chapter 2 quadratic expressions and equations

8/3/2019 Chapter 2 Quadratic Expressions and Equations

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Created By: Mohd Said B Tegoh



Identifying Quadratic Expressions

A quadratic expression is an expression

of the form ax2 + bx + c where a, b and c

are constants, a 0 and x is unknown

Examples: 3x2 + 13x + 4 and 7n2 ± 2n + 3




A quadratic expression must satisfyboth of the following conditions:

(a) There is only one unknown(b) The highest power of the unknown

is 2




When the value of b and c is 0, the

quadratic expression has the general

form of ax2, e.g. 3x2



Forming Quadratic Expressions

A quadratic expression is formed when two

linear expressions with the same variable

(unknown) are multiplied together

Example

(x ± 1)(2x + 3) = 2x2 + x - 3



Able to simplify algebraic expressions

Able to expand and factorise algebraic

expressions

Able to simplify algebraic fractions



How to expanda single pair of brackets?



A B2m

m 2

2m

Area A = 2m x m

2m2= Area B = 2m x 2

4m=

Area A + Area B = ?

2m2 4m

+



A B2m

m 2+

By multiplying the width with the total length of thetwo rectangles (A and B), write an expression for their

total area.

2m

( )

m + 2



A B2m

m 2+

2m(m + 2)

=

2m2 + 4m

m 2

(2m x m)

+

(2m x 2)

=

m 2

To find the total area of the two rectangles, we can also findthe area of each rectangle and add them up.



Example

)3(2 x

= 2x _ 6



)4(2 mm

= _ mmx2 )4)2( xm

= mm 82 2

Example



)33(

3

2a

= +ax3

3

2

=

3

3

2x

2a 2

1

1

1

1

Example



How to expandtwo pairs of brackets?



Cards with specification

x

x

x

1

1

1

Card A Card B Card C

A rea A = A rea B = A rea C =x2 x 1



Arrange All the Cards

into One Rectangle

A rea A = A rea B = A rea C =x2 x 1



x

(x + 2)

x 2

(x + 1)

1

Area of theRectangle ?

A + 3B + 2C

= x2 + 3x + 2

A B B

B CC

PROVED !!!!

(x + 2)(x + 1)

= x2

+ 3x + 2

Length of theRectangle ?

(x + 2)Width of theRectangle ?

(x + 1)

A rea A =

A rea B =

A rea C =

x2

x

1



(x+ 2)

x 2

(x + 1) A B B

B C C

(x + 2)(x + 1) x (x + 1)=+ 2 (x + 1)= x2 + x + 2x + 2= x2 + 3x + 2

To find the total area of the two rectangles, we can also find thearea of each rectangle and add them up.

The rectangle of length (x + 2) and width (x + 1)can be divided into two rectangular sections



(x + 3)(2x +1) x(2x + 1)

=

+ 3(2x + 1)= 2x2 + x + 6x + 3=

2x2 + 7x + 3

Common Method Used to Carry Out

The Expansion of Algebraic Expressions

Multiply each term within the first pair of

brackets by every term within the second

pair of brackets



Short Process : To find (x + 3)(2x + 1), we

must ensure that each term in the firstbracket multiplies each term in the second.

The arrows in the figure below help us to see

that all terms have been taken into account:

(x + 3)(2x + 1) = 2x2 + x + 6x + 3

= 2x2 + 7x + 3



( x + 3 ) ( 2x - 1 )

= 2x2

6x + (-x) = 5x

+ 5x _ 3



(x + 3)(2x ² 1) = ?

x

2x

+3

-1

1. A rrange the

the expressions

given in two rows

2. Multiply vertically

to get ax2

4. Perform cross

multiplication 3. Multiply vertically

to get c

6x

-x5x-32x2

5. A dd them up to get bx

Thus,

(x + 3)(2x ² 1)

= 2x2 + 5x - 3



(x + y)2

= (x + y)(x + y)

