1. science
2. mathematics
3. algebra

NSM Enhanced 10 5.1–5.3
Chapter Focus
Quadratic
Equations
2
In this chapter, students learn to solve
quadratic equations by factorising,
completing the square or using the
quadratic formula. An exercise on choosing
the best method for a given quadratic helps
develop a deeper understanding of the
three methods. Problems that involve real
situations are then provided so students
can see the relevance of solving quadratic
equations in everyday life.
... I just hope
it’s easy!
Learning Outcomes
x2 – 8x + 7 = 0
(x – 7)(x – 1) = 0
x – 7 = 0 or x – 1 = 0
x = 7 or 1
PAS5.3.2 Solves linear, quadratic and
ge
Chapter Contents
s
simultaneous equations, solves
and graphs inequalities, and
rearranges literal equations.
pa
2:01 Solution using factors
2:02 Solution by completing the square
2:03 The quadratic formula
Investigation: How many solutions?
2:04 Choosing the best method
Fun Spot: What is an Italian referee?
PAS5·3·2
PAS5·3·2
PAS5·3·2
PAS5·3·2
2:05 Problems involving quadratic
equations
PAS5·3·2
Investigation: Temperature and altitude
Fun Spot: Did you know that 2 = 1?
Maths Terms, Diagnostic Test, Revision
Assignment, Working Mathematically
Learning Outcomes
PAS5·3·2 Solves linear, quadratic and simultaneous equations, solves and graphs inequalities, and
rearranges literal equations.
Sa
m
pl
e
Working Mathematically Stage 5.3
1 Questioning, 2 Applying Strategies, 3 Communicating, 4 Reasoning, 5 Reflecting
Vocabulary Preview
28
Teacher’s notes
coefficient
completing the square
factorise
quadratic equation
quadratic formula
28
New Signpost Mathematics Enhanced 10 5.1–5.3 ``TEACHER EDITION
Chapter 2
Learning Outcomes
2:01 Solution Using Factors
Outcome PAS5·3·2
PAS5.3.2 Solves linear, quadratic and
simultaneous equations, solves
and graphs inequalities, and
rearranges literal equations.
Prep Quiz 2:01
Factorise:
Solve:
x2 + 4x + 3
6x2 − 3x
x+2=0
5x = 0
1
4
7
9
2
5
8
10
x2 − 5x + 4
x2 − 9
3x − 1 = 0
2x + 3 = 0
3 x2 + 5x
6 4x2 − 25
Knowledge and Skills
Students learn about:
A quadratic equation is one in which the highest power of the unknown
pronumeral is 2. So equations such as:
x2 + 4x + 3 = 0, x2 + 5x = 0,
x2 − 25 = 0 and 2x2 − 3x + 7 = 0
are all quadratic equations.
■ A quadratic
equation is an
equation of the
‘second degree’.
The solving of a quadratic equation depends on the following observation
(called the Null Factor Law).
•
solving equations of the form
ax2 + bx + c = 0 using:
– factors
– completing the square
– the quadratic formula
•
solving a variety of quadratic equations
such as
If ab = 0, then at least one of a and b must be zero.
3x2 = 4
x2 – 8x – 4 = 0
x(x – 4) = 4
(y – 2)2 = 9
Worked examples
1 a If (x − 1)(x + 7) = 0
then either
x − 1 = 0 or x + 7 = 0
∴ x = 1 or
x = −7
b If 2x(x + 3) = 0
then either
2x = 0 or x + 3 = 0
∴ x = 0 or
x = −3
c If (2x − 1)(3x + 5) = 0
then either
2x − 1 = 0 or 3x + 5 = 0
2x = 1 or
3x = −5
∴ x = 1--- or
x = –-----52
A quadratic equation can have two solutions.
s
c (2x − 1)(3x + 5) = 0
c 2x2 + 9x − 5 = 0
c 6x2 = 5x + 6
•
ge
Solutions
b 2x(x + 3) = 0
b x2 − 49 = 0
b 5x2 = 2x
pa
Solve the quadratic equations:
1 a (x − 1)(x + 7) = 0
2 a x2 + 4x + 3 = 0
3 a x2 + x = 12
3
Sa
m
pl
e
2 To solve these equations, they are factorised first so they look like the equations in
example 1.
b
x2 − 49 = 0
or x2 − 49 = 0
a
x2 + 4x + 3 = 0
(x + 3)(x + 1) = 0
(x − 7)(x + 7) = 0
So x2 = 49
So
x+3=0⎫
So
x−7=0⎫
∴
x = 7 or −7
⎬
⎬
or
x+1=0⎭
or
x+7=0⎭
ie
x = ±7
∴ x = −3 or −1
∴
x = 7 or −7
To factorise an
expression like
c
2x2 + 9x − 5 = 0
2x2 ⫹ 9x ⫺ 5,
(2x − 1)(x + 5) = 0
you can use the
⎫
So
2x − 1 = 0
METHOD
CROSS METHOD.
