 3 + 4𝑖, 𝑧2 = 2 + 𝑖, 𝑧3 = −1 + 2𝑖 and 𝑤 = 1 + 𝑖 are complex numbers.

i) Plot 𝑧1, 𝑧2and 𝑧3on an Argand diagram. ii) Find 𝑧1 + 𝑤, 𝑧2 + 𝑤 and 𝑧3 + 𝑤. iii) Plot the answers from part (ii) on an Argand diagram.

Solution

i)

ii) 𝑧1 + 𝑤 = (3 + 4𝑖) + (1 + 𝑖) = 4 + 5𝑖 𝑧2 + 𝑤 = (2 + 𝑖) + (1 + 𝑖) = 3 + 2𝑖 𝑧3 + 𝑤 = (−1 + 2𝑖) + (1 + 𝑖) = 0 + 3𝑖

iii)


Question 11.1

𝑧1 = 2 + 2𝑖, 𝑧2 = 6 + 𝑖, 𝑧3 = −1 − 2𝑖 and 𝑤 = −2 − 𝑖 are complex number.

i) Plot 𝑧1, 𝑧2and 𝑧3on an Argand diagram.

ii) Find 𝑧1 + 𝑤, 𝑧2 + 𝑤 and 𝑧3 + 𝑤.

iii) Plot the answers from part (ii) on an Argand diagram and comment on what the addition has caused.

Solution

i)

ii)

iii)


Type 2: Multiplication of a complex number by a normal number.

Multiplying a complex number by a normal number can “stretch” or “contract” the complex number.

Example 2 𝑧1 = 2 + 4𝑖 ,𝑧2 = −1 + 2𝑖 and 𝑎 = 2.

i) Plot z1 and z2 on an Argand diagram. ii) Evaluate az1 and az2. iii) Plot the answer from part (ii) on an Argand diagram and comment on your findings

Solution

i)

ii) 𝑎𝑧1 = 2(2 + 4𝑖) = 4 + 8𝑖 𝑎𝑧2 = 2(−1 + 2𝑖) = −2 + 4𝑖

iii)

z1 and z2 have been stretched by a factor of 2.


Question 11.2

𝑧1 = 3 + 𝑖, 𝑧2 = −1 − 2𝑖 and 𝑎 = −2.

i) Plot z1 and z2 on an Argand diagram.

ii) Evaluate az1 and az2.

iii) Plot the answer from part (ii) on an Argand diagram and comment on the diagram.

Solution

i)

ii)

iii)

Note: If you multiple by a negative number it will also reverse the direction of the complex number, as seen above.


Type 3: Multiplying a complex number by an imaginary number

Rule: i will rotate it anticlockwise by 900

−i will rotate it clockwise by 900

Tip for remembering which one is which: Draw a minus sign. Which way did the pen go …

Example 3 𝑧1 = 2 + 𝑖

i) Find z2, if z2 = iz1 ii) Plot z1 and z2 on an Argand diagram. iii) Describe the transformation that maps z1 and z2.

Solution

i) 𝑧2 = 𝑖(2 + 𝑖) = 2𝑖 + 𝑖2

∴ 𝑧2 = −1 + 2𝑖

ii)

iii) z1 has rotated 900 anticlockwise.


Question 11.3

𝑧1 = 2 + 𝑖

i) Find z2, if z2 is found by multiplying z1 by −i

ii) Plot z1 and z2 on an Argand diagram.

iii) Describe the transformation that maps z1 and z2.

Solution


12. z, z2, z3, z4, z5, z6, etc.

