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CONTENTS Mathematical Formulae Form

iv

4

r

te

ap

Ch

1

Circular measure

1.1 Functions 1.2 Composite Functions 1.3 Inverse Functions SPM Practice 1

1 2 14 19 27

r

te

ap

Ch

6

6.1 Linear and Non-Linear Relations 6.2 Linear Law and Non-Linear Relations 6.3 Applications of Linear Law SPM Practice 6

Ch

2

Quadratic Functions

29

2.1 Quadratic Equations and Inequalities 2.2 Types of Roots of Quadratic Equations 2.3 Quadratic Functions SPM Practice 2

30 37 39 52

3

Systems of Equations

3.1

Systems of Linear Equations in Three Variables 3.2 Simultaneous Equations involving One Linear Equation and One Non-Linear Equation SPM Practice 3

55

56

er

7

128

7.1 Divisor of a Line Segment 7.2 Parallel Lines and Perpendicular Lines 7.3 Areas of Polygons 7.4 Equations of Loci SPM Practice 7

129 132 137 142 145

4

68 70 75 80 85

5

r

9

5.1 Arithmetic Progressions 5.2 Geometric Progressions SPM Practice 5

87 88 96 104

Solution of Triangles

179

Sine Rule 180 Cosine Rule 184 Area of a Triangle 188 Application of Sine Rule, Cosine Rule and Area of a Triangle 191 SPM Practice 9 194

Ch

r

te

ap

Progressions

149 155 166 173

9.1 9.2 9.3 9.4

r

te

ap

148

te

Indices, Surds and Logarithms 67

Laws of Indices Laws of Surds Laws of Logarithms Applications of Indices, Surds and Logarithms SPM Practice 4

Vectors

8.1 Vectors 8.2 Addition and Subtraction of Vectors 8.3 Vectors in a Cartesian Plane SPM Practice 8 ap

4.1 4.2 4.3 4.4

Ch

8

Ch

t

ap

Ch

r

te

ap

Ch

60 65

er

109 114 121 124

Coordinate Geometry

r

te

ap

Ch

108

pt

a Ch

er

t ap

Linear Law

10

Index Numbers

10.1 Index Numbers 10.2 Composite Index SPM Practice 10

196 197 200 211 iii

Form

5

r

te

ap

Ch

6

er

t ap

Ch

1

Circular Measure

216

1.1 Radian 1.2 Arc Length of a Circle 1.3 Area of Sector of a Circle 1.4 Application of Circular Measures SPM Practice 1

217 219 225 233 236

r

te

ap

Ch

2

Differentiation

3

Integration as the Inverse of Differentiation 3.2 Indefinite Integral 3.3 Definite Integral 3.4 Application of Integral SPM Practice 3

265 266 271 287 290

r

4

Permutation and Combination

4.1 Permutation 4.2 Combination SPM Practice 4

293 294 300 304

er

pt

a Ch

5

Probability Distribution

5.1 Random Variable 5.2 Binomial Distribution 5.3 Normal Distribution SPM Practice 5

iv

Linear Programming

7.1 Linear Programming Model 7.2 Application of Linear Programming SPM Practice 7

Ch

351 352 354 360

r

te

8

Kinematics of Linear Motion 364

8.1

3.1

te

7

ap

264

ap

er

a Ch

r

Integration

Ch

Positive Angles and Negative Angles 330 Trigonometric Ratios of any Angle 331 Graphs of Sine, Cosine and Tangent Functions 336 6.4 Basic Identities 341 6.5 Addition Formulae and Double Angle Formulae 342 6.6 Application of Trigonometric Functions 344 SPM Practice 6 348 pt

te

ap

329

6.1 6.2 6.3

241

2.1 Limit and its Relation to Differentiation 242 2.2 The First Derivative 244 2.3 The Second Derivative 249 2.4 Application of Differentiation 250 SPM Practice 2 261