⎬
or
x+5=0⎭
1
∴
x = --- or −5
2
Homework 2:01
Answers
Prep Quiz 2:01
1 (x + 1)(x + 3)
2 (x − 4)(x − 1)
3 x(x + 5)
4 3x(2x − 1)
5 (x − 3)(x + 3)
7 −2
6 (2x − 5)(2x + 5)
8 1---
9 0
10 −1 1---
Working Mathematically
Students learn to:
•
solve quadratic equations and discuss
the possible number of roots for any
quadratic equation (Applying Strategies,
Communicating)
Facts
Students sometimes express their solutions
using ‘and’ such as x = 2 and x = 5. The
correct word to be used is ‘or’. The reason
comes from set theory.
continued ➜➜➜
Chapter 2 Quadratic Equations
checking the solutions of quadratic
equations
29
The intersection of two sets A and B, written
A ∩ B, is the set of elements that are in A
and B.
For example, if A = {1, 2, 3, 4} and
B = {2, 4, 6, 8}, then A ∩ B = {2, 4}.
The union of two sets A and B, written
A ∪ B, is the set of elements that are in
A or B (or both).
For example,
{1, 2, 3, 4} ∪ {2, 4, 6, 8} = {1, 2, 3, 4, 6, 8}.
The solution of a quadratic equation is a
union of two sets.
3
2
Chapter 2 Quadratic Equations
29
NSM Enhanced 10 5.1–5.3
Teaching Strategies
and Ideas
3 Before these equations are solved, all the terms are gathered to one side of the equation,
letting the other side be zero.
b
5x2 = 2x
c
6x2 = 5x + 6
a
x2 + x = 12
5x2 − 2x = 0
6x2 − 5x − 6 = 0
x2 + x − 12 = 0
(x + 4)(x − 3) = 0
x(5x − 2) = 0
(3x + 2)(2x − 3) = 0
So
x+4=0⎫
So
x=0⎫
So
3x + 2 = 0 ⎫
⎬
⎬
⎬
or
x−3=0⎭
or
5x − 2 = 0 ⎭
or
2x − 3 = 0 ⎭
2
2
3
∴
x = −4 or 3
∴
x = 0 or --∴
x = − --- or ---
There are a number of methods for
factorising quadratics. The main method
taught is the cross method (presented in
the Year 9 Student Book), the product sum
and factor method can also be taught.
5
The product, sum and factor method
b = 11
Now find two numbers h and k that have a
product of 10 and a sum of 11
hk = 10
h + k = 11
10 × 1 = 10
10 + 1 = 11
1
ge
Exercise 2:01
Find the two solutions for each equation.
Check by substitution to ensure your answers are correct.
a x(x − 5) = 0
b x(x + 7) = 0
d 5a(a − 2) = 0
e 4q(q + 5) = 0
g (x − 2)(x − 1) = 0
h (x − 7)(x − 3) = 0
j (y + 3)(y + 4) = 0
k (t + 3)(t + 2) = 0
m (a − 6)(a + 6) = 0
n (y + 8)(y − 7) = 0
p (a + 1)(2a − 1) = 0
q (3x + 2)(x − 5) = 0
s (4x − 1)(2x + 1) = 0
t (3a − 4)(2a − 1) = 0
v 6x(5x − 3) = 0
w (9y + 1)(7y + 2) = 0
pa
The two numbers are 10 and 1
Write the x term in the expression as a sum
of 10x and x
5x2 + 11x + 2 = 5x2 + x + 10x + 2
Foundation Worksheet 2:01
Quadratic equations PAS5·3·2
1 Factorise
a x2 − 3x
b x2 + 3x + 2
2 Solve
a x(x − 4) = 0 b (x − 1)(x + 2) = 0
c
f
i
l
o
r
u
x
2x(x + 1) = 0
6p(p − 7) = 0
(a − 5)(a − 2) = 0
(x + 9)(x + 5) = 0
(n + 1)(n − 1) = 0
2x(3x − 1) = 0
(6y − 5)(4y + 3) = 0
(5x − 1)(5x + 1) = 0
e
Factor in pairs
After factorising the left-hand side of each equation, solve the following.