The Examiner can ask you what happens to a complex number when you repeatedly multiply it by itself:

Example 1 z is a complex number 1 + i, where i2 = − 1

i) Find z2 ii) Find z3 iii) Show z4 = − 4 iv) Find z8 v) Verify z12 = − 64 vi) Find z41

Solution

i) 𝑧2 = (𝑧)(𝑧) = (1 + 𝑖)(1 + 𝑖) = 1 + 1𝑖 + 1𝑖 + 𝑖2 = 1 + 2𝑖 + (−1) = 1 + 2𝑖 − 1 = 2𝑖

ii) From the rules of indices in the Algebra chapter we should know that 𝑧3 = 𝑧2 × 𝑧1 (1.e. when we multiply powers we add the powers) 𝑧3 = (2𝑖)(1 + 𝑖) = 2𝑖 + 2(−1) = −2 + 2𝑖

iii) 𝑧4 can be written as 𝑧2 × 𝑧2 𝑧4 = 𝑧2 × 𝑧2 = (2𝑖) × (2𝑖) = 4𝑖2 = 4(−1) = −4

iv) 𝑧8 = 𝑧4 × 𝑧4 = (−4) × (−4) = 16

v) 𝑧12 = 𝑧4 × 𝑧8 = (−4) × (16) = −64

vi) 𝑧41 = 𝑧8 × 𝑧8 × 𝑧8 × 𝑧8 × 𝑧8 × 𝑧1 = (16) × (16) × (16) × (16) × (16) × (1 + 𝑖) = (1,048,576)(1 + 𝑖) = 1,048,576 + 1,048,576𝑖


Comment:

13. Past and probable exam questions

Question 1


Question 2


Question 3


Question 4


Question 5


Question 6


Question 7


Question 8

A big Trigonometry element in this question. Really nasty by the Examiner.


14. Solutions to Complex Numbers

Question 1.1

i)

√−25 = √25 × −1

= √25√−1 = 5𝑖

ii)

√−49 = √49 × −1

= √49√−1 = 7𝑖

iii)

√−100 = √100 × −1

= √100√−1 = 10𝑖

iv)

√−144 = √144 × −1

= √144√−1 = 12𝑖

Question 2.1


i) −3+9i −3 9

ii) ¼ + 5i ¼ 5

iii) 6i 0 6

iv) 4+i 4 1

v) −7 −7 0

vi) a + bi a b

Question 3.1

i) (3 + 9𝑖) + (5 − 3𝑖) = 3 + 9𝑖 + 5 − 3𝑖

= 8 + 6𝑖 ii)

(7 − 4𝑖) − (2 + 3𝑖) = 7 − 4𝑖 − 2 − 3𝑖 = 5 − 7𝑖

Question 3.2

i) 𝑧1 + 𝑧2 = (7 − 3𝑖) + (2 + 5𝑖)

= 7 − 3𝑖 + 2 + 5𝑖 = 9 + 2𝑖

ii) 𝑧1 − 𝑧2 = (7 − 3𝑖) − (2 + 5𝑖)

= 7 − 3𝑖 − 2 − 5𝑖 = 5 − 8𝑖


Question 4.1

i) 3i(5 − 2i) = 15i − 6𝑖2

= 15𝑖 − 6(−1) = 15𝑖 + 6

= 6 + 15𝑖 ii)

7(2 − i) + 9i(1 − 2i) = 14 − 7i + 9i − 18𝑖2 = 14 + 2𝑖 − 18(−1)

= 14 + 18 + 2𝑖 = 32 + 2𝑖

Question 4.2

i) 𝑧1

2 = (4 + 7𝑖)2 = (4 + 7𝑖)(4 + 7𝑖)

= 16 + 28𝑖 + 28𝑖 + 49𝑖2 = 16 + 56𝑖 + 49(−1)

= 16 − 49 + 56𝑖 = −33 + 56𝑖

ii) 𝑧1𝑧2 = (4 + 7𝑖)(2 + 3𝑖)

= 8 + 12𝑖 + 14𝑖 + 21𝑖2 = 8 + 26𝑖 + 21(−1)

= 8 + 26𝑖 − 21 = −13 + 26𝑖

iii) 3𝑖𝑧1 + 5𝑧2 = 3𝑖(4 + 7𝑖) + 5(2 + 3𝑖)

