QUADRATIC FUNCTIONS AND INEQUALITIES

Integrated Programme/mainstreamSecondary Three Mathematics

Name : _____________( ) Class:______

Date : ___________ to _____________

Term 1: Unit 4 Notes

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 solution set on the number line.

 

Solving Quadratic Equations by FactorisingA quadratic equation is an equation like:

y = x2

y = x2 + 2y = x2 + x – 4y = x2 + 2x – 3

Contains a x2 termContains a x2 term

There are several methods of solving these but one methods that you must know is called FACTORISING

Sec 2 Revision

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)

3 x 2 = 6 0 x 2 = 0 3 x 0 = 0

x + 3 = 0

x = -3

x - 2 = 0

x = 2or

(x + 3)(x + 2)

x(x + 2) + 3(x + 2)

x X (x + 2) + 3 X (x + 2)

x X x + x X 2 + 3 X x + 3 X 2

x2 + 2x + 3x + 6

x2 + 5x + 6

You try

(x + 5)(x + 2)

(x – 2)(x + 3)

(x + 2)(x – 4)

(x – 3)(x – 2)

Solve by factorising: 0 = x2 + 7x + 12 1 x 12 =

122 x 6 = 123 x 4 = 12

Write down all the factor pairs of 12.

From this list, choose the pair that adds up to 7

3 + 4 = 7

Put these numbers into brackets

0 = (x + 3)(x + 4)

x = – 3 and – 4

(x )(x )What goes with the x?

(x )(x )What goes with the x?

Solve by factorising: 0 = x2 + x - 6

1 x -6 = -6 2 x -3 = -6 3 x -2 = -6 6 x -1 = -6

Write down all the factor pairs of – 6

From this list, choose the pair that adds up to 1

(3) + (-2) = 1

3 – 2 = 1Put these numbers into brackets

0 = (x + 3)(x - 2)

x = – 3 and 2

Copy and fill in the missing values when you factorise x2 + 8x + 12 = 0

Find all the factor pairs of 12 1 x 12 = 12

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

Solve by factorising

1. x2 + 3x + 2 = 0

2. x2 + x – 12 = 0

3. x2 – 12x – 20 = 0

4. x2 – x – 12 = 0

1 x2 + 5x + 6 = 0

2 x2 - x – 6 = 03 x2 + 8x + 12 =

04 x2 + x – 12 = 05 x2 - 8x + 15 =

06 x2 + 3x – 21 =

07 x2 - 3x – 18 =

08 x2 - 10x – 24 =

09 x2 + 8x + 16 =

010 x2 - 4x – 60 =

0

1 x2 + 5x + 6 = 0

(x + 3)(x + 2)

2 x2 - x – 6 = 0 (x – 3)(x + 2) 3 x2 + 8x + 12 =

0(x + 2)(x + 6)

4 x2 + x – 12 = 0 (x – 3)(x + 4) 5 x2 - 8x + 15 =

0(x – 3)(x – 5)

6 x2 + 3x – 21 = 0

(x + 7)(x – 4)

7 x2 - 3x – 18 = 0

(x – 6)(x + 3)

8 x2 - 10x – 24 = 0

(x - 12)(x + 2)

9 x2 + 8x + 16 = 0

(x + 4)(x + 4)

10 x2 - 4x – 60 = 0

(x - 10)(x + 4)

1 x2 + 5x + 6 = 0

(x + 3)(x + 2)

-3 and -2

2 x2 - x – 6 = 0 (x – 3)(x + 2) 3 and -23 x2 + 8x + 12 =

0(x + 2)(x + 6)

-2 and -6

4 x2 + x – 12 = 0 (x – 3)(x + 4) 3 and -45 x2 - 8x + 15 =

0(x – 3)(x – 5) 3 and 5

6 x2 + 3x – 21 = 0

(x + 7)(x – 4) -7 and 4

7 x2 - 3x – 18 = 0

(x – 6)(x + 3) 6 and -3

8 x2 - 10x – 24 = 0

(x - 12)(x + 2)

12 and -2

9 x2 + 8x + 16 = 0

(x + 4)(x + 4)

-4 and -4

10 x2 - 4x – 60 = 0

(x - 10)(x + 4)

10 and -4

x2 – 4

x2 + 0x – 4

(x – 2)(x + 2)

-1 x 4 = -4

-2 x 2 = -4

4 x -1 = -4

-2 + 2 = 0Notice that x2 – 4 could be written as

x2 – 22

(x – 2)(x + 2)

This is often called the difference between two

squaresx2 – 25

(x + 5)(x – 5)

This is often called the difference between two

squaresx2 – 25

(x + 5)(x – 5)

1 x2 - 92 x2 - 1003 x2 - 36 4 x2 - 495 x2 - 816 x2 - 647 x2 - 188 x2 - 24

1 x2 - 9 (x + 3)(x – 3)2 x2 - 100 (x + 10)(x – 10)3 x2 - 36 (x + 6)(x – 6)4 x2 - 49 (x + 7)(x – 7)5 x2 - 81 (x + 9)(x – 9)6 x2 - 64 (x + 8)(x – 8)7 x2 - 18 (x + √18)(x –

√18)8 x2 - 24 (x + √24)(x –

√24)

For quadratic equations that are not expressed as an equation between two squares, we can always express them as

If this equation can be factored, then it can generally be solved easily.

