Chapter Learning Area: Algebra
1 Functions
Fungsi
1.1 Functions / Fungsi
Smart Tip
Function involves two sets, domain and codomain. The elements in a domain are called objects. Images are found in
codomain. A function is a relation such that each object has only one image.
Fungsi melibatkan dua set, domain dan kodomain. Unsur dalam domain dikenali sebagai objek. Imej didapati daripada kodomain. Fungsi ialah
satu hubungan dengan keadaan setiap objek hanya mempunyai satu imej.
Exercise 1 Determine whether each of the following relations is a function or not. Give your reason.
Tentukan sama ada setiap hubungan yang berikut ialah fungsi atau bukan. Berikan alasan anda.
TP 1 Mempamerkan pengetahuan asas tentang fungsi.
Example 1 a 1 2 3 15
b 4 20
1 c 2 6 7 24
2 d 8 28
3 5 10
4 35
7
Solution This relation is not a function This relation is a function
This relation is a function because because there is object that has because each object has only
each object has only one image. more than one image. one image.
Hubungan ini ialah satu fungsi kerana
setiap objek mempunyai satu imej.
3 4 2 5 p 14
4 q 21
6 3 5 0 r 28
6 2
9 5
20 10
This relation is not a function This relation is not a function This relation is a function
because there is object that has because there is object that has because each object has only
more than one image. no image. one image.
Exercise 2 Determine whether each of the following graphs is a function or not. Give your reason.
Tentukan sama ada setiap graf yang berikut ialah fungsi atau bukan. Berikan alasan anda.
TP 1 Mempamerkan pengetahuan asas tentang fungsi.
Example 2 1 y 2 y
y Solution Ox Ox
The graph
is a function The graph is not a function The graph is not a function
because when tested with because when tested with
O x because when vertical line, there is more than vertical line, there is more than
tested with one point that cuts the graph. one point that cuts the graph.
vertical line, there is only one
point that cuts the graph.
Graf ini ialah satu fungsi kerana
apabila diuji dengan garis
mencancang, hanya wujud satu titik
sahaja yang memotong graf itu.
1
3 y 4y 5 y
Ox Ox Ox
The graph is a function The graph is not a function The graph is a function
because when tested with because when tested with because when tested with
vertical line, there is only one vertical line, there is more vertical line, there is only one
point that cuts the graph. than one point that cuts the point that cuts the graph.
graph.
Exercise 3 State the domain, codomain and the range of the following functions.
Nyatakan domain, kodomain dan julat bagi fungsi berikut.
TP 1 Mempamerkan pengetahuan asas tentang fungsi.
Example 3
x 4x Smart Tip
1• •4 1 Domain contains the objects only.
2• •6 Domain mengandungi objek sahaja.
3• •8
•10 2 Codomain contains the images and non-images.
Set P •12 Kodomain mengandungi imej serta bukan imej.
3 Range contains the images only.
Set Q
Julat mengandungi imej sahaja.
Solution Codomain/Kodomain = {4, 6, 8, 10, 12}
Domain = {1, 2, 3} Range/Julat = {4, 8, 12}
Domain = {1, 2, 3}
Common Error
{(1, 2), (–2, 5), (3, 8)}
Domain = (–2, 1, 3)
Codomain/Kodomain = (2, 5, 8)
Range/Julat = (2, 5, 8)
This is wrong because the set notation of domain, codomain and range is { }. The correct answer for domain is {–2, 1, 3}.
Salah kerana tatatanda set bagi domain, kodomain dan julat ialah { }. Jawapan yang betul bagi domain ialah {–2, 1, 3}.