Ch

Trigonometric Functions

306 307 311 316 325

Displacement, Velocity and Acceleration as a Function of Time 365 8.2 Differentiation in Kinematics of Linear Motion 373 8.3 Integration in Kinematics of Linear Motion 379 8.4 Application of Kinematics of Linear Motion 382 SPM Practice 8 387

SPM Model Paper

390

Answers

399

Mathematical Formulae Form 4 r

te

ap

Ch

2

Quadratic Functions

–b ± √b2 – 4ac 2a

x=

er

pt

a Ch

4

Indices, Surds and Logarithms

am × an = am + n am ÷ an = am – n (am)n = amn √a × √b = √ab a √a ÷ √b = b loga mn = loga m + loga n m loga = loga m – loga n n loga mn = n loga m loga b = er

t

ap

Ch

logc b logc a

5

Progressions

Arithmetic progressions, Tn = a + (n – 1)d n Sn = [2a + (n – 1)d] 2 n Sn = [a + l ] 2 Geometric progressions, Tn = arn – 1 a(r n – 1) Sn = for | r | . 1 or r–1 a(1 – r n) for | r | , 1 Sn = 1–r a S∞ = for | r | , 1 1–r

r

=

_ +

te

ap

Ch

7

Coordinate Geometry

Divisor of a line segment, nx1 + mx2 ny1 + my2 (x, y) = , m+n m+n

1

2

Area of triangle 1 = u(x1y2 + x2y3 + x3y1) – (x2y1 + x3y2 + x1y3)u 2 Area of quadrilateral 1 = u(x1y2 + x2y3 + x3y4+ x4y1) 2 – (x2y1 + x3y2 + x4y3+ x1y4)u r

te

ap

Ch

8

Vectors

u ~r u = √x2 + y2 ^ ~r ~r = u ~r u r

te

ap

Ch

9

Solution of Triangles

a b c = = sin A sin B sin C a2 = b2 + c2 – 2bc cos A b2 = a2 + c2 – 2ac cos B c2 = a2 + b2 – 2ab cos C Area of triangle 1 1 1 = ab sin C = bc sin A = ac sin B 2 2 2 Heron’s Formula a+b+c = √s(s – a)(s – b)(s – c), s = 2

I=

r

te

ap

Ch

10

Index Numbers

Q1 – ∑Ii wi × 100 I = Q0 ∑wi v

Form 5 r

te

ap

Ch

1

Circular Measure

Arc length, s = rq 1 2 rq 2 Heron’s Formula = √s(s – a)(s – b)(s – c) , a+b+c s = 2 Area of sector, A =

r

C

4

2

Differentiation

dy dv du =u +v dx dx dx du dv v –u dx dx u dy y= , = v2 v dx dy dy du = × dx du dx y = uv,

n! (n – r)! n! n Cr = (n – r)! r! n!

P= a! b! c!…

Pr =

r

te

Ch

5

Mean, µ = np s = √npq Z=

Ch

3

Integration

Area under a curve = =

b

y dx

a b

x dy

a

Volume of revolution = =

b a b a

πy2 dx πx2 dy

vi

X–µ s

r

te

ap

er

Probability Distribution

P(X = r) = nCr prqn – r, p + q = 1

Ch t ap

Permutation and Combination

n

ap

te

p ha

r

te

ap

Ch

6

Trigonometric Functions

sin2 A + cos2 A = 1 sec2 A = 1 + tan2 A cosec2 A = 1 + cot2 A sin 2A = 2 sin A cos A cos 2A = cos2 A – sin2 A = 2 cos2 A – 1 = 1 – 2 sin2 A 2 tan A tan 2A = 1 – tan2 A sin (A ± B) = sin A cos B ± cos A sin B ± cos (A ± B) = cos A cos B sin A sin B tan A ± tan B tan (A ± B) = 1 tan A tan B

p

a Ch

Learning Area : Algebra

r te

2

R KEYWO

Quadratic Functions

DS

• Axis of symmetry – Paksi simetri • Completing the square – Penyempurnaan kuasa dua • Discriminant – Pembezalayan • Factorisation – Pemfaktoran • General form – Bentuk am • Imaginary root – Punca khayalan • Intersection – Persilangan • Maximum point –Titik maksimum • Minimum point – Titik minimum • Parabola – Parabola • Quadratic equation – Persamaan kuadratik • Quadratic function – Fungsi kuadratik • Quadratic inequality – Ketaksamaan kuadratik • Real root – Punca nyata • Root – Punca • Vertex form – Bentuk verteks

