Chapter 2 - Quadratic Expression and Equation - Exercise

Mathematics Form 4 F4AM00401FLesson 03: Quadratic expressions.

Learning Area: Identifying quadratic expressions. 1 / 4

ACTIVITY SHEET

QUESTION 1Question:State if the expression given is a quadratic expression in one variable.

x2 + 2x + 1Solution:

QUESTION 2

Question:State if the expression given is a quadratic expression in one variable. 4x2 – 2x + 1Solution:

QUESTION 3

Question:State if the expression given is a quadratic expression in one variable.

96

12 ++ xx

Solution:

QUESTION 4

Question:State if the expression given is a quadratic expression in one variable. p3 + 2p + 1Solution: .

QUESTION 5

Question:State if the expression given is a quadratic expression in one variable. s4 – 2s4 – 3Solution:

Pupil’s copy



QUESTION 6Question:State if the expression given is a quadratic expression in one variable. x – 2x2 + 3Solution:

QUESTION 7

Question:State if the expression given is a quadratic expression in one variable. w2 – 2w2 + 3Solution:

QUESTION 8

Question:State if the expression given is a quadratic expression in one variable. y – 2y3 + 9Solution:

QUESTION 9

Question:State if the expression given is a quadratic expression in one variable. 2x2 – 4x – 16Solution:

QUESTION 10

Question:State if the expression given is a quadratic expression in one variable. 3y2 – 2x2 + 4xSolution:



QUESTION 1Question:State if the expression given is a quadratic expression in one variable.

x2 + 2x + 1Solution:Yes.

QUESTION 2

Question:State if the expression given is a quadratic expression in one variable. 4x2 – 2x + 1Solution:Yes.

QUESTION 3

Question:State if the expression given is a quadratic expression in one variable.

96

12 ++ xx

Solution:No.

QUESTION 4

Question:State if the expression given is a quadratic expression in one variable. p3 + 2p + 1Solution:No.

QUESTION 5

Question:State if the expression given is a quadratic expression in one variable. s4 – 2s4 – 3Solution:No.

ACTIVITY SHEETTeacher’s copy



QUESTION 6Question:State if the expression given is a quadratic expression in one variable. x – 2x2 + 3Solution:Yes.

QUESTION 7

Question:State if the expression given is a quadratic expression in one variable. w2 – 2w2 + 3Solution:Yes.

QUESTION 8

Question:State if the expression given is a quadratic expression in one variable. y – 2y3 + 9Solution:No.

QUESTION 9

Question:State if the expression given is a quadratic expression in one variable. 2x2 – 4x – 16Solution:Yes.

QUESTION 10

Question:State if the expression given is a quadratic expression in one variable. 3y2 – 2x2 + 4xSolution:No.

Mathematics Form 4 Lesson 04: Quadratic expressions from two linear expressions..

Learning Area: Form quadratic expressions by multiplying any two linear expressions. 1 / 4

Activity SheetPupils’ copy

QUESTION 1Question:Form quadratic expressions by multipliying the following expressions:

a) (x + 2)(x + 3)b) (s + 3)(s + 9)c) (y + 2)(y + 5)d) (2y + 3)(y + 1)e) (3y + 1)(3y + 2)

Solution:

QUESTION 2

Question:Form quadratic expressions by multipliying the following expressions:

a) (x + 3)(x – 3)b) (2x + 1)(2x – 1)c) (2y + 2)(y – 2)d) (2r + 1)(r – 2)e) (2s + 3)(2s – 5)

Solution:



QUESTION 3


a) (x – 3)(x – 3)b) (3x – 1)(2x – 1)c) (4y – 2)(y – 2)d) (2r – 1)(6r – 2)e) (2s – 3)(9s – 5)

Solution:

QUESTION 4


a) (x – 3)(x + 3)b) (3r – 1)(r + 1)c) (4t – 2)(t + 2)d) (3a – 1)(a + 2)e) (4e – 2)(4e + 2)

Solution: .



QUESTION 1Question:Form quadratic expressions by multipliying the following expressions:

a) (x + 2)(x + 3)b) (s + 3)(s + 9)c) (y + 2)(y + 5)d) (2y + 3)(y + 1)e) (3y + 1)(3y + 2)

Solution:a) x2 + 5x + 6b) s2 + 12s + 27c) y2 + 7y + 10d) 2y2 + 5y + 3e) 9y2 + 9y + 2

QUESTION 2


a) (x + 3)(x – 3)b) (2x + 1)(2x – 1)c) (2y + 2)(y – 2)d) (2r + 1)(r – 2)e) (2s + 3)(2s – 5)

Solution:a) x2 – 9b) 4x2 – 1c) 2y2 – 2y – 4d) 2r2 – 3r – 2e) 4s2 – 4s – 15




QUESTION 3


a) (x – 3)(x – 3)b) (3x – 1)(2x – 1)c) (4y – 2)(y – 2)d) (2r – 1)(6r – 2)e) (2s – 3)(9s – 5)

Solution:a) x2 – 6x + 9b) 6x2 – 5x + 1c) 4y2 – 10y + 4d) 12r2 – 10r + 2e) 18s2 – 37s + 15

QUESTION 4


a) (x – 3)(x + 3)b) (3r – 1)(r + 1)c) (4t – 2)(t + 2)d) (3a – 1)(a + 2)e) (4e – 2)(4e + 2)

Solution:a) x2 – 9b) 3r2 + 2r – 1c) 4t2 + 6t – 4d) 3a2 + 5a – 2e) 16e2 – 4

Mathematics Form 4 Lesson 05: Quadratic expressions from everyday life situations.