1 • 49 2
7• • 30 4• •16
• 25 3• •12
5• • 24
–5 • Set Q 2• •9
Set P –3• •8
–4• •4
Domain = {–5, 5, 7} Set G Set H
Domain = {–4, –3, 2, 3, 4}
Codomain/Kodomain = {24, 25, 30, 49} Codomain/ Kodomain = {4, 8, 9, 12, 16}
Range/Julat = {25, 49} Range/Julat = {4, 9, 16}
2
3 {(1, a), (2, b), (3, a), (4, c)} 4 {(a, p), (b, p), (c, p), (d, q)}
Domain = {1, 2, 3, 4} Domain = {a, b, c, d}
Codomain/Kodomain = {a, b, c} Codomain/Kodomain = {p, q}
Range/Julat = {a, b, c} Range/Julat = {p, q}
5 Set Q 6 Set G
8 Set P 4 Set F
0 3
–1 0
–1 0 2 –1 0 2
Domain = {–1, 0, 2} Domain = {–1, 0, 2}
Codomain/Kodomain = {–1, 0, 8}
Range/Julat = {–1, 0, 8} Codomain/Kodomain = {0, 3, 4}
Range/Julat = {0, 3, 4}
Exercise 4 Find the image for each of the following functions.
Cari imej bagi setiap fungsi berikut.
TP 2 Mempamerkan kefahaman tentang fungsi.
Example 4 1 f : x → 3x + 2; x = 5 2 f:x→ 3x – 2 ;x=4
f(5) = 3(5) + 2 2
2x + 1 = 15 + 2
f:x→ 3x + 4 ;x=2 = 17 f(4) = 3(4) – 2
The image of 5 is 17. 2
Solution = 12 – 2
2
f(2) = 2(2) + 1 10
3(2) + 4 = 2
= 5 = 5
10
The image of 4 is 5.
= 1
2
21.
The image of 2 is
Imej bagi 2 ialah 12.
3 h : x → 6 + 8; x = –3 4 f : x → 2x2 + 3x – 4; x = –2 5 h:x→ x+5 ;x=3
x f(–2) = 2(–2)2 + 3(–2) – 4 x–1
= 2(4) – 6 – 4
h(–3) = 6 + 8 = 8 – 6 – 4 h(3) = (3) + 5
(–3) = –2 (3) – 1
The image of –2 is –2.
= –2 + 8 = 8
2
= 6
The image of –3 is 6. = 4
The image of 3 is 4.
3
Exercise 5 Solve each of the following.
Selesaikan setiap yang berikut.
TP 2 Mempamerkan kefahaman tentang fungsi.
Example 5 1 It is given that f : x → x + 3, 2 It is given that f : x → 3x – 10,
find the object when the image find the object when the image
It is given that f : x → 2x – 5, is 7. is 5.
find the object when the
image is 1. Diberi bahawa f : x → x + 3, cari Diberi bahawa f : x → 3x – 10, cari
Diberi bahawa f : x → 2x – 5, cari objek apabila imej ialah 7. objek apabila imej ialah 5.
objek apabila imej ialah 1.
f(x) = x + 3 f(x) = 3x – 10
Solution 7 = x + 3 5 = 3x – 10
f(x) = 2x – 5 7 – 3 = x 5 + 10 = 3x
1 = 2x – 5 4 = x 15 = 3x
1 + 5 = 2x 5 = x
6 = 2x The object is 4.
3=x The object is 5.
The object is 3.
Objek ialah 3.
3 It is given that f : x → x + 4, 4 It is given that f : x → x + 8, 5 It is given that f : x → x+ 1 ,
2 x x– 1
find the object when the image x ≠ 0, find the object when the x ≠ 1, find the object when the
is 6.
image is 3. x + 8 , image is 5. x + 1
Diberi bahawa f : x → x Diberi bahawa f : x → x – 1 , x ≠ 1,
x + 4
Diberi bahawa f : x → 2 , x ≠ 0, cari objek apabila imej ialah 3. cari objek apabila imej ialah 5.
cari objek apabila imej ialah 6. f(x) = x + 8 f(x) = x+ 1
x x– 1
f(x) = x + 4
2 x+1
3 = x + 8 5 = x–1
x
6 = x + 4 5(x – 1) = x + 1
2
3x = x + 8 5x – 5 = x + 1
12 = x + 4 3x – x = 8 5x – x = 1 + 5
12 – 4 = x 2x = 8 4x = 6
8 = x x = 4 6
4
The object is 8. x =
The object is 4. 3
2
x =
The object is = 32.
