SOLVING QUADRATICS
SOLVING QUADRATICS
•General Form:
y  ax  bx  c
2
•Where a, b and c are constants
To solve a quadratic equation, the
equation must be expressed in the form:
ax  bx  c  0
2
That is, all variables and constants must be
on one side of the equals sign with zero on
the other.
Methods for Solving Quadratic equations
ax  bx  c  0
2
•Method 1 : Factorisation
•Method 2 : Quadratic Formula
 b  b 2  4ac
x 
2a
•Method 3 : Completing the Square
•Method 4 : Graphic Calculator
•Method 1 : Factorisation
Example 1 : x  12x  20  0
2
Step 1: Equation
in the right form
Step 2: Factorise
x  12x  20  0
2
x 10x  2
Step 3: Separate
either ( x  10)  0
the two parts of
the product
Step 4: Solve
each equation
x  10
or
or
 0
( x  2)  0
x2
Some points to note
Example 1 :
x 2  12x  20  0
x 2  12x  20  0
x 10x  2
 0
Need to remember your
factorisation skills
The equation is expressed as the product of two factors being
equal to zero, therefore, one (or both) of the factors must be
zero.
either x  10  0
or
x2 0
x  10
or
x2
To check your answers are correct you can substitute them
one at a time into the original equation.
Checking solutions
Example 1 :
x 2  12x  20  0
x  10
10  12(10)  20
 100 120 20
 0
x2
2  12(2)  20
 4  24  20
 0
2
2
•Method 1 : Factorisation
Example 2 :
Step 1: Equation
in the right form
Step 2: Factorise
Step 3: Separate
the two parts of
the product
Step 4: Solve
each equation
x  3x  28
2
x  3x  28  0
2
x  7x  4
either x  7  0
x7
 0
or
or
x40
x4
•Method 2 : Quadratic Formula
Example 1 :
x  8x  5  0
 b  b 2  4ac
x 
2a
2
Step 1: Determine the values of a, b and c
a = 1 b = – 8 and c = – 5
Step 2: Substitute the
values of a, b and c
Step 3: Simplify
This step is optional.
Students need to have
covered surds
8  (8) 2  4(1)(5)
x 
2(1)
8  64  20
x 
2
8  84
x 
2
8  2 21
x 
See next slide
2
Note that 2 is a factor
of the numerator
8  2 21
x 
2
Once 2 is factored out, it
can be cancelled with the
2 in the denominator
2 4  21
x 
2
x  4  21
x 
4  21 


Exact answer
8.58
4  21   0.58
Decimal
approximation
•Method 2 : Quadratic Formula
Example 2 :
3x  8 x  2  0
 b  b 2  4ac
x 
2a
2
Step 1: Determine the values of a, b and c
a = 3 b = 8 and c = 2
Step 2: Substitute the
values of a, b and c
Step 3: Simplify
This step is optional.
Students need to have
covered surds
 8  82  4(3)(2)
x 
2(3)
 8  64  24
x 
6
 8  40
x 
6
 8  2 10
x 
6
See next slide
Note that 2 is a factor
of the numerator
Once 2 is factored out, it
can be cancelled with the
6 in the denominator
 8  2 10
x 
6
4  10
x 
3
x 

2  4  10
x 
6
 4  10

3
 4  10

3

Exact answer
 0.28
 2.39
Decimal
approximation
•Method 2 : Quadratic Formula
Example 3 :
5x  7 x  6  0
 b  b 2  4ac
x 
2a
2
Step 1: Determine the values of a, b and c
a = 5 b = – 7 and c = 6
Step 2: Substitute the
values of a, b and c
7  (7) 2  4(5)(6)
x 
2(5)
Step 3: Simplify
7  49  120
x 
10
Problem - you cannot
find the square root of
a negative number
7   71
x 
10
2 Imaginary Solutions
That was a lot of work to find
that there was no solution!
It would be useful to be able to
“test” the equation before we
start.
For this we use the
DISCRIMINANT.
The Discriminant   b  4ac
2
The discriminant is a quick way to
check how many real solutions exist
for a given quadratic equation.
As shown above the symbol for the
discriminant is  and it is
2
calculated using b  4ac .
Summary of Results using Discriminant
  b  4ac
2
The equation has two
 > 0 real solutions
=0

The equation has one
real solutions
The
equation
has
no
<0
real solutions
Relating the Discriminant to graphs
  b  4ac
2
>0
y
y
y)
x
x
x
The graph cuts the x-axis in two places.
These are the 2 real solutions to the
quadratic equation.
Relating the Discriminant to graphs
  b  4ac
2
= 0
y
y)
y
x
x
x
These graphs have their turning point
on the x-axis and hence there are 2
equal solutions.
Relating the Discriminant to graphs
  b  4ac
2
< 0
y)
y
y
x
x
There are 2 imaginary solutions in
this case because the graphs do not
intersect with the x-axis.
x
•Method 3: Completing the Square Technique
Example 1 :
x  6x  3  0
2
( x  3)  9  3  0
2
This value
is half b
Subtract the
square of the
number in
the bracket
( x  3)  6  0
2
Add 6 to both sides
( x  3)  6
Take the square root
x3  6
of both sides
Subtract 3 from both
sides
2
x  3 
6
•Method 3: Completing the Square Technique
x  3 
6
This result gives us the exact answers.
x  3 
6
or
x  3  6
Use your calculator to find decimal
approximations accurate to two decimal places.
x  0.55
or
x  5.45
•Method 3: Completing the Square Technique
Example 2 :
x  5x  8  0
2
5 2 25
(x  )   8  0
2
4
This value
is half b
5 2 57
(x  ) 
0
2
4
5 2 57
(x  ) 
Add to both sides
2
4
Subtract the
square of the
number in
the bracket
57
4
Take the square root
of both sides
Subtract 2.5 from
both sides
57
x  2.5  
2
 5  57
x
2
Example 3: Solve

 x  7 x  10  0
2

 x  7 x  10  0
2
 ( x  3.5)  12.25  10  0
This value
is half b
 [( x  3.5)  2.25]  0
2
 ( x  3.5)  2.25  0
Factor out the
coefficient of x²
Subtract the
square of the
number in
the bracket
2
 ( x  3.5)   2.25
2
( x  3.5)  2.25
2
See next slide
( x  3.5)  2.25
2
x  3.5   2.25
x  3.5 
i.e. x  3.5 
2.25
2.25 or x  3.5 
2.25
•Method 4 Graphic Calculator
Graph the function
and find the value of
the x-intercepts
Use a solver function for a
polynomial of degree 2
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