Learning Area: Form quadratic expressions based on specific situations.

QUESTION 1Question:A rectangular shaped aquarium has the width of 3 cm more than its height and the length of 20 cm.If the height is x cm, can we form the quadratic expression for the volume of the aquarium?

Solution:

QUESTION 2

Question:Aishah has a piece of rectangular shaped carpet in her bedroom. Its measurements are as shown inthe diagram. Can we help her find the area of the carpet?

Solution:

ACTIVITY SHEETPupils’ copy

2x m

(3x + 1) m



QUESTION 3

Question:Mr Ahmad drives to his office every morning. He drive at the speed of (s − 2) metre/minute for 4sminutes. Form the quadratic expression for the distance between his house and his office.

Solution:

QUESTION 4

Question:Yusof Music organise a concert in which some local artist to performing. There are 20n numbers ofVIP tickets sold for RM n and 5000 normal tickets sold at RM 7 each. Form a quadratic expressionfor the collection for the concert if all the tickets were sold.

Solution:

QUESTION 5

Question:Mustapha inherits two pieces of land from his grandfather. One piece of land is square in shapewith sides of 50x metres and the area of the other piece is 2000 square metres. Find the total area ofthe two pieces of lands.

Solution:



QUESTION 1

Question:A rectangular shaped aquarium has the length of 20 cm and the width of 3 cm more than its height.If the height is x cm, can we form the quadratic expression for the volume of the swimming pool?

Solution:Formula for volume is:

Volume = length × width × height= 20 cm × (x + 3) cm × x cm= [20 × (x2 + 3x)] cm3

= (20x2 + 60x) cm3

QUESTION 2

Question:Aishah has a piece of rectangular shaped carpet in her bedroom. Its measurements are as shown inthe diagram. Can we help her find the area of the carpet?

Solution:The formula for the area is:

Area = width × length= 2x × (3x + 1)= 6x2 + 2x

QUESTION 3

Question:Mr Ahmad drives to his office every morning. He drive at the speed of (s − 2) km/minute for 4sminutes. Form the quadratic expression for the distance between his house and his office.

Solution:The formula for distance is:

Distance = speed × time= (s – 2) km/minute × 4s minutes= 4s2 – 8s km


2x m

(3x + 1) m



QUESTION 4

Question:Yusof Music organise a concert in which some local artist to performing. There are 20m numbersof VIP tickets sold for RM m and 5 000 normal tickets sold at RM 7 each. Form a quadraticexpression for the collection for the concert if all the tickets were sold.

Solution:The collection for all the tickets = RM (20m × m) + RM (5 000 × 7)

= RM (20m2 + 35 000)

QUESTION 5

Question:Mustapha inherits two pieces of land from his grandfather. One piece of land is square in shapewith sides of 50t metres and the area of the other piece is 2000 square metres. Find the total area ofthe two pieces of lands.

Solution:The total area of the two pieces of land = (50t × 50t) m2 + 2000 m2

= (2500t2 + 2000) m2

Mathematics Form 4 Lesson 06: Quadratic expressions of the form ax2 + bx + c, where b = 0 or c = 0

Learning Area: Factorise quadratic expressions of the form ax2 + bx + c,where b = 0 or c = 0. 1 / 4

QUESTION 1Question:Factorise the given quadratic expression.

6x2 + 9Solution:

QUESTION 2

Question:Factorise the given quadratic expression.

12 – 3r2

Solution:

QUESTION 3


5w2 – 5Solution:

QUESTION 4


16y2 + 8Solution: .

QUESTION 5


6s2 + 18Solution:




QUESTION 6Question:Factorise 8x2 + 6y.

Solution:

QUESTION 7

Question:Factorise 12r2 – 2r.

Solution:

QUESTION 8

Question:Factorise 10w – 4w2

Solution:

QUESTION 9

Question:Factorise 7y2 + 28y.

Solution:

QUESTION 10

Question:Factorise 9s2 – 27s.

Solution:




6x2 + 9Solution:6x2 + 9 = 3(2x2 + 3)

QUESTION 2


12 – 3r2

Solution:12 – 3r2 = 3(4 – r2)

QUESTION 3


5w2 – 5Solution: 5w2 – 5 = 5(w2 – 1)

QUESTION 4


16y2 + 8Solution: . 16y2 + 8 = 8(2y2 + 1)

QUESTION 5


6s2 + 18Solution: 6s2 + 18 = 6(s2 + 3)




QUESTION 6Question:Factorise 8x2 + 6y.

Solution: 8x2 + 6y = 2(4x2 + 3y)

QUESTION 7

Question:Factorise 12r2 – 2r.

Solution: 12r2 – 2r = 2(6r2 – r)

QUESTION 8

Question:Factorise 10w – 4w2

Solution: 10w – 4w2 = 2(5w – 2w2)

QUESTION 9

Question:Factorise 7y2 + 28y.

Solution: 7y2 + 28y = 7(y2 + 4y)

QUESTION 10

Question:Factorise 9s2 – 27s.