= 12𝑖 + 21𝑖2 + 10 + 15𝑖 = 12𝑖 + 21(−1) + 10 + 15𝑖

= −11 + 27𝑖

Question 5.1

i) −3 + 7𝑖 −3 − 7𝑖 ii) 1 − 5𝑖 1 + 5𝑖 iii) 8 − 2𝑖 8 + 2𝑖

Question 6.1

11 − 7𝑖

2 + 𝑖=

11 − 7𝑖

2 + 𝑖×

2 − 𝑖

2 − 𝑖

=(11 − 7𝑖)(2 − 𝑖)

(2 + 𝑖)(2 − 𝑖)

=22 − 11𝑖 − 14𝑖 + 7𝑖2

4 − 2𝑖 + 2𝑖 − 𝑖2

=22 + 7(−1) − 11𝑖 − 14𝑖

4 − 1(−1)

=15 − 25𝑖

5

= 3 − 5𝑖

Question 7.1

Real parts:

3𝑥 = 15

=> 𝑥 = 5

Imaginary parts:

4𝑦 = 16

=> 𝑦 = 4


Question 7.2

First multiply out the left hand side:

𝑥(1 + 2𝑖) + 𝑦(2 + 3𝑖) = 𝑥 + 2𝑥𝑖 + 2𝑦 + 3𝑦𝑖

= (𝑥 + 2𝑦) + (2𝑥 + 3𝑦)𝑖

=> (𝑥 + 2𝑦) + (2𝑥 + 3𝑦)𝑖 = 8 − 14𝑖

Compare the real parts and the imaginary parts

Real parts

𝑥 + 2𝑦 = 8

Imaginary parts

2𝑥 + 3𝑦 = −14

We now have two equation to solve simultaneously

(1) 𝑥 + 2𝑦 = 8

(2) 2𝑥 + 3𝑦 = −14

(1) × (−2) −2𝑥 − 4𝑦 = −16

(2) 2 𝑥 + 3𝑦 = −14

−𝑦 = −30

𝑦 = 30

Put y = 30 into (1):

𝑥 + 2𝑦 = 8

𝑥 + 2(30) = 8

𝑥 + 60 = 8

𝑥 = 8 − 60

=> 𝑥 = −52

Question 8.1

So, a = 1 b = −8 c = 25 Using the formula

𝑧 =−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎

=−(−8) ± √(−8)2 − 4(1)(25)

2(1)

=8 ± √64 − 100

2

=8 ± √−36

2

Remember what we learned in section 1:

=8 ± 6𝑖

2

=8

2±

6𝑖

2

= 4 ± 3𝑖

𝑧 = 4 + 3𝑖 𝑜𝑟 𝑧 = 4 − 3𝑖


Question 8.2

Let z = −2 – 4i. then substitute it into the equation z2 + 4z + 20 = 0.

(−2 − 4𝑖)2 + 4(−2 − 4𝑖) + 10 = 0

(−2 − 4𝑖)(−2 − 4𝑖) + 4(−2 − 4𝑖) + 10 = 0

4 + 8𝑖 + 8𝑖 + 16𝑖2 − 8 − 16𝑖 + 20 = 0

4 + 16𝑖 + 16(−1) − 12 − 24𝑖 + 20 = 0

4 − 16 − 8 + 20 + 16𝑖 − 16𝑖 = 0

0 = 0

=> Therefore -2 - 4i is a root and the other root is the conjugate -2 + 4i.

Question 8.3

If 5 + 4𝑖 is a root we substitute it into the equation

(5 + 4𝑖)2 + 𝑘(5 + 4𝑖) + 𝑙 = 0

25 + 20𝑖 + 20𝑖 + 16𝑖2 + 5𝑘 + 4𝑘𝑖 + 𝑙 = 0

Next bring k and l to the right hand side of the equation and put the real parts in brackets and the imaginary parts in

brackets.

