Section 2: Quadratic Functions

Introduction

This section will cover how to:

Each topic is introduced with a theory section including examples and then some practice questions. At the end of the page there is an exercise where you can test your understanding of all the topics covered in this page.

Remember that you are NOT allowed to use calculators in this topic.

 

Quadratic Graphs

Definitions
quadratic
expression
an expression of the form ax² + bx + c where a, b and c are numbers and x is the variable
(note that b and c can be zero, but a must be non-zero so that there is always an x² term)
e.g. x² + 5x + 6,   3x² – 3x + 17,   8x²,   4x² – 25,   12x² – 30x
quadratic
equation
an equation which can be written with a quadratic expression on one side and zero on the other
e.g. x² + 5x + 6 = 0,   3x² – 3x + 17 = 0,   4x² – 25 = 0
note that quadratic equations sometimes need rearranging to get zero on one side
e.g. x² = 3x – 2 can be rearranged to give x² – 3x + 2 = 0
quadratic
function
a name normally given to a function of the form y = ax² + bx + c which is used to calculate y coordinates from given x coordinates so that these can be plotted on a graph
roots of a
quadratic
the roots of a quadratic equation are the solutions to that equation
e.g. the roots of the equation x² + 5x + 6 = 0 are x = –2 and x = –3
the roots of a quadratic expression are the values which make it equal zero
e.g. the roots of the expression x² + 5x + 6 are x = –2 and x = –3
on a graph, the roots correspond to where the curve crosses the x-axis
e.g. the curve of y = x² + 5x + 6 crosses the x-axis at (–2,0) and (–3,0)
Plotting a Quadratic Graph and Identifying Roots
To plot the graph of y = x² – 2x – 3 for values of x between –2 and 4, we start by calculating a table of values:
For example for the x coordinate of –2 we calculate the y coordinate as follows:
y = (–2)² – (2 × –2) – 3 = 4 – 4 – 3 = 5
You need to be especially careful when calculating the x² term. The two main things to watch out for are:
  • squaring a negative number gives a positive answer (e.g. –2 × –2 = +4)
  • if the x² term has a coefficient, you square x first and then multiply by the coefficient (e.g. 3x² means x² × 3)
Now that we have a table of values, we can plot each ( x , y ) pair as coordinates on a set of axes, as shown below:
Finally we can use the graph to identify the roots of the quadratic expression x² – 2x – 3. They are x = –1 and x = 3.
These are the solutions of the quadratic equation x² – 2x – 3 = 0 and are the points where the curve crosses the x-axis.
Notes
  • Quadratics can have two, one or zero roots depending on how many times their curve would cross the x-axis.
  • The shape of a quadratic curve is always the same except that it is upside-down if the coefficient of x² is negative.
Practice Question
Work out the answer to each part and then click on the button marked
to see if you are correct.
For the graph, you can plot the curve yourself on paper and then check your curve against the one on the screen.
Plot the quadratic curve y = ½x² – 4x + 6 for x = 0 to 6 and use it to identify the solutions of ½x² – 4x + 6 = 0

(a) Start by completing a table of values for x = 0 up to x = 6

(b) Now plot the points on an appropriate set of axes and join them with a smooth curve

(c) Finally, use your curve to identify the solutions of ½x² – 4x + 6 = 0

The curve crosses the x-axis at ( 2 , 0 ) and ( 6 , 0 ) so the solutions are x = 2 and x = 6

 

