Transcript x 2

Integrated Programme/mainstream
Secondary Three Mathematics
Name : _____________(
) Class:______
Date : ___________ to _____________
Term 1: Unit 4
Notes
QUADRATIC FUNCTIONS AND INEQUALITIES
Factoring
Completing Square General Quadratic Formula
Discriminant and Nature of roots
Quadratic Graphs
Quadratic Inequalities
At the end of the unit, students should be able to
•
solve quadratic equations
(1) by factorization (recall Sec 2 work)
(2) by completing the square,
(3) by formula .
•
understand relationships between the roots and coefficients of the quadratic equation
•
form quadratic equations in the product form given two roots and ,
•
apply substitution to solve some higher order algebraic equations,
•
understand and use discriminant to determine the nature of roots of a quadratic
equation,
•
use discriminant to determine when is always positive (or always negative)
•
solve intersection problems between line and curve and discuss the nature of roots.
•
find the maximum or minimum value by using completing the square,
•
sketching of graphs of quadratic functions given in the form
(1) y = a(x-h)2 + c ,a > 0 or a < 0
(2) y = a(x – b)(x – c) , a > 0 or a < 0 .
•
solve quadratic inequalities using algebraic and graphical methods, representing the
Solving Quadratic
Equations by Factorising
Sec 2 Revision
A quadratic equation is an equation like:
y = x2
2 + 2
y
=
x
Contains a
2 + x – 4
2
y
=
x
x term
y = x2 + 2x – 3
There are several methods of solving
these but one methods that you must
know is called FACTORISING
3x2=6
0x2=0
3x0=0
A x B = 0 What can you say about A or B
(x + 3)(x – 2) = 0 means (x + 3) x (x – 2)
What can you say about (x + 3) or (x – 2)
x+3=0
x = -3
or
x-2=0
x=2
(x + 3)(x + 2)
You try
(x + 5)(x + 2)
x(x + 2) + 3(x + 2)
(x – 2)(x + 3)
(x + 2)(x – 4)
(x – 3)(x – 2)
 x X (x + 2) + 3 X (x + 2)
xXx+xX2+3Xx+3X2


x2 + 2x + 3x + 6
x2 + 5x + 6
Solve by factorising: 0 = x2 + 7x + 12
Write down all
the factor pairs
of 12.
(x )(x
)
What goes
From this list,
choose
the pairthe
that x?
with
adds up to 7
Put these numbers
into brackets
1 x 12 = 12
2 x 6 = 12
3 x 4 = 12
3+4=7
0 = (x + 3)(x + 4)
x = – 3 and – 4
Solve by factorising: 0 = x2 + x - 6
Write down all the
factor pairs of – 6
From this list,
choose the pair
that adds up to 1
Put these numbers
into brackets
1 x -6 = -6
2 x -3 = -6
3 x -2 = -6
6 x -1 = -6
(3) + (-2) = 1
3–2=1
0 = (x + 3)(x - 2)
x = – 3 and 2
Copy and fill in the missing
values when you factorise
x2 + 8x + 12 = 0
Solve by factorising
1. x2 + 3x + 2 = 0
2. x2 + x – 12 = 0
Find all the factor pairs of 12
2 – 12x – 20 = 0
3.
x
1 x 12 = 12
2 – x – 12 = 0
4.