Concept Map

ure mon feat t. A com e as n e a m a o s t the wn in which is he eing thro t h b t n ll a e a p h b t t d e e , curate a curv a bask c d a w n o t a ll o o lt n f u e is tap ents the curv ows a ca ir movem above sh sition of that the o m p is ra e g s h t ia t c d e d j The e an o ob the shap these tw raph. If g between n o ti c fun quadratic e missed. b l il w t e targ

29

Additional Mathematics SPM Chapter 2 Quadratic Functions

Equations and 2.1 Quadratic Inequalities Solving quadratic equations using the method of completing the square and using the quadratic formula

A

1. Certain quadratic equations can be solved by factorisation.

Form 4

For example: x2 + x – 2 = 0 (x – 1)(x + 2) = 0 x – 1 = 0 or x + 2= 0 x =1 x = –2 2. Quadratic equations such as x2 = 16, (x – 2)2 = 6 can be solved easily because the left hand side of the equation is a complete square.

For example: (x – 2)2 = 6 x – 2 = ± 6 x = 2 ± 6 = 2 +  6 or 2 – 6 = 4.449 or –0.4495 [to 4 significant figures] 3. For quadratic equations that are difficult to factorise or cannot be factorised easily such as x2 + 6x – 2 = 0 and 2x2 + 4x – 1 = 0, these equations can be solved using the (a) completing the square method, –b ±  b2 – 4ac (b) quadratic formula method, .

2a Derivation of the Quadratic Formula

INFO

4. It is easier to use the quadratic formula method to solve the quadratic equation ax2 + bx + c = 0 when the value of a, b and c are known.

Solution (a) x2 + 6x – 2 = 0 x2 + 6x = 2

x2 + 6x +

 62  = 2 +  62  2

2

Express the equation in the form of ax2 + bx + c with a = 1.

2 x2 + 6x + 32 = 2 + 9 Add  b  on 2a 2 (x + 3) = 11 both sides of the

x + 3 = ±  11 equation. 

x = –3 ± 11 = –3 +  11 or –3 –  11

= 0.3166 or –6.3166

SPM Tips Use the calculator to check your answer by substituting the answer into the quadratic equation involved.

(b) For the case when the coefficient of x2 is not 1, change the coefficient to 1 by dividing each term in the equation by that coefficient before completing the square.

2x2 + 4x – 1 = 0 Divide each term 2 2 4 1 x + x– =0 by the coefficient 2 2 2 of x2.

1 x2 + 2x = 2 2 2 1 2 2 x2 + 2x + = + 2 2 2 1 2 x + 2x + 1 = + 1 2 3 (x + 1)2 = 2 3 x+1 =± 2 x = –1 ± 1.2247

= 0.2247 or –2.2247

 

 

Try Question 1 in ‘Try This! 2.1’

1 Determine the roots for each of the following quadratic equations by using the completing the square. (a) x2 – 2 + 6x = 0 (b) 2x2 + 4x – 1 = 0

30

2 Solve each of the following quadratic equations by using the quadratic formula method. (a) 3x2 = 2x + 5 (b) 6 – 2x = 5x2

Solution (a) Rearrange 3x2 = 2x + 5 to the general form of ax2 + bx + c = 0.

3x2 – 2x – 5 = 0 Therefore, a = 3, b = –2 and c = –5.