b m2 − 5m = 0
c y2 + 2y = 0
a x2 + 3x = 0
e 9n2 − 3n = 0
f 4x2 + 8x = 0
d 6x2 + 12x = 0
h a2 − 49 = 0
i y2 − 36 = 0
g x2 − 4 = 0
2
2
k n − 100 = 0
l m2 − 64 = 0
j a −1=0
n a2 − 5a + 6 = 0
o y2 + 12y + 35 = 0
m x2 + 3x + 2 = 0
2
2
q x − 10x + 16 = 0
r m2 − 11m + 24 = 0
p a − 10a + 21 = 0
t x2 + 2x − 35 = 0
u a2 − 4a − 45 = 0
s h2 + h − 20 = 0
w y2 − 8y + 7 = 0
x a2 + 9a − 10 = 0
v x2 + x − 56 = 0
m
pl
2
Sa
Quadratic equations 1
s
Of course, you can always check your solutions by substitution. For example 3a above:
Substituting x = −4
Substituting x = 3
x2 + x = 12
x2 + x = 12
■ L.H.S. = left-hand side
L.H.S. = (3)2 + (3)
L.H.S. = (−4)2 + (−4)
R.H.S. = right-hand side
= 16 − 4
=9+3
= 12
= 12
= R.H.S.
= R.H.S.
∴ Both x = −4 and x = 3 are solutions.
ac = 5 × 2
= 10
Technology
2
To solve a quadratic equation:
• gather all the terms to one side of the equation
• factorise
solve the two resulting simple equations.
Given the quadratic trinomial 5x2 + 11x + 2
and equating it to the standard form
ax2 + bx + c
(5x2 + x) + (10x + 2) = x(5x + 1) + 2(5x + 1)
= (5x + 1)(x + 2)
3
30
New Signpost Mathematics Enhanced 10 5.1–5.3
Foundation Worksheets
Technology
Quadratic equations 2
2:01 Quadratic equations
Class Tutorial:
Solving quadratics by completing
the square
Interactive:
Factorising quadratic trinomials
30
New Signpost Mathematics Enhanced 10 5.1–5.3 ``TEACHER EDITION
Chapter 2
4
Factorise and solve the following.
b
a 2x2 + x − 1 = 0
d
c 3x2 + 17x + 10 = 0
f
e 2x2 − x − 10 = 0
h
g 4x2 + 21x + 5 = 0
j
i 4x2 − 21x + 5 = 0
l
k 2x2 + 13x − 24 = 0
n
m 4x2 − 4x − 3 = 0
p
o 9x2 + 9x + 2 = 0
r
q 12x2 − 7x + 1 = 0
Learning Outcomes
+ 7x + 2 = 0
2x2 − 11x + 12 = 0
2x2 − 11x − 21 = 0
4x2 − 19x − 5 = 0
5x2 + 16x + 3 = 0
7x2 + 48x − 7 = 0
6x2 − x − 1 = 0
10x2 + 9x + 2 = 0
10x2 − 13x + 4 = 0
3x2
These are harder to
factorise!
PAS5.3.2 Solves linear, quadratic and
simultaneous equations, solves
and graphs inequalities, and
rearranges literal equations.
Knowledge and Skills
Students learn about:
Gather all the terms to one side of each equation and then solve.
b m2 = 8m
c x2 = −5x
a x2 = 3x
e a2 = 2a + 15
f y2 = 3y − 2
d x2 = 5x − 4
h n2 = 7n + 18
i h2 = 4h + 32
g m2 = 9m − 18
k y2 + 2y = 3
l x2 − 7x = −10
j x2 + x = 2
n t2 + 3t = 28
o y2 + 2y = 15
m y2 + 3y = 18
q 2x2 − x = 15
r 4m2 − 3m = 6
p 2x2 + x = 1
t 5p2 = 17p − 6
u 2x2 = 11x − 5
s 3x2 = 13x − 14
•
solving equations of the form
ax2 + bx + c = 0 using:
■ Check answers
by substitution.