x
2 x _ = 12
3 x 4 = 12
From these choose the pair
that add up to 8
_+6=8
Put these values into the
brackets (x + _)(x + _) = 0
x = -2 and - 6
1 x2 + 5x + 6 = 0
2
3
4
5
6
7
8
9
10
x2 - x – 6 = 0
x2 + 8x + 12 = 0
x2 + x – 12 = 0
x2 - 8x + 15 = 0
x2 + 3x – 21 = 0
x2 - 3x – 18 = 0
x2 - 10x – 24 = 0
x2 + 8x + 16 = 0
x2 - 4x – 60 = 0
1 x2 + 5x + 6 = 0
(x + 3)(x + 2)
2
3
4
5
6
7
8
9
(x – 3)(x + 2)
(x + 2)(x + 6)
(x – 3)(x + 4)
(x – 3)(x – 5)
(x + 7)(x – 4)
(x – 6)(x + 3)
(x - 12)(x + 2)
(x + 4)(x + 4)
(x - 10)(x + 4)
10
x2 - x – 6 = 0
x2 + 8x + 12 = 0
x2 + x – 12 = 0
x2 - 8x + 15 = 0
x2 + 3x – 21 = 0
x2 - 3x – 18 = 0
x2 - 10x – 24 = 0
x2 + 8x + 16 = 0
x2 - 4x – 60 = 0
1 x2 + 5x + 6 = 0
(x + 3)(x + 2)
-3 and -2
2
3
4
5
6
7
8
9
(x – 3)(x + 2)
(x + 2)(x + 6)
(x – 3)(x + 4)
(x – 3)(x – 5)
(x + 7)(x – 4)
(x – 6)(x + 3)
(x - 12)(x + 2)
(x + 4)(x + 4)
(x - 10)(x + 4)
3 and -2
-2 and -6
3 and -4
3 and 5
-7 and 4
6 and -3
12 and -2
-4 and -4
10 and -4
10
x2 - x – 6 = 0
x2 + 8x + 12 = 0
x2 + x – 12 = 0
x2 - 8x + 15 = 0
x2 + 3x – 21 = 0
x2 - 3x – 18 = 0
x2 - 10x – 24 = 0
x2 + 8x + 16 = 0
x2 - 4x – 60 = 0
x2 – 4
x2 + 0x – 4
(x – 2)(x + 2)
Notice that x2 – 4
could be written as
x 2 – 22
(x – 2)(x + 2)
-1 x 4 = -4
-2 x 2 = -4
4 x -1 = -4
-2 + 2 = 0
This is often called
the difference
between two
squares
x2 – 25
(x + 5)(x – 5)
1
2
3
4
5
6
7
8
x2 - 9
x2 - 100
x2 - 36
x2 - 49
x2 - 81
x2 - 64
x2 - 18
x2 - 24
1
2
3
4
5
6
7
8
x2 - 9
x2 - 100
x2 - 36
x2 - 49
x2 - 81
x2 - 64
x2 - 18
x2 - 24
(x + 3)(x – 3)
(x + 10)(x – 10)
(x + 6)(x – 6)
(x + 7)(x – 7)
(x + 9)(x – 9)
(x + 8)(x – 8)
(x + √18)(x – √18)
(x + √24)(x – √24)
Completing Square
For quadratic equations that are not
expressed as an equation between two
squares, we can always express them as
ax  bx  c  0
2
If this equation can be factored, then it
can generally be solved easily.
If the equation can be put in the form
k ( x  m)  n
2
2
then we can use the square root method
described previously to solve it.
“Can we change the equation from the
2
form ax  bx  c  0
to the form ( x  m )2  n 2 ?”
The procedure for changing ax 2  bx  c  0
is as follows. First, divide by a , this
gives
b
c
x  x 0
a
a
2
c
Then subtract from both sides. This
a
gives
b
c
x  x
a
a
2
Recall that
( x  r )  x  2rx  r
2
If we let
2
b
 2r
a
we can solve for r to get
b
r
2a
2
b
Substituting r 
in ( x  r )2  x 2  2rx  r 2
2
a
we get
2
b 2
b
b
2
(x  )  x  x  2
2a
a
4a
Using the symmetric property of
equations to reverse this equation we get
2
b
b
b 2
x  x  2  (x  )
a
4a
2a
2
Now we will return to where we left our
original equation.
2
b
c
b
2
If we add
to
both
sides
of
x  x
2
a
a
4
a
we get
2
2
b
b
c
b
x2  x  2    2
a
4a
a 4a
b 2  4ac

4a 2
or
b 2 b 2  4ac
(x  ) 
2a
4a 2
We can now solve this by taking the
square root of both sides to get
b
b  4ac
x

2a
2a
2
b
b  4ac
x

2a
2a
2
b  b  4ac
x
2a
2
Example:
3 x 2  12 x  20  0 [By completing square method]
20
2
3( x  4 x  )  0
3
20
2
3[( x  2)  4  ]  0
3
2
3( x  2)  32  0
3( x  2) 2  32
32
2
( x  2) 
3
x  1.27, x  5.27
Quadratic Formula
 b  b  4ac
x
2a
2
Song :
The Quadratic formula allows you to find the roots of a quadratic
equation (if they exist) even if the quadratic equation does not
factorise.