Solution: 9s2 – 27s = 9(s2 – 3s)

Mathematics Form 4 Lesson 07: Quadratic expressions of the form px2 - q where p and q are perfect squares

Learning Area: Factorise quadratic expressions of the form px2 – q,where p and q are perfect squares. 1 / 5


4x2 – 16Solution:

QUESTION 2


25 – r2

Solution:

QUESTION 3


9w2 – 1Solution:

QUESTION 4


4y2 – 16Solution: .

QUESTION 5


64s2 – 25Solution:




QUESTION 6Question:

Factorise 9

4 – 49

2x

Solution:

QUESTION 7

Question:

Factorise 25

1 – x2

Solution:

QUESTION 8

Question:Factorise 100 – 4w2

Solution:

QUESTION 9

Question:Factorise 81y2 – 1.

Solution:

QUESTION 10

Question:Factorise 144 – 121s2.

Solution:




4x2 – 16Solution:4x2 – 16 = (2x)2 – 42

= (2x – 4)(2x + 4)

QUESTION 2


25 – r2

Solution: 25 – r2 = 52 – r2

= (5 – r)(5 + r)

QUESTION 3


9w2 – 1Solution: 9w2 – 1 = (3w)2 – 12

= (3w – 1)(3w + 1)

QUESTION 4


4y2 – 16Solution: 4y2 – 16 = (2y)2 – 42

= (2y – 4)(2y + 4)




QUESTION 5

Question:Factorise the given quadratic expression.64s2 – 25Solution: 64s2 – 25 = (8s)2 – 52

= (8s – 5)(8s + 5)

QUESTION 6

Question:

Factorise 9

4 - 49

2x

Solution:

9

4 - 49

2x=

22

73

2

−

x

=

−73

2 x

+73

2 x



QUESTION 7

Question:

Factorise 25

1 – r2

Solution:

−25

1 r_ =

2

5

1

– r2

=

+

− rr5

1

5

1

QUESTION 8

Question:Factorise 100 – 4w2

Solution: 100 – 4w2 = 102 – (2w)2

= (10 – 2w)(10 + 2w)

QUESTION 9

Question:Factorise 81y2 – 1.

Solution: 81y2 – 1 = (9y)2 – 12

= (9y – 1)(9y + 1)

QUESTION 10

Question:Factorise 144 – 121s2.

Solution: 144 – 121s2 = 122 – (11s)2

= (12 – 11s)(12 + 11s)

Mathematics Form 4 Lesson 08: Quadratic expressions of the form ax2 + bx + c, where a,b and c ≠ 0

Learning Area: Factorise quadratic expressions of the form ax2 + bx + c,where a, b and c are not equal to zero. 1 / 4


4x2 + 22x + 28Solution:

QUESTION 2


10 – 14r – 12r2

Solution:

QUESTION 3


2w2 + w – 1Solution:

QUESTION 4


2y2 + 5y – 12Solution: .

QUESTION 5


24s2 + 49s + 15Solution:




QUESTION 6Question:Factorise 9 – 9x – 4x2.

Solution:

QUESTION 7

Question:Factorise 16 – 6r – r2.

Solution:

QUESTION 8

Question:Factorise 10 – w – 3w2

Solution:

QUESTION 9

Question:Factorise 7y2 + 10y – 8.

Solution:

QUESTION 10

Question:Factorise 2s2 + 7s + 6.

Solution:




4x2 + 22x + 28Solution:4x2 + 22x + 28 = (2x + 4)(2x + 7)

QUESTION 2


10 – 14r – 12r2

Solution: 10 – 14r – 12r2 = (5 – 3r)(2 + 4r)

QUESTION 3


2w2 + w – 1Solution: 2w2 + w – 1 = (2w – 1)(w + 1)

QUESTION 4


2y2 + 5y – 12Solution: 2y2 + 5y – 12 = (2y – 3)( y + 4)

QUESTION 5


24s2 + 49s + 15Solution: 24s2 + 49s + 15 = (8s + 3)(3s + 5)




QUESTION 6Question:Factorise 9 – 9x – 4x2.

Solution: 9 – 9x – 4x2 = (3 – 4x)(3 + x)

QUESTION 7

Question:Factorise 16 – 6r – r2.

Solution: 16 – 6r – r2 = (2 – r)(8 + r)

QUESTION 8

Question:Factorise 10 – w – 3w2

Solution: 10 – w – 3w2 = (5 – 3w)(2 + w)

QUESTION 9

Question:Factorise 7y2 + 10y – 8.

Solution: 7y2 + 10y – 8 = (7y – 4)( y + 2)

QUESTION 10

Question:Factorise 2s2 + 7s + 6.

Solution: 2s2 + 7s + 6 = (2s + 3)(s + 2)

Mathematics Form 4 Lesson 09: Quadratic expressions with common factor coefficients

Learning Area: Factorise quadratic expressions containing coefficients with common factors. 1 / 5


4x2 + 28x + 24Solution:

QUESTION 2


12t2 – 14t – 10Solution:

QUESTION 3


18u2 + 6u – 24Solution:

QUESTION 4


21y2 − 36y – 12Solution: .

QUESTION 5


28r2 − 14r − 84Solution:




QUESTION 6Question:Factorise 30a2 + 55a − 50.

Solution:

QUESTION 7

Question:Factorise 8b2 − 32.

Solution:

QUESTION 8

Question:Factorise 30 – 3w – 9w2.

Solution:

QUESTION 9

Question:Factorise 18y2 – 6y – 24.

Solution:

QUESTION 10

Question:Factorise 24 − 56x − 16x2.