2:02 Solution by Completing
the Square
•
Outcome PAS5·3·2
pa
Solutions
x2 − 5x + (− --5- )2 = (x −
2
--5- )2
2
the quadratic formula
checking the solutions of quadratic
equations
1 (x + 1)2
2 (x – 1)2
3 (x + 3)2
4 (x – 5)2
1
m
pl
Now, to solve a quadratic equation using this technique, we follow the steps in the example
below.
x2 + 4x − 21 = 0
■ Move the constant to the R.H.S.
= 21
x2 + 4x
Add (half of x coefficient)2 to both sides.
x2 + 4x + 22 = 21 + 22
2
∴ (x + 2) = 25
x + 2 = ± 25
x = −2 ± 5
∴ x = 3 or −7
Sa
–
Expand and simplify:
e
x2 + 8x + 42 = (x + 4)2
completing the square
Prep Quiz
Because (x + a)2 = x2 + 2ax + a2, the coefficient of the x term must be halved to give the value
of a.
2 x2 − 5x + . . .
1 x2 + 8x + . . .
Half of 8 is 4, so the perfect square is:
Half of −5 is − 5--- , so the perfect square is:
2
–
3x2 = 4
x2 – 8x – 4 = 0
x(x – 4) = 4
(y – 2)2 = 9
•
What must be added to the following to make perfect squares?
1 x2 + 8x
2 x2 − 5x
factors
solving a variety of quadratic equations
such as
ge
This method depends upon completing an algebraic expression to form a perfect square, that is, an
expression of the form (x + a)2 or (x − a)2.
Worked examples
–
s
3
1
5 (x + 2 )2
6 (x – 4 )2
7 (2x + 3)2
8 (3x – 4)2
1
1
9 (3x + 3 )2
10 (2x – 5 )2
Answers
1 x2 + 2x + 1
2 x2 – 2x + 1
3 x2 + 6x + 9
4 x2 – 10x + 25
31
Chapter 2 Quadratic Equations
6 x2 – 2 x +16
7 4x2 + 12x + 9
8 9x2 – 24x + 16
4
1
9 9x2 + 2x + 9
Answers
Exercise 2:01
1 a
d
g
j
m
0, 5
0, 2
2, 1
−3, −4
6, −6
p −1,
s
v
1
--4
,
0,
1
--2
− 1--2
3
--5
0, −3
b
5, 0
d 0, −2
e
0,
1
--3
f
0, −2
2, −2
1, −1
−1, −2
7, 3
4, −5
7, −8
h
k
n
q
t
w
7, −7
10, −10
2, 3
8, 2
5, −7
7, 1
i
l
o
r
u
x
6, −6
8, −8
−5, −7
8, 3
9, −5
1, −10
1
--2
, −1
b
− 1--- , −2
c − 2--- , −5
4 a
d
g
j
m
d 4, 1 1---
e
2 1--- , −2
f
7, −1 1---
h
− 1--4
i
1
--4
2 a
b
e
h
k
n
0, −7
0, −5
7, 3
−3, −2
−8, 7
c
f
i
l
o
q
− 2--- , 5
r 0,
t
1 1--3
w
− 1--9
3
,
1
--2
,
− 2--7
0, −1
0, 7
5, 2
−9, −5
−1, 1
u
5
--6
x
1
--5
1
--3
,
− 3--4
,
− 1--5
g
j
m
p
s
v
3 a
2
g
− 1--4
, −5
3
2
5,
c 0, −2
3
2
5,
1
1
1
5 x2 + x + 4
j
− 1--- , −3
m
1 1--2
p
− 2--5
5
,
− 1--2
,
− 1--2
k
1 1--- , −8
n
1
--2
2
,
− 1--3
,
1
--4
q
1
--3
0, 3
4, 1
6, 3
1, −2
3, −6
b
e
h
k
n
0, 8
5, −3
9, −2
1, −3
4, −7
p
1
--2
, −1
q
3, −2 1---
s
2 1--- , 2
t
3,
3
1
10 4x2 – 5 x + 25
2
2
--5
Chapter 2 Quadratic Equations
l
1
--7
o
− 1--3
, − 2---
r
4
--5
1
--2
c
f
i
l
o
0, −5
2, 1
8, −4
5, 2
3, −5
, −7
,
3
r 2, − 3--4
u 5,
1
--2
31
NSM Enhanced 10 5.1–5.3
Note that the previous example could have been factorised to give (x − 3)(x + 7) = 0, which,
of course, is an easier and quicker way to find the solution. The method of completing the square,
however, can determine the solution of quadratic equations that cannot be factorised. This can be
seen in the examples below.
