“Teach A Level Maths”
Vol. 1: AS Core Modules
5: Solving Equations
© Christine Crisp
Solving Equations
Module C1
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Expressions, Equations and Identities
e.g. 3 x 2  2 x  1 is an expression
The value of the expression can be found for any
value of the unknown, x
e.g. x  2
 3 x 2  2 x  1  12  4  1  7
e.g. 2 x  1  0 is a linear equation
2
e.g. 3 x  2 x  1  0 is a quadratic equation
These equations can be solved. There is one value
satisfying the 1st equation and two values which
satisfy the 2nd equation.
e.g. 3 x  2 x  1  (3 x  1)( x  1)
2
is an identity
An identity is true for all values of the unknown.
( Identities are sometimes written with
 instead of = )
Solving Equations
Solving Linear Equations
 Collect the terms containing the unknown on
one side of the equation and the constants on
the other
e.g.
3x  4  x  7
 3x 


Linear equations only have
xconstants
 7  4 and x-terms
without powers.
2 x  11
11
x
2
Solving Equations
Solving Quadratic Equations
x2  x
e.g. 1
Get zero on one side

x x0
2
Try to factorise

x ( x  1)  0
Do NOT cancel x as
a solution will then
be lost.
( Common factor )
Two factors multiplied together = 0,
so one must be zero.

x0
or
x 1  0
x 1

Solving Equations
Solving Quadratic Equations
e.g. 2
Zero on one side

x  7x  6
2
x2  7x  6  0
Try to factorise

( Trinomial )
3 2
or
6 1
( x  6)( x  1)  0
Two factors multiplied together = 0, so one factor
must equal zero.


x6  0
x6
or
or
x 1  0
x 1
Solving Equations
Solving Quadratic Equations
 x2  7x  8  0
e.g. 3
4 2
or
8 1
Multiply by -1

x  7x  8  0

( x  8)( x  1)  0
2
Trinomial
Try to factorise
Two factors multiplied together = 0, so one factor
must be zero.

x8
or
x  1
Solving Equations
Solving Quadratic Equations
4
x
x
e.g. 4
Multiply by x

x2  4
In this example there is no linear term.
Instead of getting 0 on the r.h.s. we can
square root directly.

x  2
N.B.
Solving Equations
Exercises
Solve the following quadratic equations
Solutions
1.
x2  5x  6  0
2.
x  5x  0
3.
x 50
2
2
2
( x  6)( x  1)  0  x  6,  1
x( x  5)  0  x  0, 5
x 5
2
 x   5,
2
x  x  12
x  x  12  0
( x  4)( x  3)  0
9
5.
x
x
2
6. 2 x  5 x  2  0
2
4.
5
x 9
 x  4,  3
 x  3
( 2 x  1)( x  2)  0  x  12 , 2
Solving Equations
A useful tip:
If a quadratic equation is written as ax 2  bx  c  0
2
then if b  4ac is a perfect square, the quadratic
will factorise
a  2, b  1, c  3
e.g. 1
2x 2  x  3
b 2  4ac  (1) 2  4(2)( 3)
 1  24  25
The quadratic factorises!
[
e.g. 2
2 x 2  x  3  (2 x  3)( x  1)
]
a  1, b  5, c  3
b 2  4ac  25  4(1)( 3)
 13
x 2  5x  3  0
The quadratic does not factorise!
Solving Equations
Solving Quadratic Equations
e.g. 5
x 2  4x  1  0
This quadratic doesn’t factorise so complete the square


( x  2) 2  4  1  0
( x  2) 2  3  0
To solve for x, we need to square root, so we isolate
the squared term on the left of the equal sign (l.h.s.)
Square
rooting



( x  2) 2  3
x2  3
x  2 3
N.B. 2 Solutions!
These answers are exact but can be given as
approximate decimals.
Solving Equations
Solving Quadratic Equations
The method used in the last example can be
generalised to give us a formula which is easier to use
2
when the coefficient of x is not 1
The formula will be proved but you don’t need
to know the proof.
However, you must memorise the result.
Solving Equations
Proof of the Quadratic Formula
Consider
ax 2  bx  c  0
b
c
Divide by a:
2
x  x 0
a
a
2
2
b
b
Complete the square:  x       c  0
2a   2a 
a