Substitute the values of a, b and c into the quadratic formula,  –(–2) ±  (–2)2 – 4(3)(–5) x = 2(3) –b ±  b2 – 4ac 2 ±  4 + 60 x= = 2a 6 2 ±  64 = 6 2±8 = 6 2+8 2–8 x = or x = 6 6 5 = or = –1 3 (b) Rearrange 6 – 2x = 5x2 to the general form of ax2 + bx + c = 0.

5x2 + 2x – 6 = 0 Therefore, a = 5, b = 2 and c = –6.

Substitute the values of a, b and c into the quadratic formula,  –2 ±  (2)2 – 4(5)(–6) x = 2(5) –b ±  b2 – 4ac –2 ±  124 x= = 2a 10

Solution Applying the Pythagoras theorem, AC2 = AB2 + BC2 (p + 4)2 = (p – 3)2 + (p + 1)2 2 p + 8p + 16 = p2 – 6p + 9 + p2 + 2p + 1 p2 – 12p – 6 = 0  –(–12) ±  (–12)2 – 4(1)(–6) p= 2(1) 12 ±  168 = 2 12 +  168 12 –  168 = or 2 2 = 12.48

or

–0.4807 (not acceptable)

Therefore, p = 12.48 Try Questions 3 – 7 in ‘Try This! 2.1’

C A L C U L A T O R Corner Solve the quadratic equation x2 – 14x – 39 = 0.

1: Press the MODE key until EQN is displayed.

2: Press 1 3: ‘Unknowns?’ is displayed, press  4: ‘Degree?’ is displayed, press 2 5: ‘a?’ is displayed, press 1 = 6: ‘b?’ is displayed, press – 1 4 = 7: ‘c?’ is displayed, press – 3 9 = 8: One of the roots is, ‘x1 = 16.38’ is displayed

–2 +  124 –2 –  124 x = or x = 10 10 = 0.9136 or = –1.314

8: Press the  key. Another root is, ‘x2 = –2.381’ is displayed

Try Question 2 in ‘Try This! 2.1’

B

1. If α and β are the roots of a quadratic equation, then

3 A (p + 4) cm (p – 3) cm

B

Forming quadratic equations from given roots

(p + 1) cm

C

The diagram above shows a right-angled triangle ABC with AB = (p – 3) cm, BC = (p + 1) cm and AC = (p + 4) cm. Find the value of p. [Give your answer correct to 4 significant figures.]

x = α and x=b (x – α) = 0 and (x – b) = 0 (x – α)(x – b) = 0 x2 – αx – bx + αb = 0 x2 – (α + b)x + αb = 0 α + b is called the sum of the roots (SOR) and ab is called the product of the roots (POR). x2 – (SOR)x + (POR) = 0

31

Form 4

Additional Mathematics SPM Chapter 2 Quadratic Functions

Additional Mathematics SPM Chapter 2 Quadratic Functions

2. One of the methods to form a quadratic equation from the given roots of α and b is by expanding the product of (x – α)(x – b) = 0.

3. The second method is by finding the sum of roots, (α + b) and the product of roots, (αb) to be substituted into the quadratic equation x2 – (SOR)x + (POR) = 0.

4. To determine the sum of roots and product of roots of a quadratic equation

Form 4

Quadratic equation ax2 + bx + c = 0 where a ≠ 0 Divide each term of the equation by a so that the coefficient of x2 becomes 1. b c x2 + a x + a = 0

Form a quadratic equation with the given roots.

2 (a) 2 and –3 (c) –4 only (b) and 5 3 Solution (a) SOR = 2 + (–3) = –1 POR = 2 × (–3) = –6 The quadratic equation is

x2 – (SOR)x + (POR) = 0 x2 – (–1)x + (–6) = 0

x2 + x – 6 = 0

 

32

(c) SOR = –4 + (–4) = –8 POR = –4 × (–4) = 16 The quadratic equation is

x2 – (SOR)x + (POR) = 0 x2 – (–8)x + (16) = 0

x2 + 8x + 16 = 0

One root only means there are two equal roots.