The formula states that for a quadratic equation of the form :
ax2 + bx + c = 0
The roots of the quadratic equation are given by :
 b  b 2  4ac
x
2a
Example :
Use the quadratic formula to solve the equation :
x 2 + 5x + 6= 0
Solution:
x 2 + 5x + 6= 0
a=1 b=5 c=6
 b  b 2  4ac
x
2a
 5  5 2  ( 4  1  6)
x
21
 5  25  ( 24)
x
2
5 1
x
2
x
51
or
2
x
 51
2
x = - 2 or x = - 3
These are the roots of the equation.
Example :
Use the quadratic formula to solve the equation :
8x 2 + 2x - 3= 0
Solution :
8x 2 + 2x - 3= 0
a = 8 b = 2 c = -3
 b  b  4ac
x
2a
2
 2  2 2  ( 4  8  3 )
x
2 8
 2  4  ( 96)
x
16
 2  100
x
16
 2  10
x
16
or
 2  10
x
16
x = ½ or x = - ¾
These are the roots of the equation.
Example :
Use the quadratic formula to solve for x to 2 d.p :
2x 2 +3x - 7= 0
Solution:
2x 2 + 3x – 7 = 0
a=2 b=3 c=-7
 b  b 2  4ac
x
2a
 3  3 2  ( 4  2  7 )
x
2 2
 3  9  ( 56)
x
4
 3  65
x
4
x
 3  8.0622
4
or
x
 3  8.0622
4
x = 1.27 or x = - 2.77
These are the roots of the equation.
Discriminant
Example :
x2 - 8x + 16 = 0
a=1;
b=-8;
c=16
b2-4ac=(-8)2-4(1)(16)
=64-64
b2-4ac=0
real, rational, equal
Example :
2x2 + 5x – 3 = 0
a= ?
b= ?
c= ?
b2-4ac=52-4(2)(-3)
=25+24
b2-4ac=49
Real, rational, unequal
Example :
x2 + 5x + 3 = 0
b2-4ac=52-4(1)(3)
=25-12
b2-4ac=13
real, irrational, unequal
Example :
x2 – x + 2 = 0
b2-4ac=12-4(1)(2)
=1-8
b2-4ac=-7
imaginary
b2 - 4ac = 0 : real, rational, equal.
b2 - 4ac > 0 : perfect square , real, rational, unequal.
b2 - 4ac > 0 : not a perfect square – real, irrational, unequal.
b2 - 4ac < 0 : imaginary, complex, no solution.
Quadratic Graphs
The graph of f ( x)  ax  bx  c is a
parabola. The graph looks like
2
if a > 0
if a < 0
Key features of the graph:
1. The maximum or minimum point on the
graph is called the vertex. The xcoordinate of the vertex is:
b
x
2a
 b 
 b
y  a     b
c
 2a 
 2a 
2
 b2 
y  c   
 4a 
2. The y-intercept; the y-coordinate of the
point where the graph intersects the y-axis.
The y-intercept is:
When x = 0, y = c
3. The x-intercepts; the x-coordinates of the
points, if any, where the graph intersects
the x-axis. To find the x-intercepts, solve
the quadratic equation
ax  bx  c  0.
2
Example:
Sketch the graph of f ( x)  x 2  2 x  8.
vertex: min. point
b
2
x

 1, f (1)  9; vertex (1, 9)
2a
2
y-intercept: f (0)  8
x-intercepts: x 2  2 x  8  ( x  4)( x  2)  0
x  4, x  2.