Solution:




4x2 + 28x + 24Solution:4x2 + 28x + 24 = 4(x2 + 7x + 6)

= 4(x + 1)(x + 6)

QUESTION 2


12t2 – 14t – 10Solution: 12t2 – 14t – 10 = 2(6t2 − 7t − 5)

= 2(2t + 1)(3t −5)

QUESTION 3


18u2 + 6u – 24Solution: 18u2 + 6u – 24 = 6(3u2 + u − 4)

= 6(3u + 4)(u − 1)

QUESTION 4


21y2 − 36y – 12Solution: 21y2 − 36y – 12 = 3(7y2 − 12y − 4)

= 3(7y + 2)(y − 2)




QUESTION 5


28r2 −14r − 84Solution: 28r2 −14r − 84 = 14(2r2 − r − 6)

= 14(2r − 3)(r + 2)



QUESTION 6Question:Factorise 30a2 + 55a − 50.

Solution: 30a2 + 55a − 50 = 5(6a2 + 11a _ 10)

= 5(2a + 5)(3a − 2)

QUESTION 7

Question:Factorise 8b2 − 32.

Solution: 8b2 − 32 = 8(b2 − 4)

= 8(b − 2)(b + 2)

QUESTION 8

Question:Factorise 30 – 3w – 9w2.

Solution: 30 – 3w – 9w2 = 3(10 − w − 3w2)

= 3(5 – 3w)(2 + w)

QUESTION 9

Question:Factorise 18y2 − 6y – 24.

Solution: 18y2 − 6y – 24 = 3(6y2 − 2y − 8)

= 3(3y – 4)( 2y + 2)

QUESTION 10

Question:Factorise 24 − 56x − 16x2.

Solution: 24 − 56x − 16x2 = 4(6 − 14x − 4x2)

= 4(3 − x)(2 − 4x)

Mathematics Form 4 Lesson 10: Quadratic equations and their general forms

Learning Area: 1. Identify quadratic equations with one unknown;2. Write quadratic equations in general form i.e. ax2 + bx + c 1 / 5

QUESTION 1


Question:Is “12k2 + k + 2 = 0” a quadratic equation with one unknown? State the reason for your answer.

Solution:

QUESTION 2

Question:Is “4 + 12t + t2 = t – 2” a quadratic equation with one unknown? State the reason for your answer.

Solution:

QUESTION 3

Question:Identify if the equation given is a quadratic equation with one unknown.

3u – 5 = 0

Solution:

QUESTION 4


3e – 4 = 5f2

Solution: .

QUESTION 5


2p2 + 9p = 15

Solution:



QUESTION 6Question:Write the given quadratic equation in general form i.e. ax2 + bx + c.

5c2 − 2c = 25

Solution: Write the given quadratic equation in general form i.e. ax2 + bx + c.

3 + 2e+ 3e2 = e – 2

QUESTION 7

Question:Write the given quadratic equation in general form i.e. ax2 + bx + c.

x2 – 5x = 3x2 + 14

Solution:

QUESTION 8


1 + 9t = 12t2 – 5

Solution:

QUESTION 9

Question:State the values of a, b and c when 5 + 4p – p2 = –3p + 17 is written in general form.

Solution:

QUESTION 10

Question:Write –2a + 8 = 2a2 – 1 in the general form.

Solution:



QUESTION 1Question:Is “12k2 + k + 2 = 0” a quadratic equation with one unknown? State the reason for your answer.

Solution:Yes, because it has one unknown and the highest power of the unknown is equal to 2.

QUESTION 2

Question:Is “4 + 12t + t2 = t – 2” a quadratic equation with one unknown? State the reason for your answer.

Solution: Yes, because it has one unknown and the highest power of the unknown is equal to 2.

QUESTION 3


3u – 5 = 0

Solution: No.

QUESTION 4


3e – 4 = 5f2

Solution:No




QUESTION 5


2p2 + 9p = 15

Solution: Yes.



QUESTION 6Question:Write the given quadratic equation in general form i.e. ax2 + bx + c.

5c2 – 2c = 25

Solution:5c2 – 2c – 25 = 0

QUESTION 7


x2 – 5x = 3x2 + 14

Solution:2x2 + 5x + 14 = 0

QUESTION 8


1 + 9t = 12t2 – 5

Solution:12t2– 4 – 9t = 0

QUESTION 9

Question:State the values of a, b and c when 5 + 4p – p2 = –3p + 17 is written in general form.

Solution: p2 – 7p + 12 = 0, a = 1, b = –7 and c = 12.

QUESTION 10

Question:Write –2a + 8 = 2a2 – 1 in the general form.

Solution:2a2 + 2a – 9 = 0

Mathematics Form 4 Lesson 11: Quadratic equations based on specific situations.

Learning Area: Form quadratic equations based on specific situations. 1 / 6

QUESTION 1Question:The area of a school field is 200 m2. Its length is 5 m longer than the width. Form a quadraticequation based on the situation.

Solution:

QUESTION 2

Question:Abu Bakar builds a rectangular shaped dining table himself. The measurements of the table top areas shown in the diagram. Can we help him find the quadratic equation to represent the situation?

Solution:


Area = 2m3a m

(2a + 4) m



QUESTION 3

Question:Ali and Din went for a cycling trip. Ali cycles at the speed of (r − 2) metre/minute for 4r minuteswhile the distance travelled by Din is 2 670 m. The total distance travelled by both of them is 4 km.Form the quadratic equation to represent the situation above.