Try Question 8 in ‘Try This! 2.1’

Given α and b are the roots of a quadratic equation 5x2 – 2x – 4 = 0. Form a quadratic equation with roots (a) (5a + 1) and (5b + 1) 2 2 (b) and a b

4

 

(3x – 2)(x – 5) = 0 3x2 – 17x + 10 = 0

5

b Sum of roots = – a c Product of roots = a

2 17 (b) SOR = + 5 = 3 3 2 10 POR = (5) = 3 3 The quadratic equation is 17 10 x2 – x+ =0 3 3 3x2 – 17x + 10 = 0

Alternative Method 2 x= and x =5 3 3x = 2 and x =5 3x – 2 = 0 and x–5 =0

Multiply each term of the equation by 3.

Solution Divide each term of the equation by 5, so that the coefficient of x2 becomes 1.

2 4 x2 – x – = 0 5 5 Given a and b are the 2 a+b= roots of the quadratic 5 equation.

4 ab = – 5 (a) The new roots are (5a + 1) and (5b + 1).

New SOR = (5a + 1) + (5b + 1)

= 5(a + b) + 2 2

=5 +2 5

=4

 

New POR = (5a + 1)(5b + 1)

= 25ab + 5a + 5b + 1

= 25ab + 5(a + b) + 1 4 2

= 25 – +5 +1 5 5

= –17



  

Additional Mathematics SPM Chapter 2 Quadratic Functions

Therefore, x2 – 4x – 17 = 0 2 2 (b) The new roots are and . a b 2 2 New SOR = + a b 2b + 2a

= ab 2(b + a)

= ab 2 2

= 5 4 – 5

= –1

 

2 2 New POR = a b 4

= ab 4

= 4 – 5

= –5

  

25 2 25 2 2a = 2 25 2 a = 4 5 a = ± ............... b 2 Substitute b into a: POR = (a)(2a) =

5 5 , k =6 2 2 = 15 5 5 When a = – , k = 6 – 2 2 = –15

 

When a =



Therefore, x2 + x – 5 = 0 Try Question 9 in ‘Try This! 2.1’

Therefore, the possible values of k are 15 and –15.

SPM Tips 1 This example can also be solved by using a and a 2 as the roots of the equation.

SPM

Highlights

The quadratic equation 4x2 + 5x – 6 = 0 has roots h and

k. Form a quadratic equation with roots

6 One root of the quadratic equation 2x2 – kx + 25 = 0 is half of the other root of the equation. Find the possible values of k.

4x2 + 5x – 6 = 0

SOR = h + k = – POR = hk = – New SOR =

Divide each term of the equation by 2.

Assume one of the roots is a.

Then, the other root is 2a. k SOR = a + 2a = – – 2 k 3a = 2 k = 6a..............



h k and .

2 2

Solution: x2 +





Try Questions 10 – 17 in ‘Try This! 2.1’

The new quadratic equation is: x2 – (SOR)x + (POR) = 0 x2 – (–1)x + (–5) = 0

Solution 2x2 – kx + 25 = 0 k 25 x2 – x + =0 2 2

Form 4

The new quadratic equation is: x2 – (SOR)x + (POR) = 0 x2 – (4)x + (–17) = 0

3 2

5 3 x– =0 4 2 5 4

h k h+k + = = 2 2 2

h New POR = 2

k hk = = 2 4

  

–

–

5 4 =– 5 2 8

3 2 =– 3 4 8

The new equation is

5

3

 8 x + – 8  = 0

x2 – –

8x2 + 5x – 3 = 0

33

Additional Mathematics SPM Chapter 2 Quadratic Functions

SPM

Highlights

7

One of the roots of the quadratic equation x2 + (n – 5)x – 2n2 = 0, where n is a constant, is the negative of the other root of the equation. Find the product of the roots.

Solution:

Assume that the roots for the equation x2 + (n – 5)x – 2n2 = 0 are a and – a.

SOR = a + (– a) = –(n – 5) 0 = –n + 5 n = 5.............

