Chapter 2 Quadratic Expressions and Equations
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8/3/2019 Chapter 2 Quadratic Expressions and Equations
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Created By: Mohd Said B Tegoh
8/3/2019 Chapter 2 Quadratic Expressions and Equations
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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
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Identifying Quadratic Expressions
A quadratic expression must satisfyboth of the following conditions:
(a) There is only one unknown(b) The highest power of the unknown
is 2
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Identifying Quadratic Expressions
When the value of b and c is 0, the
quadratic expression has the general
form of ax2, e.g. 3x2
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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
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Able to simplify algebraic expressions
Able to expand and factorise algebraic
expressions
Able to simplify algebraic fractions
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8/3/2019 Chapter 2 Quadratic Expressions and Equations
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How to expanda single pair of brackets?
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A B2m
m 2
2m
Area A = 2m x m
2m2= Area B = 2m x 2
4m=
Area A + Area B = ?
2m2 4m
+
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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
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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.
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Example
)3(2 x
= 2x _ 6
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)4(2 mm
= _ mmx2 )4)2( xm
= mm 82 2
Example
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)33(
3
2a
= +ax3
3
2
=
3
3
2x
2a 2
1
1
1
1
Example
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How to expandtwo pairs of brackets?
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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
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Arrange All the Cards
into One Rectangle
A rea A = A rea B = A rea C =x2 x 1
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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
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(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
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(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
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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
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( x + 3 ) ( 2x - 1 )
= 2x2
6x + (-x) = 5x
+ 5x _ 3
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(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
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(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
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(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
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(x - y)(x + y)
= x2 + xy - xy - y2
= x2 - y2
(2x - 5)(2x + 5) = (2x)2 - 52
= 4x2 - 25
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( 5x + 3 ) ( 3x - 2 )
= 15x2
9x + (-10x) = -x
_ x _ 6
Example
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( 2x - 3 ) ( 3x - 7 )
= 6x2
-9x + (-14x) = -23x
_ 23x + 21
Example
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(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
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(3x - y)(3x + y) = (3x)2 - y2
= 9x2 ² y2
(2m - n)(2m + n) = (2m)2 - n2
= 4m2 ² n2
Example
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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
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(a + b)2 a2 + 2ab + b2
Expansion
Factorization
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Factors, Common Factors,
Highest Common Factor?
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15= x1 15=3 5x
The factors of 15 is 1, 3, 5,and 15
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3xy
= x1 3xy
=3 xyx
=
=
x x 3y
yx 3xThe factors of 3xy is 1, 3, x, y,
3x, 3y, xy and 3xy
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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
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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
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Factorisation of
Quadratic Expressions
4p + 6 = 2 3( )
2 is the common factor of 4p and 6
2p +
4p 6,2
2p 3,
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Factorisation of
Quadratic Expressions
2e + 3ef = e 3f ( )
e is the common factor of 2e and 3ef
2 +
2e 3ef ,e
2 3f ,
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4p2q , 6pq32p
q
2p2q , 3pq3
2pq , 3q32p , 3q2
HCF = 2pq
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6x3y , 9xy23x
y
2x3y , 3xy2
2x2y , 3y22x2 , 3y
HCF = 3xy
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Factorisation of
Quadratic Expressions
4p2 q + 6pq3 = 2pq 3q2( )
2pq is the common factor of 4p2q and 6pq3
2p +
6x3y - 9xy2 = 3xy 3y( )2x2 -
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4p2 - 9 = (2p)2 ² 32
=(2p ² 3)(2p + 3)
Use the identitya2 ² b2 = (a + b)(a ² b)
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x+
x
+
15 =
8 =
-45 =
4 =
(a)
(b)
3 5
3 5
-5 9
-5 9
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x+
x
+
18 =
-11 =
40 =
13 =
(c)
(d)
-9 -2
-9 -2
8 5
8 5
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x+
x
+
-15 =
2 =
-21 =
-4 =
(e)
(f)
-3 5
-3 5
-7 3
-7 3
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x+
x
+
-21 =
-4 =
-30 =
-1 =
(g)
(h)
-7 3
-7 3
-6 5
-6 5
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x+
x
+
45 =
-14 =
210=
29 =
(i)
(j)
-9 -5
-9 -5
15 14
15 14
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x+
x
+
-30 =
-7 =
-100=
-21 =
(k)
(l)
-10 3
-10 3
-25 4
-25 4
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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
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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
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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
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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
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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
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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)
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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)
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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)
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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)
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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)
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(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
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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
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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
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(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
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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
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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
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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
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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.
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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
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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
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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
LHS = ¶left-hand-side·
RHS = ¶right-hand-side·
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
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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
LHS = ¶left-hand-side·
RHS = ¶right-hand-side·
x = -2 satisfies equation x 2 + 5x + 6 = 0
x = -2 is a root of the equation x 2 + 5x + 6 = 0
= RHS
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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
LHS = ¶left-hand-side·
RHS = ¶right-hand-side·
x = -3 satisfies equation x 2 + 5x + 6 = 0
x = -3 is a root of the equation x 2 + 5x + 6 = 0
= RHS
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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.
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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
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(
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
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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
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3x2 = 2x ± 2 + 7
3x2 ± 2x ± 5 = 0
( )( ) = 0
x = 5 , - 1
3
5
3
_
x
+
1x
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«««««««««.
( )( ) = 0
x = 4 , - 35 2
5x - 4 2x + 3
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«««««««««.
( )( ) = 0
x = 3 , - 12 4
2x - 3 4x + 1
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«««««««««.
( )( ) = 0
x = -1 , - 33 4
3x + 1 4x + 3
2
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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
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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
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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
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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
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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
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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
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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
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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
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