Technology
Completing the square
Worked examples
Solve:
1 x2 + 6x + 1 = 0
2 x2 − 3x − 5 = 0
3 3x2 − 4x − 1 = 0
Solutions
1 x2 + 6x + 1 = 0
x2 + 6x
= −1
x2 + 6x + 32 = −1 + 32
(x + 3)2 = 8
x+3=± 8
∴ x = −3 ± 8
ie x = −3 + 2 2 or −3 − 2 2
(x −0·17 or −5·83)
Completing the square
2
2
(x −
x
Technology
32
=
+ (− 2--- )2 =
3
2 2
--- )
3
=
1
--3
1
--3
7
--9
4
= ± 7 1--4
s
When the coefficient
of x2 is not 1,
we first of all
divide each term
by that coefficient.
3
e
32
= 7 1---
+ (− 2--- )2
2
7
x − --- = ± ------3
3
2
7
∴ x = --- ± ------3 3
2– 7
2+ 7
ie x = ---------------- or ---------------3
3
(x 1·55 or −0·22)
Sa
Completing the Square activity on the
Companion Website requires students to
select an equation from a list and calculate
the third term required to complete the
square. It also allows students to match the
equation with its turning point form.
Students can view a variety of different
graphs and match one with the original
equation.
3
m
pl
ICT
4
--- x
3
(x −
Class Tutorial:
Solving quadratics using formulae
Homework 2:02
3
4
--- x
3
x2 −
x2 −
ge
3x2 − 4x − 1 = 0
x2 − 4--- x − 1--- = 0
pa
3
2
3 2
--- )
2
− 3--2
3
29
∴ x = --- ± ---------2 2
3 – 29
3 + 29
ie x = ------------------- or ------------------2
2
(x 4·19 or −1·19)
■ Note that the solution involves
a square root, ie the solution is
irrational. Using your calculator,
approximations may be found.
x2 − 3x − 5 = 0
x2 − 3x
=5
2
x − 3x + (− 3--- )2 = 5 + (− 3--- )2
■ Note:
You can use the following fact to check your
answers.
For the equation:
ax2 + bx + c = 0
the two solutions must add up to equal – --b
a
– 6In example 1, (−0·17) + (−5·83) = −6 [or ----]
New Signpost Mathematics Enhanced 10 5.1–5.3
Teacher’s notes
New Signpost Mathematics Enhanced 10 5.1–5.3 ``TEACHER EDITION
1
In example 3, 1·55 + (−0·22) = 1·33 [ 4--- ]
3
Chapter 2
Learning Outcomes
Exercise 2:02
PAS5.3.2 Solves linear, quadratic and
1
What number must be inserted to complete the square?
b x2 + 8x + . . .2 = (x + . . .)2
a x2 + 6x + . . .2 = (x + . . .)2
c x2 − 2x + . . .2 = (x − . . .)2
d x2 − 4x + . . .2 = (x − . . .)2
e x2 + 3x + . . .2 = (x + . . .)2
f x2 − 7x + . . .2 = (x − . . .)2
g x2 + 11x + . . .2 = (x + . . .)2
h x2 − x + . . .2 = (x − . . .)2
5x
2x
i x2 + ------ + . . .2 = (x + . . .)2
j x2 − ------ + . . .2 = (x − . . .)2
2
3
2
Solve the following equations, leaving your answers in surd form.
b (x + 1)2 = 2
c (x + 5)2 = 5
a (x − 2)2 = 3
e (x − 3)2 = 7
f (x + 2)2 = 11
d (x − 1)2 = 10
h (x + 10)2 = 12
i (x − 3)2 = 18
g (x + 3)2 = 8
j (x + --1- )2 = 5
k (x − --2- )2 = 3
l (x + 1 --1- )2 = 12
2
3
2
1
1
2
2
m (x − 1) = 2 --n (x + 3) = 4 --o (x − 1--- )2 = 5---
•
x=
9
Solve the following equations by completing the square. Also find approximations for your
answers, correct to two decimal places.