Form 4

POR = –2n = –2(5)2 = –50 2

SPM

Highlights

The quadratic equation px2 – 4x + q = 0, where p and q are constants, has roots a and 3a. Express p in terms of q.

Solution: px2 – 4x + q = 0 4 q =0 x2 – x + p p 4 SOR = a + 3a = – – p 4 4a = p 1 a = . ............. p q POR = (a)(3a) = p q 2 3a = ............. p Substitute  into .

1 2 q 3 = p p 3 q = p2 p 3p = p2q 3p – p2q = 0 p(3 – pq) = 0 p = 0 or 3 – pq = 0 (unacceptable) pq = 3 3 p= q



Determine the range of values of x that satisfy each of the following quadratic inequalities by using graph sketching method, number line method or table method. (a) x2 + x – 6  0 (b) 6 + x – x2 > 0 (c) 2x2 – 5x + 2 . 0 Solution (a) x2 + x – 6  0 Graph sketching method When f(x) = 0, x2 + x – 6 = 0 (x – 2)(x + 3) = 0 x – 2 = 0 or x + 3 = 0 x =2 x = –3 2 The coefficient of x , a = 1 . 0 ⇒ the shape of the graph is .

The graph sketch:

f(x)

–3



x

2

Therefore, the range of values of x that satisfy the quadratic inequality x2 + x – 6  0 is –3  x  2. (b) 6 + x – x2 > 0 Number line method Factorise the quadratic equation, 6 + x – x2 = 0 (2 + x)(3 – x) = 0 (2 + x) = 0 or (3 – x) = 0 x = –2 x =3

 

(2 + x) ≥ 0 when x ≥ –2 + +

–

+

–2

(3 – x) ≥ 0 when x ≤ 3

+

3

–

+ (+) × (+) = (+)

C

Solving quadratic inequalities

1. The range of values of x that satisfy a quadratic inequality can be determined by using • graph sketching method • number line method • table method INFO

34

Therefore, the range of values of x that satisfy the quadratic inequality 6 + x – x2 > 0 is –2 < x < 3.

SPM Tips

Use the graph sketching method to check the answer.

Additional Mathematics SPM Chapter 2 Quadratic Functions

(c) 2x2 – 5x + 2 . 0 Table method 2x2 – 5x + 2 = 0 (2x – 1)(x – 2) = 0

Example of HOTS Question 6.6 m xm

(2x – 1)

–

0

+

+

+

(x – 2)

–

–

–

0

+

(2x – 1)(x – 2)

+

0

–

0

+

From the table, it shows that (2x – 1)(x – 2) . 0 1 when x  or x . 2.

2 Try Question 18 in ‘Try This! 2.1’

The diagram above shows a rectangular piece of land with a length of 6.6 m and a width of 4.8 m.

Amin wants to lay square tiles with side x m around the land to build a walkway. If the area of the region where the tiles are laid is 12.24 m2, find the value of x.

Solution

8 Find the range of values of x for (2x – 1)(x + 4) < 4 + x.

Solution (2x – 1)(x + 4) < 4 + x 2x2 + 8x – x – 4 < 4 + x 2x2 + 7x – 4 – 4 – x < 0 2x2 + 6x – 8 < 0 x2 + 3x – 4 < 0 a = 1 . 0 ⇒ the shape of the graph is

Area of the region where the tiles are laid = 2 × 6.6 × x + 2 × (4.8 – 2x) × x = 13.2x + 9.6x – 4x2 = 22.8x – 4x2 Hence, 22.8x – 4x2 = 12.24 4x2 – 22.8x + 12.24 = 0 400x2 – 2 280x + 1 224 = 0 50x2 – 285x + 153 = 0

  –(–285) ±  (–285)2 – 4(50) (153) 2(50) 285 ± 225 = 100

x =

.

x + 3x – 4 = 0 (x – 1)(x + 4) = 0 x – 1 = 0 or x+4 =0 x = 1 or x = –4 The graph intersects the x-axis at x = 1 and x = –4 When

4.8 m

xm

2

510 100

or

x=

x = 5.1

or

x = 0.6

x =

60 100

5.1 m is longer than the width of the land which is 4.8 m. So, x = 5.1 is unacceptable.

