1. Simplify √128
Simplify 4√45
Rationalise and simplify 2 5√2

2. 8√2
12√5
√2 5

3. Quadratic graphs and equations
Difference of two squares
Solve the equation 𝑥 2 – 49 = 0 and hence sketch the graph 𝑦 = 𝑥 2 – 49.
Chapter 3.1
𝑥 2 – 49 can be written as 𝑥 2 +0𝑥– 49
Notice that this is 𝑥 2 – 7 2  
Quadratic
Perfect square
Difference of two squares
𝑥 2 +0𝑥– 49
= (𝑥 + 7)(𝑥 − 7)
−7 + 7 = 0 and
(−7) x 7 = −49 
(𝑥 + 7)(𝑥 − 7) = 0
⇒ 𝑥 = −7 or 𝑥 = 7

4. Quadratic graphs and equations
Factorise 6 𝑥 2 + 𝑥 − 12.
(ii) Solve the equation 6 𝑥 2 + 𝑥 − 12 = 0.
(iii) Sketch the graph 𝑦 = 6 𝑥 2 + 𝑥 − 12.
Chapter 3.1
Quadratic
Perfect square
Difference of two squares

5. Quadratic graphs and equations
(𝑖) The technique for finding how to split the middle term is now adjusted. Start by multiplying the two outside numbers together:
6 × (−12) = −72.
Now look for two numbers which add to give +1 (the coefficient of x) and multiply to give −72 (the number found above).
(+9) + (−8) = (+9) × (−8) = −72
Splitting the middle term gives
Chapter 3.1
Quadratic
Perfect square
Difference of two squares
6 𝑥 𝑥 − 8𝑥 − 12
= 3𝑥 (2𝑥 + 3) − 4(2𝑥 + 3)
3𝑥 is the largest factor
of both 6 𝑥 2 and 9𝑥.
−4 is the largest factor
of both −8𝑥 and −12.
= (3𝑥 − 4)(2𝑥 + 3)

6. Quadratic graphs and equations
(𝑖𝑖) (3𝑥 − 4)(2𝑥 + 3) = 0
⇒ 3𝑥 − 4 = 0 or 2𝑥 + 3 = 0
⇒𝑥= or 𝑥=− 3 2
(𝑖𝑖𝑖) The graph is a ∪-shaped curve which
crosses the x axis at the points where
𝑥= 4 3 and where 𝑥=− 3 2
When 𝑥 = 0, 𝑦 = −12,
so the curve crosses the
y axis at (0, −12).
Chapter 3.1
Quadratic
Perfect square
Difference of two squares

7. Quadratic graphs and equations
(i) Factorise 𝑥 − 𝑥 2 .
(ii) Hence sketch the graph of 𝑦 = 𝑥 − 𝑥 2 .
Chapter 3.1
Quadratic
Perfect square
Difference of two squares

8. Quadratic graphs and equations
The coefficient of 𝑥 2 is −1 so here the outside numbers are 15 and −1
(𝑖) Start by multiplying the two outside numbers together: 15 × −1 = −15
Now look for two numbers which add to give +2 (the coefficient of 𝑥) and
multiply to give −15 (the number found above).
(+5) + (−3) = +2 (+5) . (−3) = −15
Splitting the middle term gives
Chapter 3.1
Quadratic
Perfect square
Difference of two squares
15+5𝑥−3 𝑥− 𝑥 2
= 5(3+𝑥) −𝑥(3+𝑥)
=(5−𝑥)(3+𝑥)

9. Quadratic graphs and equations
(ii) For numerically large positive or negative
values of x, the value of
𝑦 = (5 − 𝑥)(3 + 𝑥) will be
negative and consequently the
curve will be ∩-shaped.
It will cross the 𝑥 axis when 𝑦 = 0,
i.e. at 𝑥 = +5 and 𝑥 = −3,
and the 𝑦 axis when 𝑥 = 0,
i.e. at 𝑦 = 15.
(i) Factorise 𝑥 − 𝑥 2 .
(ii) Hence sketch the graph of 𝑦 = 𝑥 − 𝑥 2 .
Chapter 3.1
Quadratic
Perfect square
Difference of two squares