= x2 + xy + xy + y2

= x2 + 2xy + y2

(2a + 3)2 = (2a)2 + 2 x 2a x 3 + 32

= 4a2 + 12a + 9



(x - y)2

= (x - y)(x - y)

= x2 - xy - xy + y2

= x2 - 2xy + y2

(2a - 3)2 = (2a)2 - 2 x 2a x 3 + 32

= 4a2 - 12a + 9



(x - y)(x + y)

= x2 + xy - xy - y2

= x2 - y2

(2x - 5)(2x + 5) = (2x)2 - 52

= 4x2 - 25



( 5x + 3 ) ( 3x - 2 )

= 15x2

9x + (-10x) = -x

_ x _ 6

Example



( 2x - 3 ) ( 3x - 7 )

= 6x2

-9x + (-14x) = -23x

_ 23x + 21

Example



(4x + 5)2 = (4x)2 + 2 x 4x x 5 + 52

= 16x2 + 40x + 25

(2x + 3y)2 = (2x)2 + 2 x 2x x 3y + (3y)2

= 4x2 + 12xy + 9y2

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

= 4x2 - 12xy + 9y2

Example



(3x - y)(3x + y) = (3x)2 - y2

= 9x2 ² y2

(2m - n)(2m + n) = (2m)2 - n2

= 4m2 ² n2

Example



Factorisation of quadratic expressions is a process of finding

two linear expressions whose product is equal to the quadratic

expression.

For Example;

(x + 2)(x + 3) = x2 + 5x + 6

x2 + 5x + 6 = (x + 2)(x + 3)

Expansion

Factorisation

The two linear expressions, (x + 2) and (x + 3), are called

the factors of the quadratic expression x2 + 5x + 6



(a + b)2 a2 + 2ab + b2

Expansion

Factorization



Factors, Common Factors,

Highest Common Factor?



15= x1 15=3 5x

The factors of 15 is 1, 3, 5,and 15



3xy

= x1 3xy

=3 xyx

=

=

x x 3y

yx 3xThe factors of 3xy is 1, 3, x, y,

3x, 3y, xy and 3xy



9pq

= x1 9pq= 3 3pqx

The factors of 9pq is

1, 3, 9, p, q, 3p, 3q, 9p,

9q, pq, 3pq and 9pq

= 9 pqx

= p 9qx

=q 9px

=3p 3qx

= 3q 3px



3xy 2xy

1 1

3 2

x xy y

3x 2x

3y 2y

xy xy

3xy 2xy

F A CTORS

3xy 2xy,xy

3,2

xy is the highestcommon factor

(HCF) of 3xy and

2xy



Factorisation of

Quadratic Expressions

4p + 6 = 2 3( )

2 is the common factor of 4p and 6

2p +

4p 6,2

2p 3,



Factorisation of


2e + 3ef = e 3f ( )

e is the common factor of 2e and 3ef

2 +

2e 3ef ,e

2 3f ,



4p2q , 6pq32p

q

2p2q , 3pq3

2pq , 3q32p , 3q2

HCF = 2pq



6x3y , 9xy23x

y

2x3y , 3xy2

2x2y , 3y22x2 , 3y

HCF = 3xy



Factorisation of


4p2 q + 6pq3 = 2pq 3q2( )

2pq is the common factor of 4p2q and 6pq3

2p +

6x3y - 9xy2 = 3xy 3y( )2x2 -



4p2 - 9 = (2p)2 ² 32

=(2p ² 3)(2p + 3)

Use the identitya2 ² b2 = (a + b)(a ² b)



x+

x

+

15 =

8 =

-45 =

4 =

(a)

(b)

3 5

3 5

-5 9

-5 9



x+

x

+

18 =

-11 =

40 =

13 =

(c)

(d)

-9 -2

-9 -2

8 5

8 5



x+

x

+

-15 =

2 =

-21 =

-4 =

(e)