6 The diagram below shows the relation between set P and set Q. (a) Using the function notation,
Rajah di bawah menunjukkan hubungan antara set P dan set Q. express f in terms of x.
Set P f(x) Set Q Menggunakan tatatanda fungsi,
1• •1 ungkapkan f dalam sebutan x.
(b) Find the value of h.
5 • • 0.2
Cari nilai h.
h • • 0.1 (c) If set P represents all the whole
20 • • 0.05 numbers, does every object has its
image? HOTS Analysing
Jika set P mewakili semua nombor bulat,
adakah setiap objek mempunyai imejnya?
(a) x = 1; 1 = 1 (b) f(h) = 1
(1) h
x = 5; 1 = 0.2 0.1 = 1
(5) 10
x = 20; 1 = 0.05 ∴ h = 10
(20)
∴ f(x) = 1 , x ≠ 0 (c) No. Image of 0 is not defined.
x
4
Exercise 6 Find the values of x for each of the following.
Cari nilai-nilai x bagi setiap fungsi berikut.
TP 3 Mengaplikasikan kefahaman tentang fungsi untuk melaksanakan tugasan mudah.
Example 6 1 It is given that f(x) = |3x|, find the object when
It is given that f(x) = |2x – 3|, find the object f(x) = 12.
when f(x) = 7.
Diberi bahawa f(x) = |2x – 3|, cari objek apabila f(x) = 7. Diberi bahawa f(x) = |3x|, cari objek apabila f(x) = 12.
|3x| = 12
Solution 3x = –12 or 3x = 12
|2x – 3| = 7 x = –4 x = 4
2x – 3 = –7 or/atau 2x – 3 = 7
2x = 7 + 3
2x = –7 + 3 2x = 10
x = 5
2x = –4
x = –2
2 It is given that f(x) = |x – 2|, find the object when 3 It is given that f(x) = |2x + 1|, find the object
f(x) = 3. when f(x) = 5.
Diberi bahawa f(x) = |x – 2|, cari objek apabila f(x) = 3. Diberi bahawa f(x) = |2x + 1|, cari objek apabila f(x) = 5.
|x – 2| = 3 |2x + 1| = 5
x – 2 = –3 or x – 2 = 3
x = –3 + 2 x = 3 + 2 2x + 1 = –5 or 2x + 1 = 5
x = –1 x = 5 2x = 5 – 1
2x = –5 – 1 2x = 4
x = 2
2x = –6
x = –3
4 It is given that f(x) = |2x – 5|, find the object 5 It is given that f(x) = 3x – 1 , find the object
when f(x) = 9. when f(x) = 6. 2
Diberi bahawa f(x) = |2x – 5|, cari objek apabila f(x) = 9. Diberi bahawa f(x) = 3x – 1 , cari objek apabila f(x) = 6.
2
|2x – 5| = 9 3x – 1
2
2x – 5 = –9 or 2x – 5 = 9 =6
2x = –9 + 5 2x = 9 + 5 3x – 1 3x – 1
2 2
2x = –4 2x = 14 = –6 or = 6
x = –2 x = 7 3x – 1 = –12 3x – 1 = 12
3x = –12 + 1 3x = 12 + 1
3x = –11 3x = 13
x = – 131 x = 13
3
5
Exercise 7 Complete the table and sketch the graph for the domain given. Then, find the range of the
function.
Lengkapkan jadual dan lakarkan graf bagi domain yang diberi. Kemudian, cari julat bagi fungsi itu.
TP 3 Mengaplikasikan kefahaman tentang fungsi untuk melaksanakan tugasan mudah.