Solution:

QUESTION 4

Question:Siew Leng brother is 3 years younger than her and her grandfather’s age is 5 times the product ofboth their ages. Her grandfather’s age is 65. Form a quadratic equation to represent the situationgiven.

Solution:



QUESTION 5

Question:The number of girls in a class is half of the number of boys. If the product of multiplicationbetween the number of girls and the number of boys is equal to 6 times the sum of the girls andboys, form a quadratic equation to represent the situation.

Solution:



QUESTION 1

Question:The area of a school field is 200 m2. Its length is 5 m longer than the width. Form a quadraticequation based on the situation.


Area = width x lengthLet the width be x, therefore the length is x +5

200 = x (x + 5)200 = x2 + 5x

x2 + 5x – 200 = 0

QUESTION 2

Question:Abu Bakar builds a rectangular shaped dining table himself. The measurements of the table top areas shown in the diagram. Can we help him find the quadratic equation to represent the situation?


Area = width x length2 = 3a (2a + 4)2 = 6a2 + 12a

6a2 + 12a – 2 = 0


Area = 2m3a m

(2a + 4) m



QUESTION 3

Question:Ali and Din went for a cycling trip. Ali cycles at the speed of (r _ 2) metre/minute for 4r minuteswhile the distance travelled by Din is 2 670 m. The total distance travelled by both of them is 4 km.Form the quadratic equation to represent the situation above.

Solution:The distance travelled by Ali + The distance travelled by Din = The total distance travelled.[(r – 2) _ 4r] m + 2 670 = 4 0004r2 – 8r + 2 670 = 4 0004r2 – 8r – 1330 = 0

QUESTION 4

Question:Siew Leng brother is 3 years younger than her and her grandfather’s age is 5 times the product ofboth their ages. Her grandfather’s age is 65. Form a quadratic equation to represent the situationgiven.

Solution:Let Siew Leng age be y.5 × (Siew Leng’s age) x (Her brother’s age) = Her grandfather’s age5 × (y) × (y – 3) = 655y (y – 3) = 655y2 – 15y = 655y2 – 15y – 65 = 0



QUESTION 5

Question:The number of girls is a class is half of the number of boys. If the product of multiplication betweenthe number of girls and the number of boys is equal to 6 times the sum of the girls and boys, form aquadratic equation to represent the situation.

Solution:Let the number of girls be g and the number of boys be 2g.The number of girls x The number of boys = 6 (The number of girls + The number of boys)

Substitute the values into the formulag x 2g = 6(g + 2g) 2g2 = 6g + 12g 2g2 = 18g

Form a quadratic equation2g2 – 18g = 0

Mathematics Form 4 Lesson 12: Roots of quadratic equation

Learning Area: Determine whether a given value is the root of a specific quadratic equation. 1 / 7


QUESTION 2

Question:Determine if r = 4 and r = −2 are the roots of the quadratic equation,

r2 − 2r − s8 = 0

Solution:

QUESTION 3

Question:Determine if w = 2 and w = −2 are the roots of the quadratic equation,

2w2 − w −1 = 0

Solution:

QUESTION 4

Question:Determine if y = 1 is a root of the quadratic equation, 2y2 + 4y – 6 = 0.

Solution: .

QUESTION 5

Question:Determine if p = 0 and p = 5 are the roots of the quadratic equation,

p2 – 5p = 0.

Solution:



QUESTION 6Question:Determine if x = 3 and x = −3 are the roots of the quadratic equation,

9 – 9x – 4x2 = 0.

Solution:

QUESTION 7

Question:

014xequation quadratic theof roots theare 2

1 xand ,

2

1 xif Determine 2 =−−==

Solution:

QUESTION 8

Question:

0116yequation quadratic theof roots theare 4

1 y and ,

4

1y if Determine 2 =−−==

Solution:

QUESTION 9

Question:

013sequation quadratic theof roots theare 3

1 s and ,

3

1s if Determine 2 =−−==

Solution:

QUESTION 10

Question:

019tequation quadratic theof roots theare 3

1 tand ,

3

1 tif Determine 2 =−−==

Solution:



QUESTION 1Question:Determine if x = 2 and x = −3 is the root of the equation, x2 + x – 6 = 0.

Solution:x2 + x – 6 = (2)2 + (2) − 6

= (4) + 2 − 6= 0

x2 + x – 6 = (−3)2 + (−3) − 6= (9) − 3 − 6= 0

Both x = 2 and x = −3 are the roots of the quadratic equation x2 + x – 6 = 0.

QUESTION 2

Question:Determine if r = 4 and r = _2 are the roots of the quadratic equation,

r2 − 2r − 8 = 0

Solution:r2 −2r − 8 = (4)2 − 2(4) − 8

= 16 − 8 − 8= 0

r2 − 2r − 8 = (−2)2 − 2(−2) − 8= 4 + 4 − 8= 0

Both r = 4 and r = −2 are the roots of the quadratic equation r2 − 2r − 8 = 0



QUESTION 3

Question:Determine if w = 2 and w = −2 are the roots of the quadratic equation,

2w2 − w −1 = 0

Solution:2w2 − w – 1 = 2(2)2 − (2) − 1

= 8 − 2 − 1= 5

2w2− w – 1 = 2(−2)2 − (−2) − 1= 8 + 2 − 1= 9

Both w = 2 and w = −2 are not the roots of the quadratic equation 2w2 − w − 1 = 0.