25 + 40𝑖 + 16(−1) = (−5𝑘 − 𝑙) − 4𝑘𝑖

9 + 40𝑖 = −1(5𝑘 + 𝑙) − 4𝑘𝑖

Now compare and the real and imaginary parts.

Real Parts:

9 = −5𝑘 − 𝑙

Imaginary Parts:

−4𝑘 = 40

=> 𝑘 = −10

9 = −5(−10) − 𝑙

9 = 50 − 𝑙

9 − 50 = −𝑙

−41 = −𝑙

=> 𝑙 = 41

Question 9.1


Question 9.2

i) 𝑢 = 5 + 3𝑖 ii) 𝑤 = −4 + 𝑖 iii) �̅� = 5 − 3𝑖 iv) �̅� − 4 − 𝑖 v) 𝑢 + 𝑤 = (5 + 3𝑖) + (−4 + 𝑖)

= 1 + 4𝑖

Question 10.1

√(12)2 + (5)2

= √144 + 25

= √169 = 13

Question 10.2

i)

|𝑧1| = √(3)2 + (−1)2

= √9 + 1

= √10 ii)

|𝑧2| = √(−5)2 + (2)2

= √25 + 4

= √29 iii)

𝑧1. 𝑧2 = (3 − 𝑖)(−5 + 2𝑖) = −15 + 6𝑖 + 5𝑖 − 2𝑖2 = −15 + 11𝑖 − 2(−1) = −15 + 2 + 11𝑖

= −13 + 11𝑖

|𝑧1. 𝑧2| = √(−13)2 + (11)2

= √169 + 121

= √290 iv)

|𝑧1|. |𝑧2| = √10. √29

= √290

|𝑧1. 𝑧2| = √290

=> |𝑧1|. |𝑧2| = |𝑧1. 𝑧2|


Question 11.1

i)

ii) 𝑧1 + 𝑤 = (2 + 2𝑖) + (−2 − 𝑖) = 𝑖 𝑧2 + 𝑤 = (6 + 𝑖) + (−2 − 𝑖) = 4 𝑧3 + 𝑤 = (−1 − 2𝑖) + (−2 − 𝑖) = −3 − 3𝑖

iii)

The points undergo the same translations.

Question 11.2

i)


ii) 𝑎𝑧1 = −2(3 + 𝑖)

= −6 − 2𝑖 𝑎𝑧2 = −2(−1 − 2𝑖) = 2 + 4𝑖

iii)

The direction of the complex number has been reversed and stretched.

Question 11.3

i) 𝑧2 = −𝑖(2 + 𝑖) = −2𝑖 − 𝑖2 = −2𝑖 − (−1)

∴ 𝑧2 = 1 − 2𝑖

ii)

iii) z1 is mapped onto z2 by a clockwise rotation of 900 about the origin.


Past and probable exam questions solutions

Question 1

(a) 𝑤 = 𝑢 − 𝑣 − 2

= (3 + 2𝑖) − (−1 + 𝑖) − 2

= 3 + 2𝑖 + 1 − 𝑖 − 2

= 2 + 𝑖

(b)

(c)

2𝑢 + 𝑣

𝑤=

2(3 + 2𝑖) + (−1 + 𝑖)

2 + 𝑖

=6 + 4𝑖 − 1 + 𝑖

2 + 𝑖

=5 + 5𝑖

2 + 𝑖

Now to multiply the top and the bottom by the complex conjugate:

5 + 5𝑖

2 + 𝑖×

2 − 𝑖

2 − 𝑖

=(5 + 5𝑖)(2 − 𝑖)

(2 + 𝑖)(2 − 𝑖)

=10 − 5𝑖 + 10𝑖 − 5𝑖2

4 + 2𝑖 − 2𝑖 − 𝑖2

=10 − 5(−1) − 5𝑖 + 10𝑖

4 − (−1)

=15 + 5𝑖

5

= 3 + 𝑖


Question 2

a)

b) Answer: Yes

Reason: z1 is further from the origin than z2.