Therefore, x = 0.6

–4

1

x

Try this HOTS question

Therefore, the range of values of x is –4 < x < 1.

SPM Tips Eliminating the common expression of (x + 4) in the inequality will cause the range of values of x obtained incomplete.

(2x – 1)(x + 4) < (4 + x) 2x – 1 < 1 2x < 1 + 1 x <1

The profit made by a factory that is producing nuts in packets is given by P(t) = 40t2 – 7t – 5 676, where t is the time, in hours, the production process is running. Find the time of production needed for the factory to get back its capital.

Answer: The time of production must be at least 12 hours.

Try Questions 20 – 21 in ‘Try This! 2.1’ Try Question 19 in ‘Try This! 2.1’

35

Form 4

Factorise the quadratic x  1 x = 1 1  x  2 x = 2 x . 2 2 2 2 equation

Additional Mathematics SPM Chapter 2 Quadratic Functions

Try This!

2.1

1. Solve each of the following quadratic equations by using the completing the square method. (a) x2 – x = 3 (b) x2 + x – 4 = 0 (c) x2 + 5x + 2 = 0 (d) x(x – 2) = 6 (e) x(x + 2) – 3 = 0 (f) 2x2 + 3x – 2 = 0 (g) 3x2 + 6x – 1 = 0 (h) 4x2 – 3x – 2 = 0

Form 4

2. Solve each of the following quadratic equations by using the formula. (a) x2 + 4x – 3 = 0 (b) 2x2 = 3x + 4 (c) 3x2 + 7x – 5 = 0 (d) 4x2 – 5x = 2 (e) 3x(x + 3) = 4 3. The sum of the squares for two consecutive positive integers is 2 521. Find the values of the two integers.

4. Given two squares with side lengths of x cm and (x – 2) cm respectively. The total area of the two squares is 340 cm2. Find the total perimeter for both squares.

5. Nesa’s age is four times her son’s age in this year.

The product of Nesa’s age and her son’s age two years later will be 270. Find Nesa’s age and her son’s age in this year.

6. The product of two consecutive odd integers is 1 763. Find the values of these two integers.

(2x – 1) cm

x cm

The diagram above shows a triangle with an area of 60 cm2. Find (a) the value of x, (b) the perimeter of the triangle.

HOTS Applying

8. Form quadratic equations which have the following roots. (a) 2 and 3 (b) – 4 and 5 2 (c) –2 and –5 (d) only 3 1 2 (e) –3 and (f) 3 +  2 and 3 –  2 9. Given a and b are the roots of a quadratic equation 2x2 – 8x – 5 = 0. Form quadratic equations with the following roots. (a) 2a and 2b

36

11. One of the roots of the quadratic equation 3x2 + hx = 6x – 25 is one third the other root. Find the possible values of h.

12. Given the quadratic equation qx2 – 12x + 8 = 0, q ≠ 0, has two equal roots. Find the possible values of q.

13. Given a and b are the roots of the quadratic equation a b 5x2 – 6x + h = 0, whereas and are the roots k k of the quadratic equation 2x2 + 3x – 5 = 0. Find the values of h and k.

14. Given the quadratic equation x2 – (p – 10)x + 5q = 0 has roots 6 and 10. Find the values of p and q.

15. Given q and 4 are the roots of the quadratic equation (2x – h)2 = 4x. Find the possible values of h and q.

16. The roots of the quadratic equation (x + 1)(x – 5) = p(q – x) – 9 are 2 and – 6. Find the possible values of p and q.

17. Given –3 and (k – 2) are the roots of the quadratic equation x2 + (h – 5)x – 12 = 0, where h and k are constants. Find the values of h and k.