10. Quadratic graphs and equations
Solve the quartic equation 4 𝑥 4 − 17 𝑥 = 0.
Replacing 𝑥 2 by 𝑧 gives the quadratic equation 4 𝑧 2 − 17𝑧 + 4 = 0.
Chapter 3.1
⇒(𝑧 − 4)(4𝑧 − 1) = 0
Remember that you replaced 𝑥 2 by 𝑧 earlier.
Quadratic
Perfect square
Difference of two squares
⇒𝑧 = 4 𝑜𝑟 4𝑧 = 1
⇒ Either 𝑥 2 = 4 or 4 𝑥 2 = 1
⇒The solution is 𝑥 =±2 or 𝑥=± 1 2

11. Quadratic graphs and equations
In a right-angled triangle, the side BC is 1 cm longer than the side AB and the hypotenuse AC is 29 cm long.
Find the lengths of the three sides of the triangle.
Chapter 3.1
Quadratic
Perfect square
Difference of two squares

12. Quadratic graphs and equations
Let the length of AB be x cm. So the length of BC is x + 1 cm.
The hypotenuse is 29 cm.
Using Pythagoras’ theorem:
𝑥 𝑥 = 29 2
Chapter 3.1
⇒ 𝑥 𝑥 𝑥 + 1 = 841
⇒ 2 𝑥 2 + 2𝑥 − 840 = 0
Quadratic
Perfect square
Difference of two squares
⇒ 𝑥 2 + 𝑥 − 420 = 0
Divide by 2 to make the factorisation easier.
⇒ (𝑥 + 21)(𝑥 − 20) = 0
Notice that −21 is
rejected since x is a length.
⇒ 𝑥 = −21 or 𝑥 = 20, giving 𝑥 + 1 = 21
The lengths of the three sides are therefore 20 cm, 21 cm and 29 cm.
Make sure that you answer the question fully.

13. Select menu MODE A: Equation/Func
Step 1: Select equation vs simultaneous equation solver.
A polynomial is an expression of the form where the are constants, for example, or .
The degree of a polynomial is the highest power, so the degree of a cubic equation, e.g. , is 3. So for quadratic equations, use 2.
“Solve ”
Choose polynomial mode, degree 3.
Enter 1, -1, 3, -4, pressing = after each number. Use down arrows to scroll through solutions.
You can use this to sketch graphs, solve equations, or check your factorisation if you are told not to use a calculator.

14. Solve the following either manually or with a calculator

15. The completed square form
Chapter 3.2
The completed square form
“It’s not that I’m so smart, it’s just that I stay with problems longer.”
- Albert Einstein

16. The completed square form
Write 𝑥 2 + 5𝑥 + 4 in completed square form.
Hence state the equation of the line of symmetry and the coordinates of the vertex of the curve 𝑦 = 𝑥 2 + 5𝑥 + 4.
𝑥 2 + 5𝑥
5÷ 2= , = 25 4
Chapter 3.2
= 𝑥 2 + 5𝑥 − 25 4
Parabola
Vertex
Completing the square
= 𝑥 − 9 4
This is the completed
square form.
The line of symmetry is
𝑥+ 5 2 = 0 , or 𝑥=− 5 2
The vertex is (− 5 2 , − 9 4 ).

17. The completed square form
Use the method of completing the square to write down the equation of the line of symmetry and the coordinates of the vertex of the curve 𝑦 = 2 𝑥 2 + 6𝑥 + 5.
Chapter 3.2
Parabola
Vertex
Completing the square

18. The completed square form
2 𝑥 2 + 6𝑥 + 5=2 𝑥 2 + 3𝑥
First take out a factor of 2:
Now proceed as before with the expression in the square brackets:
=2 𝑥 −
Chapter 3.2
=2 𝑥
Parabola
Vertex
Completing the square
=2 𝑥
The least value of the expression in the brackets is when 𝑥 = 0, so the equation of the line of symmetry is 𝑥= − The vertex is − 3 2 ,

19. The completed square form
Use the method of completing the square to find the equation of the line of symmetry and the coordinates of the vertex of the curve 𝑦 = − 𝑥 2 + 6𝑥 + 5.
Chapter 3.2
Start by taking out a factor of (−1):
− 𝑥 2 +6𝑥+ 5𝑥
= − 𝑥 2 −6𝑥 − 5
Completing the square for the expression in square brackets gives:
Parabola
Vertex
Completing the square
− 𝑥 2 −6𝑥 − 5 =− 𝑥−3 2 − 3 2 −5
= − 𝑥−3 2 −14
= − 𝑥−
The least value of the expression in the brackets is when 𝑥 − 3 = 0, so the equation of the line of symmetry is 𝑥 = 3.
The vertex is (3, 14). This is a maximum point.
It is a maximum since the
coefficient of 𝑥 2 is negative.