Example:
Sketch the graph of f ( x)   x  4 x  4
2
4
Vertex: x  
 2, f (2)  0; (2,0)
2(1)
y-intercept:
f (0)  4
x-intercept(s):  x  4 x  4  0
2
x  4 x  4  0; ( x  2)  0; x  2
2
Example:
Sketch the graph of
Vertex:
f ( x)  x  4 x  5
4
x
 2,
2(1)
y-intercept:
2
f (2)  1; (2,1)
f (0)  5
x-intercept(s): x2  4 x  5  0 has no real solutions.
Quadratic Inequalities
What do they look like?
Here are some examples:
x  3x  7  0
2
3x 2  4 x  4  0
x  16
2
Quadratic Inequalities
When solving inequalities we are trying to find all possible
values of the variable which will make the inequality true.
Consider the inequality
x2  x  6  0
We are trying to find all the values of x for which the
quadratic is greater than zero or positive.
Solving a quadratic inequality
We can find the values where the quadratic equals zero
by solving the equation,
x2  x  6  0
x  3x  2  0
x  3  0 or x  2  0
x  3 or x  2
Solving a quadratic inequality
You may recall the graph of a quadratic function is a parabola
and the values we just found are the zeros or x-intercepts.
2
The graph of y  x  x  6 is
(-2,0)
(3,0)
Solving a quadratic inequality
From the graph we can see that in the intervals around the
zeros, the graph is either above the x-axis (positive) or below
the x-axis (negative). So we can see from the graph the
interval or intervals where the inequality is positive.
But how can we find this out without graphing the quadratic?
We can simply test the intervals around the zeros in the
quadratic inequality and determine which make the inequality
true.
Solving a quadratic inequality
2
x
 x6  0
For the quadratic inequality,
we found
x = 3 and x = –2 by solving the equation . x 2  x  6  0
Put these values on a number line and we can see three
intervals that we will test in the inequality. We will test one
value from each interval.
-2
3
Solving a quadratic inequality
Interval
Test Point
Evaluate in the inequality
x2  x  6  0
True/False
 ,2
x  3
 32   3  6  9  3  6  6  0
True
 2, 3
x0
02  0  6  0  0  6  6  0
False
 3, 
x4
42  4  6  16  4  6  6  0
True
Solving a quadratic inequality
Thus the intervals  ,2 or 3,  make up the solution
set for the quadratic inequality, x 2  x  6  0 .
In summary, one way to solve quadratic inequalities is to find
the x-intercept/s and test a value from each of the intervals
surrounding the zeros to determine which intervals make the
inequality true.
Example : Solve 2 x  3x  1  0
2
Step 1: Solve
2 x 2  3x  1  0
Step 2:
Sketch the quadratic graph
2x 1x 1  0
2 x  1  0 or x  1  0
1
x  or x  1
2
0.5  x  1
Quadratic with linear
Solve: x2 – 8x + 16 > 2x +7
y = 2x + 10
Estimate ?
x<1
x>9
y = x2 – 8x +16
Example:
Solve: x2 – 8x + 16 > 2x +7
Algebraically:
1. Rearrange first
2. Solve like the others
x2 – 8x + 16 > 2x +7 (-2x)
x2 – 10x + 16 > 7
(-7)
x2 – 10x + 9 > 0 Like the ones we did
(x-9)(x-1) > 0
x>9 or x<1
Try this one
Solve: x2 + x + 4 > 4x +14
First: try a sketch
Algebraically:
1. Rearrange first
2. Solve like the others
x2 + x + 4 > 4x +14
(-4x)
x2 – 3x + 4 > 14
(-14)
x2 – 3x - 10 > 0
(x+2)(x-5) > 0
x<-2 or x>5
Summary
In general, when solving quadratic inequalities
1. Find the zeros by solving the equation you get when you
replace the inequality symbol with an equals.
2. Find the intervals around the zeros using a number line
and test a value from each interval in the number line.
3. The solution is the interval or intervals which make the
inequality true.
Practice Problems
5 x 2  13x  6  0
9  x2  0
0.219  x  2.28
No solution
2 x 2  5x  1  0
 0.4  x  3
x 2  5 x  4
 4  x  1
x2  2x  4
x  3, x  3