c)

|𝑧1| = √(3)2 + (−4)2

= √9 + 16

= √25

= 5

|𝑧2| = √(1)2 + (2)2

= √1 + 4

= √5

=> |𝑧1| > |𝑧2|

d)

3 − 4𝑖

1 + 2𝑖×

1 − 2𝑖

1 − 2𝑖

=(3 − 4𝑖)(1 − 2𝑖)

(1 + 2𝑖)(1 − 2𝑖)

=3 − 6𝑖 − 4𝑖 + 8𝑖2

1 − 2𝑖 + 2𝑖 − 4𝑖2

=3 + 8(−1) − 6𝑖 − 4𝑖

1 − 4(−1)

=−5 − 10𝑖

5

= −1 − 2𝑖


Question 3

a) 𝑧1 = (3 + 2𝑖)2 − 5𝑖)

= 6 − 15𝑖 + 4𝑖 − 10𝑖2

= 6 − 10(−1) − 11𝑖

= 6 + 10 − 11𝑖

=> 𝑧1 = 16 − 11𝑖

𝑧2 = (5 + 4𝑖)(17 − 13𝑖) − (5 + 3𝑖)(17 − 13𝑖)

= 85 − 65𝑖 + 68𝑖 − 52𝑖2 − 85 + 65𝑖 − 51𝑖 + 39𝑖2

= −52𝑖2 + 39𝑖2 + 17𝑖

= −52(−1) + 39(−1) + 17𝑖

=> 𝑧2 = 13 + 17𝑖

𝑧3 = (5

2+

7

2𝑖)

2

− (5

2+

1

2𝑖)

2

= (5

2+

7

2𝑖) (

5

2+

7

2𝑖) − (

5

2+

1

2𝑖) (

5

2+

1

2𝑖)

=25

4+

35

4𝑖 +

35

4𝑖 +

49

4𝑖2 −

25

4−

5

4𝑖 −

5

4𝑖 −

1

4𝑖2

=60

4𝑖 +

48

4𝑖2

= 15𝑖 + 12𝑖2

= 12(−1) + 15𝑖

=> 𝑧3 = −12 + 15𝑖

𝑧4 = 1 + 𝑖 + 𝑖2 + 𝑖3

= 1 + 𝑖 + 𝑖2 + 𝑖 × 𝑖2

= 1 + 𝑖 + (−1) + 𝑖(−1)

=> 𝑧4 = 0

b)

|𝑧1| = √(16)2 + (−11)2

= √256 + 121

= √377

|𝑧2| = √(13)2 + (17)2

= √169 + 289

= √458

=> |𝑧1| > |𝑧2|

=> 𝑧1 𝑖𝑠 𝑓𝑎𝑟𝑡ℎ𝑒𝑟 𝑓𝑟𝑜𝑚 0 𝑜𝑛 𝑎𝑛 𝐴𝑟𝑔𝑎𝑛𝑑 𝑑𝑖𝑎𝑔𝑟𝑎𝑚


Question 4

a) i)

6 − 4𝑖 + 3𝑢 = 5𝑖 3𝑢 = 5𝑖 + 4𝑖 − 6

3𝑢 = −6 + 9𝑖

𝑢 =−6 + 9𝑖

3

=> 𝑢 = −2 + 3𝑖

ii)

b) i)

|𝑧| = √(1)2 + (1)2

= √2

ii)

𝑧̅ = 1 − 𝑖

𝑧2 + 𝑧̅2 = (1 + 𝑖)2 + (1 − 𝑖)2

= (1 + 𝑖)(1 + 𝑖) + (1 − 𝑖)(1 − 𝑖)

= 1 + 𝑖 + 𝑖 + 𝑖2 + 1 − 𝑖 − 𝑖 + 𝑖2

= 2 + 2𝑖 − 2𝑖 + 2𝑖2

= 2 + 2𝑖2

= 2 + 2(−1)

= 0

iii)