b x2 − 2x − 5 = 0
c x2 − 4x − 8 = 0
a x2 + 2x − 1 = 0
e x2 − 6x + 2 = 0
f x2 + 4x + 1 = 0
d x2 + 6x − 8 = 0
h x2 + 2x = 4
i x2 − 12x = 1
g x2 + 10x = 5
k x2 + 7x − 3 = 0
l x2 + x − 3 = 0
j x2 + 5x + 2 = 0
n x2 + 3x − 5 = 0
o x2 − 11x + 5 = 0
m x2 + 9x + 3 = 0
2
2
q x + 3x = 2
r x2 − 5x = 1
p x −x=3
t 2x2 + 3x − 4 = 0
u 2x2 − 8x + 1 = 0
s 2x2 − 4x − 1 = 0
2
2
w 5x − 4x − 3 = 0
x 4x2 − x − 2 = 0
v 3x + 2x − 3 = 0
solving equations of the form
ax2 + bx + c = 0 using:
– factors
– completing the square
– the quadratic formula
•
solving a variety of quadratic equations
such as
pa
•
3x2 = 4
x2 – 8x – 4 = 0
x(x – 4) = 4
(y – 2)2 = 9
checking the solutions of quadratic
equations
Outcome PAS5·3·2
Working Mathematically
Students learn to:
e
As we have seen in the previous section, a quadratic equation is one involving a squared term.
In fact, any quadratic equation can be represented by the general form of a quadratic equation:
ax2 + bx + c = 0
−b ± b2 − 4 ac
2a
•
Completing the square
2:03 The Quadratic Formula
developing the quadratic formula
s
3
3
Knowledge and Skills
Students learn about:
ge
2
2
simultaneous equations, solves
and graphs inequalities, and
rearranges literal equations.
•
m
pl
where a, b, c are all integers, and a is not equal to zero.
If any quadratic equation is arranged in this form, a formula using the values of a, b and c can be
used to find the solutions.
solve quadratic equations and discuss
the possible number of roots for any
quadratic equation (Applying Strategies,
Communicating)
The quadratic formula for ax2 + bx2 + c = 0 is:
Sa
– b ± b 2 – 4ac
x = ------------------------------------2a
Chapter 2 Quadratic Equations
33
Answers
Exercise 2:02
1 a 3
e 3--2
i
5
--4
2 a 2± 3
b 4
f 7--2
j
c 1
g 11
-----2
d 2
h 1--2
1
--3
b –1± 2
c
–5± 5
e
3± 7
f
– 2 ± 11
g –3±2 2
i
3±3 2
j
– 1--2- ± 5
k
2
--3
± 3
d 1 ± 10
h – 10 ± 2 3
l
– 1 1--2- ± 2 3
5
m 1 ± --2
3 a
e
i
m
q
u
0·41, −2·41
5·65, 0·35
12·08, −0·08
−0·35, −8·65
0·56, −3·56
3·87, 0·13
3 2
n – 3 ± ---------2
b
f
j
n
r
v
3·45, −1·45
−0·27, −3·73
−0·44, −4·56
1·19, −4·19
5·19, −0·19
0·72, −1·39
5
± ------3
o
1
--3
c
g
k
o
s
w
5·46, −1·46
0·48, −10·48
0·41, −7·41
10·52, 0·48
2·22, −0·22
1·27, −0·47
d
h
l
p
t
x
1·12,
1·24,
1·30,
2·30,
0·85,
0·84,
Chapter 2 Quadratic Equations
−7·12
−3·24
−2·30
−1·30
−2·35
−0·59
33
NSM Enhanced 10 5.1–5.3
Prep Quiz
Using your calculator find the value of the
following correct to two decimal places
1 25 – 9
PROOF OF THE QUADRATIC
FORMULA
ax2 + bx + c = 0
b
c
x 2 + --x + -- = 0
a
a
b
x 2 + --x = – --c
a
a
2
b
b
b 2 c
x 2 + --x + ⎛ ------⎞ = ⎛ ------⎞ – -⎝
⎠
⎝
a
2a
2a⎠
a
2
b 2 – 4ac
b-⎞
⎛ x + ----= ------------------⎝
2a⎠
4a 2
This formula is very
useful if you can’t
factorise an expression.