2 0ax bx c

Completing Square

If the equation can be put in the form

then we can use the square root method described previously to solve it.

2 2( )k x m n

“Can we change the equation from the formto the form ?”

2 0ax bx c 2 2( )x m n

The procedure for changingis as follows. First, divide by , this gives

Then subtract from both sides. This gives

2 0ax bx c a

2 0b c

x xa a

c

a

2 b cx x

a a

Recall that

If we let

we can solve for to get

2 2 2( ) 2x r x rx r

2b

ra

r

2

br

a

Substituting in we get

Using the symmetric property of equations to reverse this equation we get

2

br

a 2 2 2( ) 2x r x rx r

22 2

2( )

2 4

b b bx x x

a a a

22 2

2( )

4 2

b b bx x x

a a a

Now we will return to where we left our original equation. If we add to both sides ofwe get

2

24

b

a2 b cx x

a a

2 22

2 24 4

b b c bx x

a a a a

2

2

4

4

b ac

a

22

2

4( )

2 4

b b acx

a a

or

We can now solve this by taking the square root of both sides to get

2 4

2 2

b b acx

a a

2 4

2 2

b b acx

a a

2 4

2

b b acx

a

Example:

27.5,27.13

32)2(

32)2(3

032)2(3

0]3

204)2[(3

0)3

204(3

020123

2

2

2

2

2

2

xx

x

x

x

x

xx

xx [By completing square method]

Quadratic Formula

a

acbbx

2

42

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 :

a

acbbx

2

42

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

a

acbbx

2

42

12

)614(55 2

x

2

)24(255 x

2

15 x

2

15

2

15

xorx

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

a

acbbx

2

42

82

)384(22 2

x

16

)96(42 x

16

1002 x

16

102

16

102

xorx

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

a

acbbx

2

42

22

)724(33 2

x

4

)56(93 x

4

653 x

4

0622.83

4

0622.83

xorx

x = 1.27 or x = - 2.77

These are the roots of the equation.

Discriminant

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

Example :

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 is a parabola. The graph looks like

if a > 0 if a < 0

2( )f x ax bx c

Key features of the graph:

1. The maximum or minimum point on the graph is called the vertex. The x-coordinate of the vertex is:

2

bx

a

a

bcy

ca

bb

a

bay

4

222

2

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

2 0.ax bx c

Example:

Sketch the graph of 2( ) 2 8.f x x x

vertex: min. point

y-intercept:

x-intercepts:

21, (1) 9; vertex (1, 9)

2 2

bx f

a

(0) 8f

2 2 8 ( 4)( 2) 0

4, 2.

x x x x

x x

Sketch the graph of

Vertex:

y-intercept:

x-intercept(s):

2( ) 4 4f x x x

42, (2) 0; (2,0)

2( 1)x f

(0) 4f

2

2

4 4 0

4 4 0; ( 2) 0; 2

x x

x x x x

Example:

Sketch the graph of

Vertex:

y-intercept:

x-intercept(s):

2( ) 4 5f x x x

42, (2) 1; (2,1)

2(1)x f

(0) 5f

2 4 5 0 has no real solutions.x x

Example:

Quadratic InequalitiesWhat do they look like?

Here are some examples: 0732 xx

0443 2 xx

162 x

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

We are trying to find all the values of x for which the

quadratic is greater than zero or positive.

062 xx

Solving a quadratic inequalityWe can find the values where the quadratic equals zero

by solving the equation, 062 xx

023 xx

02or03 xx

2or3 xx

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.

The graph of is

62 xxy

(-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

For the quadratic inequality, we found

x = 3 and x = –2 by solving the equation .

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.

062 xx

062 xx

-2 3

Solving a quadratic inequalityInterval Test Point Evaluate in the inequality True/False

2,

3,2

,3

06639633 2

06600600 2

066416644 2

3x

0x

4x

True

True

False

062 xx

Solving a quadratic inequalityThus the intervals make up the

solution

set for the quadratic inequality, .

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.

062 xx

,3or2,

Example : Solve

Step 1: Solve

0132 2 xx

0132 2 xx

0112 xx

01or012 xx

1or2

1 xx

Step 2:

Sketch the quadratic graph

15.0 x

Quadratic with linear

y = x2 – 8x +16

y = 2x + 10

Solve: x2 – 8x + 16 > 2x +7

Estimate ?

x<1

x>9

Example:Solve: x2 – 8x + 16 > 2x +7

Algebraically:1. Rearrange first2. Solve like the others

x2 – 8x + 16 > 2x +7

x2 – 10x + 9 > 0

(x-9)(x-1) > 0

x2 – 10x + 16 > 7

(-2x)

(-7)

Like the ones we did

x>9 or x<1

Try this oneSolve: x2 + x + 4 > 4x +14

Algebraically:1. Rearrange first2. Solve like the others

x2 + x + 4 > 4x +14

x2 – 3x - 10 > 0

(x+2)(x-5) > 0

x2 – 3x + 4 > 14

(-4x)

(-14)

x<-2 or x>5

First: try a sketch

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

06135 2 xx

09 2 x

0152 2 xx

452 xx

422 xx

No solution

34.0 x

3,3 xx

28.2219.0 x

14 x