Example 7 1 f(x) = |x2 – 4|; – 2 x 3 2 f(x) = |x + 1|; –2 x 2
f(x) = |2x – 3|; –1 x 3 x –2 0 2 3 x –2 –1 2
Solution f(x) 0 4 0 5 f(x) 1 0 3
x –1 3 3 f(x)
f(x) 5 2 3
f(x)
0 Range 5
Julat 4
3
x
f(x) –3 –2 –1 0 1 2 3 1
5 –2 –1 0
x
12
3
Range/Julat: 0 f(x) 5 Range/Julat: 0 f(x) 3
–1 0 x
33
2
Range/Julat: 0 f(x) 5
3 h(x) = |2x + 3|; –2 x 1 4 h(x) = |x2 – 1|; –2 x 2 h(x)
x –2 –1 0 1 2
5 f(x) = |x32 – 4|; – 2 x 3
h(x) 3 0 1 0 3
x –2 –23 0 1 x –2 023
h(x) 1 0 3 5 405
f(x) 1
x
–2 –1 0 12
0
h(x)
5
3 h(x) f(x)
3 5
4
1 1
–2 –1 0
–2 3 –1 0 1 x x x
2 23
– 12 –2 0
Range/Julat: 0 h(x) 5 Range/Julat: 0 h(x) 3 Range/Julat: 0 f(x) 5
f(x)
5
4
–2 0 x
6 23
1.2 Composite Functions / Fungsi Gubahan
Exercise 8 The diagrams below show the relationship between set P, set Q and set R. Complete the
function in the box provided.
Rajah di bawah menunjukkan hubungan antara set P, set Q dan set R. Lengkapkan fungsi dalam petak yang
disediakan.
TP 2 Mempamerkan kefahaman tentang fungsi.
a Example 8 1 2
gf
fg gf
P Q R P Q R P Q R
Solution fg f
gf
3 4 5
fg gf fg
P Q R P Q R P Q R
f gg
Exercise 9 Complete each of the following.
Lengkapkan setiap yang berikut.
TP 3 Mengaplikasikan kefahaman tentang fungsi untuk melaksanakan tugasan mudah.
Example 9
g(x) = 5x – 9 Solution (b) hg(x) = h(5x – 9) (c) gh(–2) = 5(–2) + 6
h(x) = x + 3 (a) gh(x) = g(x + 3) = (5x – 9) + 3
Find/Cari = 5(x + 3) – 9 = 5x – 9 + 3 = –10 + 6
(a) gh(x) = 5x + 15 – 9 = 5x – 6
(b) hg(x) = 5x + 6 = –4
(c) gh(–2)
1 g(x) = x – 8 (a) gh(x) = g(2x + 1) (b) hg(x) = h(x – 8) (c) gh(5) = 2(5) – 7
h(x) = 2x + 1 = (2x + 1) – 8 = 2(x – 8) + 1 = 10 – 7
= 2x + 1 – 8 = 2x – 16 + 1 = 3
= 2x – 7 = 2x – 15
2 g(x) = x + 2 (a) g2(x) = g(x + 2) (b) hg(x) = h(x + 2) (c) hg(2) = 3(2) – 1
h(x) = 3x – 7 = (x + 2) + 2 = 3(x + 2) – 7 = 6 – 1
= x + 4 = 3x + 6 – 7 = 5
= 3x – 1
7
3 g(x) = 3x ( )(a) gh(x) = g ( )(b) (3) + 6
6 h2(x) = h 6 (c) h2(3) = (3) + 7
6 x+6 x+6
+ ( ) 9
h(x) = x 6 6 6 = 10
=3 x+6 ( ) =
6
= 18 x+6 +6
x+6
6
( ) =
6 + 6x + 36
x+6
6(x + 6)
= 6x + 42
= 6(x + 6)
6(x + 7)
x + 6
= x + 7
4 g(x) = x2 + 1 (a) gh(x) = g(x – 3) (b) hg(x) = h(x2 + 1) (c) hg(–4) = (–4)2 – 2
h(x) = x – 3 = (x – 3)2 + 1 = (x2 + 1) – 3 = 16 – 2
= (x2 – 6x + 9) + 1 = x2 – 2 = 14
= x2 – 6x + 10
5 g( ) = x + 4 (a) g2(x) = g(x + 4) (b) hg(x) (c) hg(−2)
h(x) = 2x2 – 9 = (x + 4) + 4 = h(x + 4) = 2(−2)2 + 16(−2) + 23
= x + 8 = 2(x + 4)2 – 9 = 2(4) − 32 + 23
= 2(x2 + 8x + 16) – 9 = −1
= 2x2 + 16x + 32 – 9
= 2x2 + 16x + 23
Exercise 10 Find the image for each of the following composite functions.