18. Find the range of values of x that satisfy the following quadratic inequalities by using graph sketching method, number line method or table method. (a) x2 + x – 6 . 0 (b) x2 – 3x – 10 < 0 (c) 4 + 3x – x2 > 0 (d) 9 + 5x – 4x2 , 0

7.

(b) (3a + 1) and (3b + 1) a b (c) and 2 2 4 4 (d) and a b

10. Given one of the roots of the quadratic equation 8x2 + 2x – m = 0 is twice the other root. Find the possible values of m.

19. Find the range of values of x that satisfy the following quadratic inequalities. (a) 3x2 + 13x < 10 (b) 7 – 2x , (x + 4)2 (c) 3x2 – x – 21 > x(2x + 3) (d) (2x + 1)(x – 5) > 3(2x + 1) 20. 300 packages can be produced when a packaging machine operates at a rate of x packages per minute. A study found out that when the rate of operation of the machine is increased to (x + 3) packages per minute, the time saved is 5 minutes for 300 packages. Determine the new operating rate of the machine.

HOTS Evaluating

21. In an experiment, a stone was thrown upwards at a speed of 15 m s–1 from a platform 5 m above the ground. The position of the stone from the ground can be represented by the function f(t) = 5 + 15t – 4.9t 2, where t represents the time, in seconds, after the stone was thrown. HOTS Analysing

Additional Mathematics SPM Chapter 2 Quadratic Functions

Ground

Find the time of the stone to reach the ground. [Give your answer correct to two significant figures.]

of Roots of Quadratic 2.2 Types Equations A

Relating the types of roots of quadratic equations to the discriminant value

1. The types of roots of a quadratic equation ax2 + bx + c = 0 depends on the value of the expression b2 – 4ac.

2. The expression b2 – 4ac is called the discriminant of the quadratic equation.

3. The relation between the types of roots and the values of the discriminant are shown in the table below.

Discriminant, b2 – 4ac

Types of roots

.0

Two different real roots

=0

Two equal real roots

0

No roots (or imaginary roots )

(a) (b) (c)

9 Determine the types of roots for each of the following quadratic equations. (a) 3x2 – 5x – 6 = 0 (b) 9x2 = 6x – 1 (c) x2 + 4x + 5 = 0 Solution (a) 3x2 – 5x – 6 = 0 Compare with ax2 + bx + c = 0. a = 3, b = –5, c = –6 b2 – 4ac = (–5)2 – 4(3)(–6) = 97 . 0 Therefore, equation 3x2 – 5x – 6 = 0 has two different real roots.

(c) x2 + 4x + 5 = 0 Compare with ax2 + bx + c = 0. a = 1, b = 4, c = 5 b2 – 4ac = 42 – 4(1)(5) = –4 , 0 Therefore, equation x2 + 4x + 5 = 0 has no real roots.

SPM Tips For the equation x2 + 4x + 5 = 0 in (c), the imaginary root can be written as x =

–b ±  b2 – 4ac –4 ±  –4 = 2a 2(1)

= –2 ±

 –1 4

2 –1 = –2 ±  = –2 ± i

The imaginary part, that is  –1, is represented by the symbol i.

Imaginary roots are widely used in the fields of electric and electronics.

Try Question 1 in ‘Try This! 2.2’

B

Solving problems involving types of roots of quadratic equations

1. The relation between the types of roots and the values of the discriminants for quadratic equations can be used to (a) find the value or values of unknown variables, (b) find a relation between two variables in the quadratic equation.

10 Find the range of values of p if the quadratic equation 3x2 + 4 = 8x + 2px2, where p is a constant, has two different real roots.

37

Form 4

5m

(b) 9x2 = 6x – 1 9x2 – 6x + 1 = 0 Compare with ax2 + bx + c = 0. a = 9, b = –6, c = 1 b2 – 4ac = (–6)2 – 4(9)(1) =0 Therefore, equation 9x2 = 6x – 1 has two equal real roots.

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