Expanding and Factorising Quadratics

Expanding Pairs of Brackets
If you expand brackets of the form (px + q)(rx + s) you will get a quadratic expression.
To expand brackets in this form, simply multiply each term in the first bracket by every term in the second bracket, then add up the resulting terms and simplify if necessary.
e.g.
(
x
+ 5
)(
x
+ 3
)
= x² + 3x + 5x + 15
= x² + 8x + 15
e.g.
(
x
+ 4
)(
x
– 6
)
= x² – 6x + 4x – 24
= x² – 2x – 24
e.g.
(
x
– 3
)(
2x
+ 7
)
= 2x² + 7x – 6x – 21
= 2x² + x – 21
e.g.
(
5x
– 2
)(
4x
– 3
)
= 20x² – 15x – 8x + 6
= 20x² – 23x + 6
Factorising Quadratics
Factorising a quadratic is the reverse of expanding a pair of brackets; not every quadratic can be factorised, but the majority of those which can will end up as two brackets multiplied together. In every case we can easily check if we have factorised correctly by multiplying out the brackets again.
(a) Simple Factorisation
The easiest quadratics to factorise are those that begin with just x².
These will normally end up in the form ( x + p )( x + q ), where:
  • p and q add to give the coefficient of x in the original quadratic
  • p and q multiply to give the number term in the original quadratic
For example, to factorise x² + 11x + 30, we need two numbers which add to make 11 and multiply to make 30.
These numbers are 5 and 6, so the factorised quadratic is ( x + 5 )( x + 6 ).
To factorise x² – x – 20, we need two numbers which add to make –1 and multiply to make –20.
We can see that one must be positive and one must be negative, so the answer will take the form ( x + ? )( x – ? )
The numbers which work are +4 and –5, so the factorised quadratic is ( x + 4 )( x – 5 ).
To factorise x² – 8x + 16, we need two numbers which add to make –8 and multiply to make 16.
The two numbers multiply to make a positive so they must be the same sign, but they add to give a negative so both numbers must be negative. The answer will take the form ( x – ? )( x – ? )
The numbers which work are –4 and –4, so the factorised quadratic is ( x – 4 )( x – 4 ), or ( x – 4 )².
(b) Harder Factorisation
Quadratics become harder to factorise when the coefficient of x² is not 1, for example 3x² +16x + 5.
The x² term tells you about the first part of each bracket. In this case the only way to get 3x² is to multiply 3x by x so the answer will take the form ( 3x + ? )( x + ? )
The number term tells you about the second part of each bracket. In this case the only way to get 5 is to multiply 5 by 1 so the answer must be either ( 3x + 5 )( x + 1 ) or ( 3x + 1 )( x + 5 ).
Multiplying these out gives the correct answer in the second case, so 3x² +16x + 5 = ( 3x + 1 )( x + 5 )
Factorising harder quadratics will always involve a bit of trial and error so common sense and practice are essential.
(c) Special Cases
If you expand the brackets in ( a + b )( ab ) you get a² – b². This is called the difference of two squares. If you see a quadratic which is formed from the difference of two squares then you can use this fact to factorise the quadratic:
e.g. x² – 25
   = (x)² – (5)²
   = ( x + 5 )( x – 5 )
e.g. 4x² – 9
   = (2x)² – (3)²
   = ( 2x + 3 )( 2x – 3 )
e.g. 64x² – 1
   = (8x)² – (1)²
   = ( 8x + 1 )( 8x – 1 )
If a quadratic has an x² term and an x term but no number term, then it is easier to factorise because there is a common factor and the answer becomes a term multiplied by a single bracket.
For example, the quadratic expression 6x² – 10x becomes 2x( 3x – 5 ).
Practice Questions
Work out the answer to each question then click on the button marked
to see if you are correct.
Expand the brackets and simplify:

(a) ( x + 5 )( x – 7 )

(
x
+ 5
)(
x
– 7
)
= x² – 7x + 5x – 35
= x² – 2x – 35

(b) ( 3x – 5 )( 4x – 1 )

(
3x
– 5
)(
4x
– 1
)
= 12x² – 3x – 20x + 5
= 12x² – 23x + 5
Factorise completely:

(a) x² + 10x + 9

( x + 1 )( x + 9 )

(b) x² – 7x – 30

( x + 3 )( x – 10 )

(c) 2x² + 7x + 5

( 2x + 5 )( x + 1 )

(d) 20x² – 117x – 50

( 4x – 25 )( 5x + 2 )

(e) 9x² – 1

( 3x + 1 )( 3x – 1 )

(f) 20x – 12x²

4x( 5 – 3x )

 