2 – 25 – 9
3 –6 – 36 – 4 × 1 × 3
4 –7 + 49 – 4 × 1 × 2
–5 + 25 – 4 × 1 × 2
2
–5 – 25 – 4 × 1 × 2
6
2
5
b 2 – 4ac
b- ±
x + ----= ---------------------------2a
2a
– b ± b 2 – 4ac
x = ------------------------------------2a
–8 + 82 – 4 × 2 × 3
7
2×2
Worked examples
–8 – 8 – 4 × 2 × 3
2×2
2
10
– 13
2
( )
Solutions
1 For the equation 2x2 + 9x + 4 = 0,
a = 2, b = 9, c = 4.
Substituting these values into the
formula:
2 ×2
–
13 2
2
– 4 ×2 ×45
2 For x2 + 5x + 1 = 0,
a = 1, b = 5, c = 1.
Substituting into the formula gives:
–9 ± 9 2 – 4 × 2 × 4
= --------------------------------------------------2×2
Answers
3 –10·90
4 –0·60
5 –0·44
6 –4·56
7 –0·42
8 –3·58
9 –0·13
10 –3·12
– 5 – 21– 5 + 21- or ----------------------x = ----------------------2
2
Approximations for these answers may
be found using a calculator. In this case
they would be given as:
x −0·21 or −4·79 (to 2 dec. pl.)
-----= − 2--- or − 16
4
4
∴ x = − 1--- or −4
Sa
2
Facts
New Signpost Mathematics Enhanced 10 5.1–5.3
Syllabus Links
•
The method of completing the square to
solve quadratic equations with positive
roots was used by ancient Babylonians
(around 400 BC) and the Chinese, but
did not have a general formula.
•
Euclid (300 BC) produced a more
abstract geometrical method. The first
mathematician known to have used the
general algebraic formula, allowing
negative as well as positive solutions,
was Brahmagupta (India, 7th century).
34
34
–5 ± 5 2 – 4 × 1 × 1
= --------------------------------------------------2×1
– 5 ± 25 – 4
= -------------------------------2
–---------------------5 ± 21=
2
Since there is no rational equivalent to
21 the answer may be left as:
– 9 ± 81 – 32
= ----------------------------------4
–---------------------9 ± 49=
4
–--------------9 ± 7=
4
e
2 –4
m
pl
1 4
– b ± b 2 – 4ac
x = ------------------------------------2a
– b ± b 2 – 4ac
x = ------------------------------------2a
2 ×2
4 2x2 + 2x + 7 = 0
s
( 132 )2 – 4 ×2 ×45
ge
9
– 13
+
2
Solve the following by using the quadratic formula.
1 2x2 + 9x + 4 = 0
2 x2 + 5x + 1 = 0
3 3x2 = 2x + 2
pa
8
NOTE:
This proof uses the
method of completing
the square.
Students who go on to study Stage 6
Extension 2 Mathematics in Year 12 will
be able to use complex numbers to find
solutions to questions where b2 – 4ac
is negative. Students also learn about the
discriminant again in 2 Unit Advanced
Mathematics.
New Signpost Mathematics Enhanced 10 5.1–5.3 ``TEACHER EDITION
ICT
Students need to be aware of how their
calculator evaluates square roots. Some
calculators only take the square root of the
first number entered and not the whole
expression as required in the quadratic
formula. Students should complete the Prep
Quiz to avoid making these careless errors.
Chapter 2
3 The equation 3x2 = 2x + 2 must
first be written in the form
ax2 + bx + c = 0,
ie 3x2 − 2x − 2 = 0
So a = 3, b = −2, c = −2.
Substituting these values gives:
– b ± b 2 – 4ac
x = ------------------------------------2a
–-------------------------------------------------------------------------( – 2 ) ± ( – 2 ) 2 – 4 × 3 × ( – 2 -)
=
2×3
2
±
4
+
24
= -----------------------------6
2
±
28
= -------------------6
2 + 28
2 – 28
So x = ------------------- or ------------------6
6
(ie x 1·22 or −0·55 to 2 dec. pl.)
Teaching Strategies
and Ideas
4 For 2x2 + 2x + 7 = 0,
a = 2, b = 2, c = 7.
Substituting these values gives:
It can be pointed out to students that the
2
sign in front of b – 4 ac comes from the
fact that there are two square roots of a
number. This results in the two solutions
for the quadratic. When using the calculator
students have to do two calculations, one
with the ‘+’ sign and the other with the ‘–’.