(f)

-3 5

-3 5

-7 3

-7 3



x+

x

+

-21 =

-4 =

-30 =

-1 =

(g)

(h)

-7 3

-7 3

-6 5

-6 5



x+

x

+

45 =

-14 =

210=

29 =

(i)

(j)

-9 -5

-9 -5

15 14

15 14



x+

x

+

-30 =

-7 =

-100=

-21 =

(k)

(l)

-10 3

-10 3

-25 4

-25 4



x 2 + 9 x + 14

=

14 = x

9 = +

( x + 2 )( x + 7)

2

7

7

2

x

x

+2

+7

2 x

7 x

9 x 14 x 2

a x 2 c b x

x

x

Factorise x2 + 9 x + 14

Solution



x 2 - 5 x + 6

=

6 = x

-5 = +

( x - 2 )( x - 3)

-2

-3

-3

-2

x

x

-2

-3

-2 x

-3 x

-5 x 6 x 2

a x 2 c b x

x

x

Factorise x2 - 5 x + 6

Solution



x 2 + 3 x - 10

=

-10 = x

3= +

( x - 2 )( x + 5)

-2

5

5

-2

x

x

-2

+5

-2 x

5 x

3 x -10 x 2

a x 2 c b x

x

x

Factorise x2 + 3 x - 10

Solution



x 2 - x - 6

=

6 = x

-1 = +

( x + 2 )( x - 3)

2

-3

-3

2

x

x

+2

-3

2 x

-3 x

- x -6 x 2

a x 2 c b x

x

x

Factorise x2 - x - 6

Solution



3 x 2 - 2 x - 5

=

-15 = x

-2 = +

(3 x - 5 )( x + 1)

-5

3

3

-5

3 x

x

-5

+1

-5 x

3 x

-2 x -53 x 2

a x 2 c b x

3 x

x

3 x (-5)

Factorise 3 x2 - 2 x - 5

Solution



Factorise 3 x2 - 10 x - 8

Solution

3 x 2 - 10 x - 8=

-24 = x

-10 = +(3 x + 2 )( x - 4)

2

-12

-12

2

3 x

x

+2

-4

2 x

-12 x

-10 x -83 x 2

a x 2 c b x

3 x

x

3 x (-8)



Factorise 2m2 + 3m - 2

Solution

2m2 + 3m - 2

=

-4 = x

3 = +

( 2m - 1 )(m

+ 2 )

-1

4

4

-1

2m

m

-1

+2

-m

4m

3m-22m2

a x 2 c b x

2m

m

2 x (-2)



Factorise 4m2 + 11m - 3

4m2 + 11m ± 3

=

Solution

-12= x

11= +-1

12

12

-1

4m

m

-1

+3

-m

12m

11m-34m2

a x 2 c b x

4 x (-3)

(m + 3)(4m ± 1)



Factorise 3m2 - 6m + 3

3m2 - 6m + 3

=

Solution

9 = x

-6 = +-3

-3

-3

-3

m

3m

-1

-3

-3m

-3m

-6m33m2

a x 2 c b x

3 x 3

(m - 1)(3m ± 3)



Factorise 10 p2 + 3 p - 4

10 p2 + 3 p - 4

=

Solution

-40 = x

3 = +-5

8

8

-5

2p

5p

-1

+4

-5 p

8 p

3 p-410 p2

a x 2 c b x

10 x(- 4)

(2 p - 1)(5 p + 4)



(a)(x + 1)(x ² 2)

(b)(2k ² 1)(k + 3)

(c) (3y ² 1)(y + 3)

(d) (4 ² 3n)(3 + n)

(e)2(x ² 2)(x ² 2)

(f) -3(2 ² y)(3 + y)

Expand each of the following.

22

x x

352 2 k k

383 2 y y

1253 2 nn

8822

x x

18332

y y



Factorise each quadratic expression below.