QUESTION 4

Question:Determine if y = 1 is a root of the quadratic equation, 2y2 + 4y – 6 = 0.

Solution: 2y2 + 4y – 6 = 2(1)2 + 4(1) − 6

= 2 + 4 − 6= 0

y = 1 is a root of the quadratic equation, 2y2 + 4y – 6 = 0.

QUESTION 5

Question:Determine if p = 0 and p = 5 are the roots of the quadratic equation,

p2 – 5p = 0.

Solution: p2 – 5p = (0)2 − 5(0)

= 0 p2 – 5p = (5)2 − 5(5)

= 25 − 25= 0

Both p = 0 and p = 5 are the roots of the quadratic equation p2 – 5p = 0.



QUESTION 6

Question:Determine if x = 3 and x = − 3 are the roots of the quadratic equation,

9 – 9x – 4x2 = 0.

Solution: 9 – 9x – 4x2 = 9 − 9(3) − 4(3)2

= 9 − 27 − 36= − 54

9 – 9x – 4x2 = 9 − 9(−3) − 4(−3)2

= 9 + 27 − 36= 0

x = 3 is not the roots of the quadratic equation 9 – 9x – 4x2 = 0, while x = −3 is the root of thequadratic equation.



QUESTION 7Question:

014xequation quadratic theof roots theare 2

1 xand ,

2

1 xif Determine 2 =−−==

Solution:

0

1-1

1)4

14(

1)2

1(414 22

=

=

−=

−=−x

0

1-1

1)4

14(

1)2

1(414 22

=

=

−=

−−=−x

014x equation, quadratic theof roots theare 2

1 xand ,

2

1Both x 2 =−−==

QUESTION 8

Question:

0116yequation quadratic theof roots theare 4

1 y and ,

4

1y if Determine 2 =−−==

Solution:

0

1-1

1)16

116(

1)4

1(16116 22

=

=

−=

−=−y

0

1-1

1)16

116(

1)4

1(16116 22

=

=

−=

−−=−y

0116y equation, quadratic theof roots theare 4

1 y and ,

4

1yBoth 2 =−−==



QUESTION 9Question:

013sequation quadratic theof roots theare 3

1 s and ,

3

1s if Determine 2 =−−==

Solution:

3

2

13

1

1)9

13(

1)3

1(313 22

−=

−=

−=

−=−s

3

2

13

1

1)9

13(

1)3

1(313 22

−=

−=

−=

−−=−s

013s equation, quadratic theof roots not the are 3

1 s and ,

3

1sBoth 2 =−−==

QUESTION 10

Question:

019tequation quadratic theof roots theare 3

1 tand ,

3

1 tif Determine 2 =−−==

Solution:

0

1-1

1)9

19(

1)3

1(919 22

=

=

−=

−=−t

0

1-1

1)9

19(

1)3

1(919 22

=

=

−=

−−=−t

019t equation, quadratic theof roots theare 3

1 tand ,

3

1Both t 2 =−−==

Mathematics Form 4 Lesson 13: Solutions for quadratic equations by trial and error.

Learning Area: Determine the solutions for quadratic equations by trial and error.1 / 7

QUESTION 1Question:Determine the solution for h2 + 17h + 16 = 0 by trial and error method.

Solution:

QUESTION 2

Question:Determine the solution for a2 − 9a + 14 = 0 by trial and error method.

Solution:

QUESTION 3

Question:Determine the solution for z2 + 7n + 10 = 0 by trial and error method.

Solution:

QUESTION 4

Question:Determine the solution for p2 + 8p − 9 = 0 by trial and error method.

Solution: .

QUESTION 5

Question:Determine the solution for b2 + 6b + 8 = 0 by trial and error method.

Solution:




QUESTION 6Question:Determine the solution for 2x2 + 7x + 3 = 0 by trial and error method.

Solution:

QUESTION 7

Question:Determine the solution for 2y2 + 8y + 6 = 0 by trial and error method.

Solution:

QUESTION 8

Question:Determine the solution for 9x2 − 12x + 4 = 0 by trial and error method.

Solution:

QUESTION 9

Question:Determine the solution for 3 − n − 2n2 = 0 by trial and error method.

Solution:

QUESTION 10

Question:Determine the solution for 12q2 + 18q + 6 = 0 by trial and error method.

Solution:



QUESTION 1Question:Determine the solution for h2 + 17h + 16 = 0 by trial and error method.

Solution:h2 + 17h + 16 = (−16)2 + 17(−16) + 16

= 256 − 272 + 16= 0

h2 + 17h + 16 = (−1)2 + 17(−1) + 16= 1−17 + 16= 0

h = −1 or h = −16

QUESTION 2

Question:Determine the solution for a2 − 9a + 14 = 0 by trial and error method.

Solution:a2 − 9a + 14 = (2)2 − 9(2) + 14

= 4 − 18 + 14= 0

a2 − 9a + 14 = (7)2 − 9(7) + 14= 49 − 63 + 14= 0

a = 2 or a = 7




QUESTION 3

Question:Determine the solution for n2 + 7n + 10 = 0 by trial and error method.

Solution:n2 + 7n + 10 = (− 2)2 + 7(−2) + 10

= 4 − 14 + 10= 0

n2 + 7n + 10 = (−5)2 + 7(−5) + 10= 25 − 35 + 10= 0

n = −2 or n = −5

QUESTION 4

Question:Determine the solution for p2 + 8p − 9 = 0 by trial and error method.