It can also be pointed out that if b2 – 4ac is
negative then there are no solutions to the
equation. This is highlighted in
Investigation 2:03 on page 36.
–2 ± 2 2 – 4 × 2 × 7
x = --------------------------------------------------2×2
– 2 ± – 52
= -------------------------4
But – 52 is not real!
So 2x2 + 2x + 7 = 0 has no real solutions.
You should
learn this
formula!
■ The solutions of the
equation ax2 + bx + c = 0
are given by:
b ± b 2 – 4acx = –-----------------------------------2a
Answers
pa
– b ± b 2 – 4ac
1 Evaluate -------------------------------------- if:
2a
a a = 1, b = 3, c = 2 b a = 2, b = 5, c = −2
2 Solve:
b x2 − 3x − 1 = 0
a x2 + 5x + 2 = 0
i
− 6x + 5 = 0
l 4x2 + 11x + 6 = 0
o 3x2 − 5x + 2 = 0
r 8x2 − 14x + 3 = 0
e
x2
Solve the following, leaving your answers in surd form. (Remember: A surd is an expression
involving a square root.)
b x2 + 3x + 1 = 0
c x2 + 5x + 3 = 0
a x2 + 4x + 2 = 0
2
2
e x + 2x − 2 = 0
f x2 + 4x − 1 = 0
d x +x−1=0
h x2 − 7x + 2 = 0
i x2 − 6x + 3 = 0
g x2 − 2x − 1 = 0
k x2 − 8x + 3 = 0
l x2 − 5x + 7 = 0
j x2 − 10x − 9 = 0
n 2x2 + 3x − 1 = 0
o 2x2 − 7x + 4 = 0
m 2x2 + 6x + 1 = 0
q 3x2 − 9x + 2 = 0
r 5x2 + 4x − 2 = 0
p 3x2 + 10x + 2 = 0
t 3x2 − 3x − 1 = 0
u 4x2 − 3x − 2 = 0
s 4x2 − x + 1 = 0
w 2x2 − 9x + 8 = 0
x 5x2 + 2x − 1 = 0
v 2x2 + 11x − 5 = 0
Foundation Worksheets
2:03 The quadratic formula
Chapter 2 Quadratic Equations
d 5, −2
e 5, −3
f
g 7, 2
h 6, 2
i
5, 1
j
− 1--- , −2
k
− 1--2
, −5
l
− 3--- , −2
m 3,
− 1--2
n 2, − 1---
o 1,
2
--3
− 1--2
1
--3
2, −6
3
4
5
p
, −1 1--2
, − 2---
r 1 1--- ,
2
3
1
--4
–4± 8
--------------------2
–3± 5
b --------------------2
c
– 5 ± 13
-----------------------2
–1± 5
d --------------------2
e
– 2 ± 12
-----------------------2
f
2 a
2± 8
g ---------------2
35
b −2, −4
c −1, −9
q
m
pl
2
Use the quadratic formula to solve the following
equations. All have rational answers.
b x2 + 6x + 8 = 0
a x2 + 5x + 6 = 0
d x2 − 3x − 10 = 0
c x2 + 10x + 9 = 0
f x2 + 4x − 12 = 0
e x2 − 2x − 15 = 0
2
h x2 − 8x + 12 = 0
g x − 9x + 14 = 0
k 2x2 + 11x + 5 = 0
j 3x2 + 7x + 2 = 0
n 5x2 − 9x − 2 = 0
m 2x2 − 5x − 3 = 0
q 6x2 + 7x − 3 = 0
p 6x2 + 7x + 2 = 0
Sa
1
1 a −2, −3
ge
Foundation Worksheet 2:03
The quadratic formula PAS5·3·2
Exercise 2:03
s
Exercise 2:03
– 4 ± 20
-----------------------2
7 ± 41
h ------------------2
i
6 ± 24
------------------2
j
10 ± 136
------------------------2
k
8 ± 52
------------------2
l
no solutions
– 6 ± 28
m -----------------------4
– 3 ± 17
n -----------------------4
7 ± 17
o ------------------4
– 10 ± 76
p --------------------------6
9 ± 57
q ------------------6
r
– 4 ± 56
-----------------------10
s no solutions
t
3 ± 21
------------------6
3 ± 41
u ------------------8
v
– 11 ± 161
-----------------------------4
9 ± 17
w ------------------4
x
– 2 ± 24
-----------------------10
Chapter 2 Quadratic Equations
35