1 + 5𝑖

3 + 2𝑖

=1 + 5𝑖

3 + 2𝑖×

3 − 2𝑖

3 − 2𝑖=

(1 + 5𝑖)(3 − 2𝑖)

(3 + 2𝑖)(3 − 2𝑖)

=3 − 2𝑖 + 15𝑖 − 10𝑖2

9 − 6𝑖 + 6𝑖 − 4𝑖2

=13 + 13𝑖

13

= 1 + 1𝑖

=>1 + 5𝑖

3 + 2𝑖= 𝑧


c)

𝑤2 − (𝑘 + 𝑡)𝑤 + 𝑡 = 0

(3 + 4𝑖)2 − (𝑘 + 𝑡)(3 + 4𝑖) + 𝑡 = 0

9 + 12𝑖 + 12𝑖 + 16𝑖2 − 3𝑘 − 4𝑘𝑖 − 3𝑡 − 4𝑡𝑖 + 𝑡 = 0

9 + 24𝑖 + 16(−1) − 3𝑘 − 4𝑘𝑖 − 2𝑡 − 4𝑡𝑖 = 0

9 − 16 − 3𝑘 + 4𝑡 + 24𝑖 − 4𝑘𝑖 − 4𝑡𝑖 = 0

(−7 − 3𝑘 − 2𝑡) + (24 − 4𝑘 − 4𝑡)𝑖 = 0

Comparing the real and imaginary parts on each side (they are both equal to zero).

−7 − 3𝑘 − 2𝑡 = 0

=> 3𝑘 + 2𝑡 = −7

24 − 4𝑘 − 4𝑡 = 0

=> −4𝑘 − 4𝑡 = −24

Now solve the simultaneous equations (1) and (2).

(1) × 2 6𝑘 + 4𝑡 = −14

(2) −4𝑘 − 4𝑡 = −24

2𝑘 = −38

𝑘 = −19

Now substitute k = -19 into equation (1)

3𝑘 + 2𝑡 = −7

3(−19) + 2𝑡 = −7

−57 + 2𝑡 = −7

2𝑡 = −7 + 57

2𝑡 = 50

𝑡 = 25

Question 5

a) 𝑧 = 1 − 4𝑖

−2𝑧 = −2(1 − 4𝑖) = −2 + 8𝑖

(1)

(2)


b)

2|𝑧| = 2 (√(1)2 + (−4)2)

= 2√17

|−2𝑧| = √(−2)2 + (8)2

= √68

= √4 × 17

= √4√17

= 2√17

=> 2|𝑧| = |−2𝑧|

c) The point −2𝑧 is twice as far away from the origin as the point 𝑧.

d)

𝑧 + 𝑘 = 1 − 4𝑖 + 𝑘

= (1 + 𝑘) − 4𝑖

|𝑧 + 𝑘| = 5

√(1 + 𝑘)2 + (−4)2 = 5

(1 + 𝑘)2 + (−4)2 = 52

(1 + 𝑘)(1 + 𝑘) + 16 = 25

1 + 𝑘 + 𝑘 + 𝑘2 + 16 = 25

17 + 2𝑘 + 𝑘2 = 25

17 + 2𝑘 + 𝑘2 − 25 = 0

𝑘2 + 2𝑘 − 8 = 0

Using the formula

𝑘 =−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎

=−2 ± √22 − 4(1)(−8)

2(1)

=−2 ± √4 + 32

2

=−2 ± √36

2

=−2 ± 6

2

= −2

2±

6

2

= −1 ± 3

𝑘 = 1 + 3 𝑜𝑟 𝑘 = 1 − 3

𝑘 = 4 𝑜𝑟 𝑘 = −2


Question 6

a) i) 𝑧2 = (1 + 𝑖)(1 + 𝑖)