Cari imej bagi setiap fungsi gubahan berikut.
TP 3 Mengaplikasikan kefahaman tentang fungsi untuk melaksankan tugasan mudah.
Example 10 1 Given f(x) = 2x and 2 Given f(x) = 6x and
g(x) = x – 3. Find fg(7). g(x) = x – 2. Find gf(5).
Given f(x) = 3x and g(x) = x + 5. Diberi f(x) = 2x dan Diberi f(x) = 6x dan
Find fg(4). g(x) = x – 3. Cari fg(7). g(x) = x – 2. Cari gf(5).
Diberi f(x) = 3x dan g(x) = x + 5.
Cari fg(4). g(7) = (7) – 3 f(5) = 6(5)
=4 = 30
Solution
g(4) = (4) + 5 fg(7) = f(4) gf(5) = g(30)
= 2(4) = 30 – 2
=9 =8 = 28
fg(4) = f(9)
= 3(9)
= 27
3 Given f(x) = x + 4 and 4 Given g(x) = x2 and 5 Given g(x) = 2x2 and
g(x) = 2x – 5. Find gf(3). h(x) = x + 3. Find hg(3). h(x) = 3x + 1. Find gh(4).
Diberi f(x) = x + 4 dan Diberi g(x) = x2 dan Diberi g(x) = 2x2 dan
g(x) = 2x – 5. Cari gf(3). h(x) = x + 3. Cari hg(3). h(x) = 3x + 1. Cari gh(4).
g(3) = (3)2
f(3) = (3) + 4 h(4) = 3(4) + 1
=7 =9 = 13
hg(3) = h(9)
gf(3) = g(7) gh(4) = g(13)
= 2(7) – 5 = (9) + 3 = 2(13)2
=9 = 12 = 338
8
Exercise 11 Find the value of x for each of the following composite functions.
Cari nilai x bagi setiap fungsi gubahan berikut.
TP 3 Mengaplikasikan kefahaman tentang fungsi untuk melaksanakan tugasan mudah.
Example 11 1 g(x) = x + 3 2 f(x) = 2x
h(x) = 2x + 4 g(x) = x − 5
f(x) = 3x + 1 gh(x) = 5 fg(x) = 6
g(x) = 2x − 1
fg(x) = 1 gh(x) = g(2x + 4) fg(x) = f(x − 5)
= (2x + 4) + 3 = 2(x − 5)
Solution = 2x + 7 = 2x − 10
fg(x) = f(2x – 1)
= 3(2x − 1) + 1 Given gh(x) = 5 Given fg(x) = 6
= 6x − 3 + 1 2x + 7 = 5
= 6x − 2 2x = 5 − 7 2x − 10 = 6
2x = −2
Given/Diberi fg(x) = 1 x = −1 2x = 6 + 10
6x − 2 = 1 2x = 16
6x = 1 + 2 x = 8
6x = 3
3 1
x = 6 = 2
3 f(x) = 4 , x ≠ 0 4 f(x) = 12 − x 5 h(x) = 6x − 7
x 2 h2(x) = 23
g(x) = 2x + 1
f 2(x) = 4
fg(x) = 15 h2(x) = h(6x − 7)
( ) = 6(6x − 7) − 7
f 2(x) = f 12 − x = 36x − 42 – 7
2 = 36x − 49
fg(x) = f(2x + 1)
( )12 − 12 − x Given h2(x) = 23
= 4 1)
(2x + = 2 36x − 49 = 23
2 36x = 72
x = 2
24 − 12 + x
Given fg(x) = 15 ( )
2
4 =
2x + 1 = 15 2
4 = 15(2x + 1) = 12 + x 1
4 = 30x + 15 2 2
–15 + 4 = 30x
–11 = 30x = 12 + x
4
x=– 11 Given f 2(x) = 4
30
12 + x = 4
4
12 + x = 16
x = 4
Exercise 12 Find the function f(x) for each of the following.