Solving Quadratic Equations by Factorisation

Basic Principles
Consider the equation ab = 0.
The only way this can be true is if a = 0 or b = 0.
Now consider the quadratic equation ( x + 2 )( x – 3 ) = 0.
In a similar way to the equation above, the only way this equation can be true is if x + 2 = 0 or if x – 3 = 0.
From x + 2 = 0 we get x = –2 and from x – 3 = 0 we get x = 3.
The two solutions of the equation ( x + 2 )( x – 3 ) = 0 are x = –2 and x = 3 (try them).
Finally consider the quadratic equation x² – 8x + 15 = 0.
By factorising we can rewrite this equation as ( x – 3 )( x – 5 ) = 0.
For this to be true, either x – 3 = 0 or x – 5 = 0.
This gives two solutions for the original equation: x = 3 or x = 5.
We can check these by substituting back in: (3)² – 8×(3) + 15 = 0 and (5)² – 8×(5) + 15 = 0
We can use the principles above to solve any quadratic equations where the quadratic will factorise.
In each case we factorise the quadratic, then set each bracket in turn equal to zero to find the solutions.
Other Notes
  • Always make sure the equation has the quadratic on one side and zero on the other.
  • Most quadratic equations have two solutions. Some have one solution and others have no solutions.
  • Not every quadratic factorises; this does not necessarily mean that there are no solutions.
Examples
Make sure you understand all the following examples before moving on:
e.g.
x² + 4x – 21 = 0
( x + 7 )( x – 3 ) = 0
x + 7 = 0 or x – 3 = 0
x = – 7 or x = 3
e.g.
x² – 49 = 0
( x + 7 )( x – 7 ) = 0
x + 7 = 0 or x – 7 = 0
x = – 7 or x = 7
e.g.
4x² – 15x + 9 = 0
( 4x – 3 )( x – 3 ) = 0
4x – 3 = 0 or x – 3 = 0
x = ¾ or x = 3
e.g.
4x² – 4x + 1 = 0
( 2x – 1 )( 2x – 1 ) = 0
2x – 1 = 0 or 2x – 1 = 0
x = ½ (only one solution -
also called equal roots)
e.g.
x² – 6x + 10 = 0
(the quadratic
does not factorise
so can't be solved
by factorisation)
e.g.
8x² = 2x
8x² – 2x = 0
2x( 4x – 1 ) = 0
2x = 0 or 4x – 1 = 0
x = 0 or x = ¼
Practice Questions
Work out the answer to each question then click on the button marked
to see if you are correct.
Solve by factorising:

(a) x² – x – 20 = 0

Start by factorising:
( x + 4 )( x – 5 ) = 0
x + 4 = 0 or x – 5 = 0
x = – 4 or x = 5

(b) 16x² – 9 = 0

Use the difference of two squares:
(4x)² – (3)² = 0
( 4x + 3 )( 4x – 3 ) = 0
4x + 3 = 0 or 4x – 3 = 0
x = – ¾ or x = ¾

(c) 4x² – 24x + 36 = 0

You can factorise this but all three terms have a common factor of 4 so it's easiest to divide through by 4 first:
x² – 6x + 9 = 0
( x – 3 )( x – 3 ) = 0
x – 3 = 0 or x – 3 = 0
x = 3

 

Solving Quadratic Equations using the Formula

The Quadratic Formula
If ax² + bx + c = 0, then x = b   b² – 4ac
2a
If a quadratic factorises then it is always quicker and easier to solve a quadratic equation that way, especially without a calculator. Nonetheless, the quadratic formula provides an alternative method for solving quadratic equations.
Using the Formula
The stages you should go through when using the formula are as follows:
  • Work out a, b and c (the coefficients of x², x and the number term)
  • Work out the values of –b, b² – 4ac and 2a
  • Substitute these values into the formula
  • Work out the two answers (the + and – in the sign give different answers)
Look at this example: x² + 4x – 21 = 0
a = 1
b = 4
c = –21
      b = – 4
b² – 4ac = (4)² – (4 × 1 × –21) = 16 – (–84) = 100
2a = 2
x = –4   100
2
x = – 4 + 10 or x = – 4 – 10
2 2
x = 3 or x = –7
Examples
e.g. 9x² – 9x + 2
a = 9
b = –9
c = 2
      b = 9
b² – 4ac = (–9)² – (4 × 9 × 2) = 81 – 72 = 9
2a = 18
x = 9   9
18
x = 9 + 3 or x = 9 – 3
18 18
x = 12 or x = 6
18 18
x = 2 or x = 1
3 3