20. The completed square form
(i) Write the equation 𝑥 2 − 6𝑥 + 2 in completed square form.
(ii) Hence solve the equation 𝑥 2 − 6𝑥 + 2 = 0.
𝑖 𝑥 2 − 6𝑥 + 2
= 𝑥 −3 2 −
Chapter 3.2
= 𝑥− 3 2 − 7
𝑖𝑖 𝑥− 3 2 − 7= 0
Parabola
Vertex
Completing the square
⇒ 𝑥 −3 2 =7
Take the square root of both sides:
⇒𝑥−3= ± 7
⇒𝑥 = 3 ± 7
This is an
exact answer.
Using your calculator to find the value of
7 ⇒𝑥 = or 0.354, to 3 decimal places. 
This is an
approximate answer.

21. Select menu MODE A: Equation/Func
“Solve ”
Choose polynomial mode, degree 3.
Enter 1, -1, 3, -4, pressing = after each number. Use down arrows to scroll through solutions.
Step 1: Select equation vs simultaneous equation solver.
A polynomial is an expression of the form where the are constants, for example, or .
The degree of a polynomial is the highest power, so the degree of a cubic equation, e.g. , is 3. So for quadratic equations, use 2.
You can use this to sketch the graph, minimum point, line of symmetry or check you have completed the square properly in a non-calculator question

22. Factorise, Solve, Complete the square and sketch each of the following labelling the roots, y-intercept, vertex and line of symmetry for each. (You can use your calculator to check your answers)

23. Simplify 3√2×5√2
Simplify√2( )
Simplify (𝑎+2 𝑏 )(𝑎−2 𝑏 )

24. 30
6+2√3
a 2 −4 b 2

25. Chapter 3.3 The quadratic formula
“It’s not that I’m so smart, it’s just that I stay with problems longer.”
- Albert Einstein

26. The quadratic formula
Use the quadratic formula to solve 3 𝑥 2 − 6𝑥 + 2 = 0.
Chapter 3.3
Comparing it to the form 𝑎 𝑥 2 + 𝑏𝑥 + 𝑐 = 0
gives 𝑎 = 3,𝑏 = −6,𝑐 = 2.
𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎
Substituting these values into the formula 
𝑥= −6± 36−24 6
gives
= or (to 3 d.p.)

27. The quadratic formula Show that 𝑥 2 − 2𝑥 + 2 cannot be factorised.
(ii) What happens when you try to use the quadratic
formula to solve 𝑥 2 − 2𝑥 + 2=0?
(iii) Draw the graph of 𝑦=𝑥 2 − 2𝑥 + 2.
(iv) How does your graph explain your answer to (ii)?
Chapter 3.3

28. The quadratic formula
The only two whole numbers which multiply to give 2 are 2 and 1
(or−2 and −1) and they cannot be added to get −2.
(ii) Comparing 𝑥 2 − 2𝑥 + 2 to the form 𝑎 𝑥 2 + 𝑏𝑥 + 𝑐 = 0
gives 𝑎 = 1, 𝑏 = −2 and 𝑐 = 2.
Chapter 3.3
𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎
Substituting these values into the formula
𝑥= 2± 4−8 2
gives
There is no real
number equal to
−4 so there is
no solution.
= 2± −4 2

29. The quadratic formula (iii) Chapter 3.3
(iv) Sketching the graph shows that it does not cut the x axis and so there is
indeed no real solutions to the equation.

30. The Discriminant The discriminant is the part inside the square root
b2-4ac
It determines how many roots an equation will have and so how many times it crosses or touches the x axis
If b2-4ac > 0 then the equation has two real roots
If b2-4ac = 0 then the equation has one repeated root
If b2-4ac < 0 then the equation has no real roots

31. How many roots does the equation 3x2 + 2x – 7 have?

33. For each equation below state the number of the roots of the equation and give solutions where possible in exact form

34. Solutions from the quadratic formula can be checked using your calculator