Solution:p2 + 8p − 9 = (1)2 + 8(1) − 9

= 1 + 8 − 9= 0

p2 + 8p − 9 = (− 9)2 + 8(−9) − 9= 81 − 72 − 9= 0

p = 1 or p = −9



QUESTION 5

Question:Determine the solution for b2 + 6b + 8 = 0 by trial and error method.Solution:b2 + 6b + 8 = (− 2)2 + 6(−2) + 8

= 4 − 12 + 8= 0

b2 + 6b + 8 = (− 4)2 + 6(−4) + 8= 16 − 24 + 8= 0

b = −2 or b = −4

QUESTION 6

Question:Determine the solution for 2x2 + 7x + 3 = 0 by trial and error method.

Solution:2x2 + 7x + 3 = 2(−1/2)2 + 7(−1/2) +3

= 1/2 − 7/2 + 3= 0

2x2 + 7x + 3 = 2(−3)2 + 7(−3) +3= 18 − 21 + 3= 0

x = −1/2 or x = −3



QUESTION 7

Question:Determine the solution for 2y2 + 8y + 6 = 0 by trial and error method.

Solution:2y2 + 8y + 6 = 2(−1)2 + 8(−1) + 6

= 2 − 8 + 6= 0

2y2 + 8y + 6 = 2(−3)2 + 8(−3) + 6= 18 − 24 + 6= 0

y = −1 or y = −3

QUESTION 8

Question:Determine the solution for 9x2 − 12x + 4 = 0 by trial and error method.

Solution:9x2 − 12x + 4 = 9(2/3)2 − 12(2/3) + 4

= 4 − 8 + 4= 0

x = 2/3QUESTION 9

Question:Determine the solution for 3 − n − 2n2 = 0 by trial and error method.

Solution:3 − n − 2n2 = 3 − (1) − 2(1)2

= 3 − 1 − 2= 0

3 − n − 2n2 = 3 − (−3/2) − 2(_3/2)2

= 3 + 3/2 − 9/2= 0

n = 1 or n = −3/2

Question:Determine the solution for 12q2 + 18q + 6 = 0 by trial and error method.Solution:12q2 + 18q + 6 = 12(−1/2)2 + 18(−1/2) + 6

= 3 − 9 + 6= 0

12q2 + 18q + 6 = 12(−1)2 + 18(−1) + 6= 12 − 18 + 6= 0

q = −1/2 or −1

Mathematics Form 4 Lesson 14: Solutions for quadratic equations by factorisation.

Learning Area: Determine the solutions for quadratic equations by factorisation.1 / 6

QUESTION 1Question:Determine the solution for p2 − 5p + 6 = 0 by factorisation.

Solution:

QUESTION 2

Question:Determine the solution for 6m2 + 5m +1 = 0 by factorisation.

Solution:

QUESTION 3

Question:Determine the solution for 3m2 + 3m − 6 = 0.

Solution:

QUESTION 4

Question:Determine the solution for b2 + 3b − 10 = 0.

Solution: .

QUESTION 5

Question:Factorise j2 + 7j + 12 and determine the solution for j2 + 7j + 12 = 0.

Solution:




QUESTION 6Question:Factorise x2 + 5x + 6 and determine the solution for x2 + 5x + 6 = 0.

Solution:

QUESTION 7

Question:Determine the solution for w2 − 13w + 36 = 0 by factorisation.

Solution:

QUESTION 8

Question:Determine the solution for x2 − 4x − 5 = 0 by factorisation.

Solution:

QUESTION 9

Question:What is the solution for t2 − 10t + 25 = 0 by factorisation?

Solution:

QUESTION 10

Question:What is the solution for 12r2 − 12r + 3 = 0 by factorisation?

Solution:



QUESTION 1Question:Determine the solution for p2 − 5p − 6 = 0 by factorisation.

Solution:p2 − 5p − 6 = (p + 1)(p − 6)

p + 1 = 0p = −1

p − 6 = 0p = 6

p = −1 or p = 6

QUESTION 2

Question:Determine the solution for 6m2 + 5m +1 = 0 by factorisation.

Solution:6m2 + 5m +1 = (2m + 1)(3m + 1)

2m + 1 = 02m = −1

m = −1/23m + 1 = 0

3m = −1m = _1/3

m = −1/2 or m = −1/3

QUESTION 3

Question:Determine the solution for 3m2 + 3m − 6 = 0.

Solution:3m2 + 3m − 6 = 3(m2 + m − 2)

= 3(m − 1)(m + 2)m − 1 = 0

m = 1m + 2 = 0

m = −2m = 1or m = −2




QUESTION 4

Question:Determine the solution for b2 + 3b − 10 = 0.

Solution:b2 + 3b − 10 = (b + 5)(b − 2)

b + 5 = 0b = −5

b − 2 = 0b = 2

b = 2 or b = −5

QUESTION 5

Question:Factorise j2 + 7j + 12 and determine the solution for j2 + 7j + 12 = 0.

Solution:j2 + 8j + 12 = (j + 2)(j +6)

j + 2 = 0j = −2

j + 6 = 0j = −6

j = −2 or j = −6

QUESTION 6

Question:Factorise x2 + 5x + 6 and determine the solution for x2 + 5x + 6 = 0.