2
b 
b2
c

 x   2 
2a 
a
4a



b
b2
c
x


2
2a
a
4a
b
b 2  4ac
x

2a
4a 2
 b  b 2  4ac
x
2a

Solving Equations
Solving Quadratic Equations
e.g. 6 Solve the equation
Solution: x   b 
3x 2  5x  4  0
b  4ac
2a
a  3, b   5, c   4



x
x
2
5
(5) 2  4(3)( 4)
5
6
25  48
5  73
x
6
6
or
5  73
x
6
Solving Quadratic Equations - SUMMARY
 Zero on one side
EXCEPTION: If there is no ‘x’ term write
the equation as x 2  c and square root.
 Try to factorise
• Common Factors
• Trinomial factors
If there are factors, factorise and solve
If there are no factors, complete the square
( if a = 1 ) or use the formula
 b  b 2  4ac
x
2a
Solving Equations
Exercises
1.
Use the most efficient method to solve
the following quadratic equations:
x2  4x  6  0
Solution: Complete the Square
( x  2) 2  4  6  0
 ( x  2) 2  10
 x  2   10  x  2  10
2
2. 2 x  5 x  1  0
Solution: Use the formula. a  2, b  5, c  1
5  25  8
5  33
 b  b 2  4ac
 x

x
4
4
2a
2
3. x  2 x  15  0
 x  5, 3
Solution: Factorise ( x  5)( x  3)  0
Solving Equations
The following slides contain repeats of
information on earlier slides, shown without
colour, so that they can be printed and
photocopied.
For most purposes the slides can be printed
as “Handouts” with up to 6 slides per sheet.
Solving Equations
Expressions, Equations and Identities
e.g. 3 x 2  2 x  1 is an expression
The value of the expression can be found for any
value of the unknown, x
e.g. x  2
 3 x 2  2 x  1  12  4  1  7
e.g. 2 x  1  0 is a linear equation
e.g. 3 x 2  2 x  1  0 is a quadratic equation
These equations can be solved. There is one value
satisfying the 1st equation and two values which
satisfy the 2nd equation.
e.g. 3 x  2 x  1  (3 x  1)( x  1)
2
is an identity
An identity is true for all values of the unknown.
( Identities can be written as  but only for emphasis.)
Solving Equations
Solving Linear Equations
Linear equations only have constants and x-terms
without powers.
 Collect the terms containing the unknown on one
side of the equation and the constants on the other
e.g.
3x  4  x  7
 3 x  x  7  4


2 x  11
11
x
2
Solving Equations
Solving Quadratic Equations - SUMMARY
 Zero on one side
EXCEPTION: If there is no ‘x’ term write
the equation as x 2  c and square root.
 Try to factorise
• Common Factors
• Trinomial factors
If there are factors, factorise and solve
If there are no factors, complete the
square ( if a = 1 ) or use the formula
 b  b 2  4ac
x
2a
Solving Equations
Solving Quadratic Equations
x2  x
e.g. 1
Get zero on one side

x x0
2
Try to factorise

x ( x  1)  0
Do NOT cancel x as
a solution will then
be lost.
( Common factor )
Two factors multiplied together = 0,
so one must be zero.

x0
or
x 1  0
x 1

Solving Equations
Solving Quadratic Equations
e.g. 2
x  7x  6
2
Zero on one side

x2  7x  6  0
Try to factorise

( x  6)( x  1)  0
( Trinomial )
3
2
or
6 1
Two factors multiplied together = 0, so one factor
must equal zero.


x6  0
x6
or
or
x 1  0
x 1
Solving Equations
Solving Quadratic Equations
 x2  7x  8  0
e.g. 3
or
Multiply by -1

x  7x  8  0

( x  8)( x  1)  0
2
4 2
8 1
Trinomial
Try to factorise
Two factors multiplied together = 0, so one factor
must be zero.

x8
or
x  1
Solving Equations
Solving Quadratic Equations
4
x
x
e.g. 4
Multiply by x

x2  4
In this example there is no linear term.
Instead of getting 0 on the r.h.s. we can
square root directly.

x  2
N.B.
Solving Equations
Solving Quadratic Equations
e.g. 5
x 2  4x  1  0
This quadratic doesn’t factorise so complete the square


( x  2) 2  4  1  0
( x  2) 2  3  0
To solve for x, we need to square root, so we isolate
the squared term on the left of the equal sign (l.h.s.)
Square
rooting



( x  2) 2  3
x2  3
x  2 3
N.B. 2 Solutions!
These answers are exact but can be given as
approximate decimals.
Solving Equations
Solving Quadratic Equations
e.g. 6 Solve the equation
3x 2  5x  4  0
 b  b  4ac
Solution: x 
2a
a  3, b   5, c   4
2



x
x
5
(5) 2  4(3)( 4)
5
6
25  48
5  73
x
6
6
or
5  73
x
6