Note that this question would have been much easier using factorisation!
e.g. 2x² – 12x + 18
a = 2
b = –12
c = 18
      b = 12
b² – 4ac = (–12)² – (4 × 2 × 18) = 144 – 144 = 0
2a = 4
x = 12   0
4
x = 12 + 0 or x = 12 – 0
4 4
x = 12 or x = 12
4 4
x = 3

Note that there will only ever be one solution when b² – 4ac = 0.

e.g. x² – 6x + 10
a = 1
b = –6
c = 10
      b = 6
b² – 4ac = (–6)² – (4 × 1 × 10) = 36 – 40 = – 4
2a = 2
x = 6   – 4
2
There are no solutions as you can not square root a negative number.
Note that there will never be any solutions when b² – 4ac is negative.
Practice Questions
Work out the answer to each question then click on the button marked
to see if you are correct.
Solve these quadratic equations using the formula:

(a) x² + 8x + 7 = 0

a = 1
b = 8
c = 7
      b = –8
b² – 4ac = (8)² – (4 × 1 × 7) = 64 – 28 = 36
2a = 2
x = –8   36
2
x = –8 + 6 or x = –8 – 6
2 2
x = –2 or x = –14
2 2
x = –1 or x = –7

(b) 3x² – 4x + 2 = 0

a = 3
b = –4
c = 2
      b = 4
b² – 4ac = (–4)² – (4 × 3 × 2) = 16 – 24 = –8
2a = 6
x = 4   –8
6
There are no solutions as you can not square root a negative number.

(c) 20x² – 9x + 1 = 0

a = 20
b = –9
c = 1
      b = 9
b² – 4ac = (–9)² – (4 × 20 × 1) = 81 – 80 = 1
2a = 40
x = 9   1
40
x = 9 + 1 or x = 9 – 1
40 40
x = 10 or x = 8
40 40
x = 1 or x = 1
4 5

 


Exercise

Work out the answers to the questions below and fill in the boxes. Click on the
button to find out whether you have answered correctly. If you have then the answer will be ticked
and you should move on to the next question. If a cross
appears then your answer is wrong. Click on
to clear the incorrect answer and have another go, or you can click on
to get some advice on how to work out the answer and then have another go. If you still can't work out the answer after this then you can click on
to see the solution.

 
In all questions please write terms in decreasing powers of x,  i.e. 3x² + 5x + 2 rather than 2+ 5x +3x².

 

Question 1

(a) Fill in the table of values below for the quadratic formula:

y = 2x² – 4x – 6

x –2 –1 0 1 2 3 4
y

Substitute each x value in turn into the formula to get the y value. Remember that if you square a negative number you get a positive result. See the boxes above for the correct answers.

(b) Plot the graph of y = 2x² – 4x – 6 on the grid on the right-hand side.

To plot the graph, click the button, then click each point you want to plot on the graph. When you have finished, click the button and a smooth line will be drawn in. You can then check if your answer is correct using the buttons below. If you need to have another go, click again.

Each pair of values for x and y should be plotted as coordinates ( x , y ). The correct graph has been plotted for you on the grid.

(c) Use your graph to identify the two roots of the quadratic equation:

2x² – 4x – 6 = 0.

x = or x =

The roots of the equation are the x-values which make the quadratic evaluate to zero. The quadratic evaluates to zero where it crosses the x-axis at (–1,0) and (3,0) so the solutions are x = –1 and x = 3.
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Question 2

Multiply out the brackets below, simplifying your answers:

(a)

( x + 5 )( x – 3 ) =
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Multiply each term in the first bracket by every term in the second bracket. x² – 3x + 5x – 15
= x² + 2x – 15

(b)

( 2x + 3 )( 2x – 3 ) =
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Multiply each term in the first bracket by every term in the second bracket. 4x² – 6x + 6x – 9
= 4x² – 9

(c)

( 2 + 3x )( x + 7 ) =
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Multiply out the brackets as normal, being careful as the terms are in a different order. Make sure you write your answer with decreasing powers of x. 2x + 14 + 3x² + 21x
= 3x² + 23x + 14

Question 3

Factorise the following expressions completely:

(a)

x² – 12x + 27 =
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Look for a solution in the form ( x – ? )( x – ? ). –3 + –9 = –12 and –3 × –9 = 27, so
the answer is ( x – 3 )( x – 9 )

(b)

4x² – 81 =
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This is an example of "difference of two squares". 4x² – 81 = (2x)² – (9)², so
the answer is ( 2x + 9 )( 2x – 9 )

(c)

10x² + 59x – 6 =
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The first values in each bracket are 10x and x respectively. The factorised version is ( 10x – 1 )( x + 6 )

(d)

20x² + 4x =
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Look for a common factor of both terms. 4x is the common factor so taking that
outside the brackets gives 4x( 5x + 1 )

Question 4

Solve the quadratic equations below by factorising, simplifying your answers where necessary.
If you think there is only one solution, place a dash (–) in the second solution box.
If you think there are no solutions then put a dash in both boxes.