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

x + 3 = 0x = −3

x + 2 = 0x = −2

x = −3 or x = −2



QUESTION 7

Question:Determine the solution for w2 15w + 36 = 0 by factorisation.

Solution:w2 − 15w + 36 = (w − 3)(w − 12)

w − 3 = 0w = 3

w − 12 = 0w = 12

w = 3 or w = 12

QUESTION 8

Question:Determine the solution for x2 − 4x − 5 = 0 by factorisation.

Solution:x2 − 4x − 5 = (x − 5)(x + 1)

x − 5 = 0x = 5x + 1 = 0x = −1

x = 5 or x = −1QUESTION 9

Question:What is the solution for t2 − 10t + 25 = 0 by factorisation?

Solution:

t2 + t − 30 = (t − 5)(t + 6)t − 5 = 0

t = 5t + 6 = 0

t = −6t = 5 or t = −6

QUESTION 10

Question:What is the solution for 12r2 − 12r + 3 = 0 by factorisation?

Solution:12r2 − 12r + 3 = 3(4r2 − 4r +1)

= 3(2r − 1)(2r − 1)2r − 1 = 0

2r = 1r = 1/2

Mathematics Form 4 Lesson 15: Quadratic equations in everyday life

Learning Area: Solve problems involving quadratic equations. 1 / 6

QUESTION 1

Question:A rectangular shaped playing field has a length which is 13 m more than its width. If the area is30 m2, find the length and width of the field.

Solution:

QUESTION 2

Question:A piece of wood was used to form a square. If the area of the square is 25 m2, determine the lengthof one side of the square.

Solution:




QUESTION 3

Question:Ali owns a few houses in the rural area. He wishes to sell some of them to invest in one of hisinvestment plans. He offers to sell n number of houses for RM600 000 and if the number of housesincreases by 1, then the price for the price of each house will be reduce by RM50 000. Determinethe number of houses he offers for sale in the first offer.

Solution:

QUESTION 4

Question:Encik Ahmad arranged 48 flower pots in a few rows in his garden. He re-arranges the flower potsand found that if he adds 4 flower pots to every row he will have 1 row less than the firstarrangement. Determine the number of flower pots per row in the first arrangement.

Solution:



QUESTION 5

Question:A swimming pool has the length of 12 m and the width of 8 m. A fence is build at the distance of xmetre around it. If the area of the swimming pool is equal to the area of the land bounded by thefence, determine the value of x.

Solution:



QUESTION 1

Question:A rectangular shaped playing field has a length which is 13 m more than its width. If the area is 30m2, find the length and width of the room.

Solution:Formula Area = width x lengthLet the width = w, therefore the length = w + 13Substitute the given values into the formulaArea = width _ length30 = w x (w + 13)30 = w2 + 13wW2 + 13w − 30 = 0(w − 2)(w + 15) = 0w − 2 = 0 w + 15 = 0 w = 2 w = −15

Therefore we can say the width is equal to 2m and the length is equal to 15m.


QUESTION 2

Question:A piece of wood was used to form a square. If the area of the square is 25 m2, determine the lengthof one side of the square.

Solution:Formula Area of a square = length x widthLength = width = sTherefore,Area = s2

25 = s2

s =√25s = 5The length of one side of the square is 5m.



QUESTION 3

Question:Ali owns a few houses in the rural area. He wishes to sell some of them to invest in one of hisinvestment plans. He offers to sell n number of houses for RM600 000 and if the number of housesincreases by 1, then the price for the price of each house will be reduce by RM50 000. Determinethe number of houses he offers for sale in the first offer.

Solution:The difference of price of a house of the 2 offers is RM50000.Therefore,

50000n2 + 50000n – 600000

We can conclude that the number of houses offer for sale in the first offer is 3.

QUESTION 4

Question:Encik Ahmad arranged 48 flower pots in a few rows in his garden. He re-arranges the flower potsand found that if he adds 4 flower pots to every row he will have 1 row less than the firstarrangement. Determine the number of flower pots per row in the first arrangement.

Solution:The difference of rows of the two arrangements is 3.Therefore,

We can conclude that the number of flower pots per row in the first arrangement is 12.

500001

000 600000 600 =−+nn

50000)1()1()1(1

000 600000 600 +=+−++

nnnnnnnn

)1(50000600000)600000600000( +=−+ nnnn

nn 5000050000600000 2 +=

0)12(50000 2 =−+ nn

0)4)(3(50000 =+− nn

4or 3 −=n

14

4848 =−+nn

1)4()4()4(4

4848 +=+−++

nnnnnnnn

)4(48)19248( +=−+ nnnn

nn 4192 2 +=

019242 =−+ nn

0)16)(12( =+− nn16or 21 −=n



QUESTION 5

Question:A swimming pool has the length of 12 m and the width of 8 m. A fence is build at the distance of xmetre around it. If the area of the swimming pool is equal to the area of the land bounded by thefence, determine the value of x.

Solution:If the area of the swimming pool is equal to the area of the land bounded by the fence,(2x + 12)(2x +8) = 96(2)4x2 + 40x + 96 = 1924x2 + 40x − 96 = 04(x2 + 10x − 24) = 04(x − 2)(x + 12) = 0x − 2 = 0 x = −12Therefore x = 2 or x = −12.

Chapter 2 - Quadratic Expression and Equation - Exercise

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Transcript of Chapter 2 - Quadratic Expression and Equation - Exercise