= 1 + 1𝑖 + 1𝑖 + 𝑖2 = 1 + 2𝑖 + (−1) = 1 + 2𝑖 − 1 = 2𝑖 From the rules of indices in the Algebra chapter we should know that 𝑧3 = 𝑧2 × 𝑧1 (i.e. when we multiply two numbers that are the same that have powers, we add the powers) 𝑧3 = (2𝑖)(1 + 𝑖) = 2𝑖 + 2(−1) = −2 + 2𝑖

ii) 𝑧4 can be written as 𝑧2 × 𝑧2 𝑧4 = 𝑧2 × 𝑧2 = (2𝑖) × (2𝑖) = 4𝑖2 = 4(−1) = −4

iii)

iv) As the complex number is being multiplied by itself it is moving further away from the origin.

b) 𝑧8 = 𝑧4 × 𝑧4

= (−4) × (−4) = 16 𝑧12 = 𝑧4 × 𝑧8 = (−4) × (16) = −64 𝑧16 = 𝑧4 × 𝑧12 = (−4) × (−64) = 256

c) 𝑧40 is positive because in the pattern above the powers that are multiples of eight are positive.

d) 𝑧40 = 𝑧8 × 𝑧8 × 𝑧8 × 𝑧8 × 𝑧8 = (16) × (16) × (16) × (16) × (16) = (16)5 = (24)5 = 220

e) 𝑧41 = 𝑧8 × 𝑧8 × 𝑧8 × 𝑧8 × 𝑧8 × 𝑧1

= (16) × (16) × (16) × (16) × (16) × (1 + 𝑖) = (1,048,576)(1 + 𝑖) = 1,048,576 + 1,048,576𝑖

f) |𝑧41| = √(1,048,576)2 + (1,048,576)2 = 1482910.40038


Question 7

a) (i) 𝑧1 − 𝑧2 = (5 − 𝑖) − (4 + 3𝑖)

= 5 − 𝑖 − 4 − 3𝑖 = 1 − 4𝑖

(ii)

|𝑧1 − 𝑧2| = √(1)2 + (−4)2

= √1 + 16

= √17

𝑧2 − 𝑧1 = (4 + 3𝑖) − (5 − 𝑖) = 4 + 3𝑖 − 5 + 𝑖 = −1 + 4𝑖

|𝑧2 − 𝑧1| = √(−1)2 + (4)2

= √1 + 16

= √17

=> |𝑧1 − 𝑧2| = |𝑧2 − 𝑧1| (iii) 𝑧 − 𝑤 and 𝑤 − 𝑧 will be the image of each other under central symmetry through the origin so they moduli will always be equal.

b) We are given:

𝑧1 =𝑧2

𝑧3

This can be rearranged to give:

𝑧3 =𝑧2

𝑧1

Which gives:

𝑧3 =4 + 3𝑖

5 − 𝑖

=4 + 3𝑖

5 − 𝑖×

5 + 𝑖

5 + 𝑖

=(4 + 3𝑖)(5 + 𝑖)

(5 − 𝑖)(5 + 𝑖)

=20 + 4𝑖 + 15𝑖 + 3𝑖2

25 + 5𝑖 − 5𝑖 − 𝑖2

=20 + 19𝑖 + 3(−1)

25 − (−1)

=20 + 19𝑖 − 3

25 + 1

=17 + 19𝑖

26

=17

26+

19

26𝑖


Question 8

(a) (i) a = 3 b = 1

(ii) c = 3 d = -1

(b) (i)

cos𝜃

2=

3

√10

𝜃

2= cos−1 (

3

√10)

𝜃

2= 18.43494882

𝜃 = 36.8699°

cos 𝜃 = cos 36.8699 = 0.79999

=> cos 𝜃 = 0.8

(ii) 𝑧1 = 3 + 𝑖 𝑧2 = 3 − 𝑖

|𝑧1| = √32 + 12

= √10

|𝑧2| = √32 + (−1)2

= √10

|𝑧1| × |𝑧2| × cos 𝜃 = 𝑎𝑐 + 𝑏𝑑

√10 × √10 × (0.8) = 3(3) + (1)(−1) => 8 = 8

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