23)(

161)(

169)(

100)(

726)(

3)(

2

2

2

2

2

2

x x f

ye

ed

xc

yb

x xa )13( x x

)12(6 2 y

)10)(10( x x

)43)(43( ee

)4

1)(4

1(y y

)2)(1( x x



Factorise each quadratic expression below.

1064)(

48119)(

1252)(

954)(

283)(

152)(127)(

2

2

2

2

2

2

2

y ym

x xl

p pk

p p j

x xi

x xh y y g )3)(4( y y

)5)(3( x x

)4)(7( x x

)1)(94( p p

)4)(32( p p

)3)(169( x x

)1)(52(2 y y



(2x ² 3) cm

(x ² 2) cm

x cm

x cm

If the area of the rectangle is equal to the area

of the square, then we can form an equation

(2x ² 3)(x ² 2) = (x)(x)2x2 ² 7x + 6 = x2

Equations of this form are knownas quadratic equations



Q uadratic equations are equations which fulfill thefollowing characteristics:

have an equal sign ¶=¶

have only one unknown

have 2 as the hig hest power of the unknown

Example

x2

+ 5x + 6 = 0Highest power

of x is 2Equal sign



GeneralForm

Characteristic Example

02

! cb xax

02!b xax

02

! cax

0,0,0 {{{ cba

0,0,0 !{{ cba

0,0,0 {!{ cba

07232

! x x

052 2! x x

0822

! x



Determine whether the following are quadraticequations in one unknown. Give reasons.

(a) x2 + 3x ² 6 = 0

(b) x2 + y2 = 4

(c) m2

+ m + 1

(d) n2 ² I = 3n

yes

No

No

No

one unknown, highest power of xis 2, has an equal sign

more than one unknown

no equal sign

highest power of n is 3



Roots of a quadratic equation are values of theunknown which satisfy the quadratic equation.

To determine whether a given value of

unknown is a root of a specific quadraticequation, substitute the given value for theunknown into the equation. If it satisfies theequation, then the value of the unknown is a

root of the equation and vice versa.



Example

Determine whether the following values of xare roots of the quadratic equation x 2 + 5x + 6 = 0

(a) x = 1

(b) x = 2

(c) x = -2

(d) x = -3



So luti on

(a) Substitute x = 1 into x 2 + 5x + 6 = 0

LHS = ( )2 + 5( ) + 61 1

= 1 + 5 + 6

= 12 0

LHS RHS

LHS = ¶left-hand-side·

RHS = ¶right-hand-side·

x = 1 does not satisfy equation x 2 + 5x + 6 = 0

x = 1 is not a root of the equation x 2 + 5x + 6 = 0



So luti on

(b) Substitute x = 2 into x 2 + 5x + 6 = 0

LHS = ( )2 + 5( ) + 62 2

= 4 + 10 + 6

= 20 0

LHS RHS



x = 2 does not satisfy equation x 2 + 5x + 6 = 0

x = 2 is not a root of the equation x 2 + 5x + 6 = 0



So luti on

(b) Substitute x = -2 into x 2 + 5x + 6 = 0

LHS = ( )2 + 5( ) + 6-2 -2

= 4 - 10 + 6

= 0

LHS = RHS



x = -2 satisfies equation x 2 + 5x + 6 = 0

x = -2 is a root of the equation x 2 + 5x + 6 = 0

= RHS



So luti on

(b) Substitute x = -3 into x 2 + 5x + 6 = 0

LHS = ( )2 + 5( ) + 6-3 -3

= 9 - 15 + 6

= 0

LHS = RHS



x = -3 satisfies equation x 2 + 5x + 6 = 0

x = -3 is a root of the equation x 2 + 5x + 6 = 0

= RHS



Roots of an equation are also called thesolution of an equation. Therefore, in the

given example, x = -2, x = -3 are solution of the equation x 2 + 5x + 6 = 0.