(a)

x² – 10x + 16 = 0
x =
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or x =
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Factorise into the form ( x – ? )( x – ? ). ( x – 2 )( x – 8 ) = 0
x – 2 = 0 or x – 8 = 0
x = 2 or x = 8

(b)

10x² – 5x = 0
x =
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or x =
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Player 8 or above. Click here.
Look for a common factor of both terms to take outside the brackets. 5x ( 2x – 1 ) = 0
5x = 0 or 2x – 1 = 0
x = 0 or x = ½

(c)

25x² + 20x + 4 = 0
x =
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Player 8 or above. Click here.
or x =
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Player 8 or above. Click here.
In this case both the brackets are the same as each other once factorised. ( 5x + 2 )( 5x + 2 ) = 0
5x + 2 = 0 or 5x + 2 = 0
x = –2 (only one solution)
5

Question 5

Solve the quadratic equations below using the formula, simplifying your answers where necessary.
If you think there is only one solution, place a dash (–) in the second solution box.
If you think there are no solutions then put a dash in both boxes.

(a)

x² + x – 2 = 0
x =
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or x =
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a = 1
b = 1
c = –2
      b = – 1
b² – 4ac = (1)² – (4 × 1 × –2) = 9
2a = 2
x = –1   9
2
x = – 1 + 3 or x = – 1 – 3
2 2
x = 1 or x = –2

(b)

2x² – 4x + 3 = 0
x =
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Player 8 or above. Click here.
or x =
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Player 8 or above. Click here.
a = 2
b = –4
c = 3
      b = 4
b² – 4ac = (–4)² – (4 × 2 × 3) = –8
2a = 4
x = 4   –8
4
There are no solutions because you can't square-root a negative number.

(c)

4x² – 1 = 0
x =
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or x =
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Player 8 or above. Click here.
There is no "x term" so b = 0.
a = 4
b = 0
c = –1
      b = 0
b² – 4ac = (0)² – (4 × 4 × –1) = 16
2a = 8
x = 0   16
8
x = 0 + 4 or x = 0 – 4
8 8
x = ½ or x = –½

Question 6

Solve the quadratic equations by any method you wish, simplifying your answers where necessary.
If you think there is only one solution, place a dash (–) in the second solution box.
If you think there are no solutions then put a dash in both boxes.

(a)

3x² + 13x = 10
x =
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Player 8 or above. Click here.
or x =
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Player 8 or above. Click here.
Start by getting everything on the left-hand side of the equation with zero on the right-hand side. 3x² + 13x – 10 = 0
( 3x – 2 )( x + 5 ) = 0
3x – 2 = 0 or x + 5 = 0
x = 2 or x = –5
3

(b)

5x² – 10x = 4x + 3
x =
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or x =
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Player 8 or above. Click here.
Get everything on the left-hand side of the equation (remember to collect like terms). 5x² – 14x – 3 = 0
( 5x + 1 )( x – 3 ) = 0
5x + 1 = 0 or x – 3 = 0
x = –1 or x = 3
5

(c)

(x – 5)² = 9
x =
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or x =
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Player 8 or above. Click here.
Multiply out the brackets as ( x – 5 )( x – 5 ) and then gather everything on the left-hand side of the equation. x² – 10x + 25 = 9
x² – 10x + 16 = 0
( x – 2 )( x – 8 ) = 0
x – 2 = 0 or x – 8 = 0
x = 2 or x = 8

(d)

y² + 3y + 2 = 0
y =
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or y =
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Treat the quadratic exactly as if the "y"s were "x"s. ( y + 2 )( y + 1 ) = 0
y + 2 = 0 or y + 1 = 0
y = –2 or y = –1


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