The factorisation method is commonly used

to find the solutions or roots of a givenquadratic equation.



Solve the equation 3 x2 = 2( x ± 1) + 7

3 x2 = 2( x ± 1) + 7

=2 x x

_

2 x 1

+ 7

= 2 x ± 2 + 7

= 2 x + 53 x2 _ 0 _

(3 x ± 5)( x + 1) = 0

3 x ± 5 = 0 or x + 1 = 0

E x a m p

l e

3

5! x

1! x



(

EXP (-)

log ln

Ans

tan

sin cos

x -1 CONST

DEL AC

9 =

+

^hyp f dx

.

CALC ¥

0

ENGM+

ab/c x2

RCL

7 8

2 + b + 0



ax2 + bx + c = 0

3x2 ² 2x ² 5 = 0

a = 3 b = -2 c = -5

MODE EQN1 1

Unknowns ?2 3

Degree?2 3

2 a ? 3 = b ? (-) 2 = c ?

(-) 5 x1 = 1.666666667Shift

x1 = 5 3

x2 = -1

ab/c

=

3x

Press



3x2 = 2x ± 2 + 7

3x2 ± 2x ± 5 = 0

( )( ) = 0

x = 5 , - 1

3

5

3

_

x

+

1x



«««««««««.

( )( ) = 0

x = 4 , - 35 2

5x - 4 2x + 3



«««««««««.

( )( ) = 0

x = 3 , - 12 4

2x - 3 4x + 1



«««««««««.

( )( ) = 0

x = -1 , - 33 4

3x + 1 4x + 3

2



Solve the equation 2m2 + 5m = 2

m + 1

2m

2

+ 5m

= 2m + 1

( )2m + 2=2m

2 + 5m 2m + 20- -2m

2 + 3m ± 2 = 0

( 2m ± 1 )( m + 2 ) = 02m - 1 = 0 or m + 2 = 0

2

1!m

2!m

Cl ed SPM



Cloned SPMSolve the equation

x

x

x

78!

Solution

08

7

2! x x

0)8)(1( ! x x

01! x or 08 ! x

8,1 ! x

Cl ed SPM



Cloned SPMSolve the equation 28)2(57 2

! x x

Solution281057 2

! x x

185 ! x

01857 2! x x

0)2)(97( ! x x

097 ! x or 02 ! x

2,7

9! x

Cl ed SPM



Cloned SPMSolve the equation 3

2

103 2

!

p

p p

Solution)2(3103 2

! p p p

63 ! p

063103 2! p p p

0673 2! p p

0)3)(23( ! p p

023 ! p or 03 ! p

3,

3

2! p

Cl ed SPM



Cloned SPMSolve the equation 4

153 2

!

k

k

k k 4153 2!

Solution

01543 2! k k

0)3)(53( ! k k

053 !k or 03 !k

3,3

5!k

S lvi g P ble



Solving Problem A photograph is mounted on a piece of card , 8 cm long

and 5 cm wide, leaving a border of constant width aroundthe photograph. If the area of the photograph is 18 cm2,find the width of the border.

S

o l u t i o nStep 1: Read and understand

8 cm

5 cm

x cm

x cm

x cm x cm

Find the width of the border which is x cm

8 cm



5 cm

x cm

x cm

x cmx cm

Step 2 : Devise a plan

Area of photo = 18

Length x breadth = 18

(8 ² 2x )(5 ² 2x ) = 1840 ² 16x ² 10x + 4x 2 = 18

Solve the quadratic equation and obtain the

value x cm

4x 2 ² 26x + 22 = 0



Step 3 : Carry out the plan

4x 2 ² 26x + 22 = 02x 2 ² 13x + 11 = 0

(2x ² 11)(x ² 1) = 0

2x ² 11 = 0 or x ² 1 = 0x = 11 or x = 1

2

chapter 2 quadratic expressions and equations

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