Cari fungsi f(x) bagi setiap yang berikut.
TP 3 Mengaplikasikan kefahaman tentang fungsi untuk melaksanakan tugasan mudah.
Example 12 1 g(x) = x − 5 2 g(x) = x + 8
fg(x) = x − 2 fg(x) = x − 6
g(x) = x + 6
fg(x) = 2x + 16 f[x – 5] = x – 2 f[x + 8] = x – 6
Let y = x – 5 Let y = x + 8
Solution
f [x + 6] = 2x + 16 x=y+5 x=y–8
Let/Biarkan y = x + 6 f(y) = x – 2 f(y) = x – 6
x=y–6 = (y + 5) – 2 = (y – 8) – 6
f(y) = 2x + 16 =y+3 = y – 14
f(x) = x + 3 f(x) = x – 14
= 2(y – 6) + 16
= 2y – 12 + 16
= 2y + 4
f(x) = 2x + 4
9
3 g(x) = 2x − 4 4 g(x) = x + 6 5 g(x) = 3x − 5
fg(x) = 2x + 1 fg(x) = 3x + 17 fg(x) = 6x − 9
f[2x – 4] = 2x + 1 f[x + 6] = 3x + 17 f[3x – 5] = 6x – 9
Let y = x + 6
Let y = 2x – 4 Let y = 3x – 5
x=y–6
2x = y + 4 f(y) = 3x + 17 3x = y + 5
y+5
x= y+4 = 3(y – 6) + 17 x= 3
2 = 3y – 1
f(x) = 3x – 1
f(y) = 2x + 1 f(y) = 6x – 9
( )= 2y+4 +1 ( )= 6y+5 –9
2 3
=y+5 = 2y + 1
f(x) = x + 5 f(x) = 2x + 1
Exercise 13 Find the function g(x) for each of the following.
Cari fungsi g(x) bagi setiap yang berikut.
TP 3 Mengaplikasikan kefahaman tentang fungsi untuk melaksanakan tugasan mudah.
Example 13 1 f(x) = x + 3 2 f(x) = x − 6
fg(x) = x − 8 fg(x) = x + 9
f(x) = 3x + 2
fg(x)= 3x − 19 f(x) = x + 3 f(x) = x − 6
then fg(x) = g(x) + 3 then fg(x) = g(x) − 6
Solution
f(x) = 3x + 2 Given Given
then/maka fg(x) = x − 8 fg(x) = x + 9
fg(x) = 3g(x) + 2
Given/Diberi g(x) + 3 = x − 8 g(x) − 6 = x + 9
fg(x) = 3x − 19 g(x) = x − 8 − 3 g(x) = x + 9 + 6
3g(x) + 2 = 3x − 19 = x − 11 = x + 15
3g(x) = 3x − 19 − 2
= 3x − 21
= 3(x − 7)
∴ g(x) = x − 7
3 f(x) = x − 12 4 f(x) = 3x − 7 5 f(x) = 2x − 3
fg(x) = 3x − 4 fg(x) = 3x + 20 fg(x) = 10x + 5
f(x) = x − 12 f(x) = 3x − 7 f(x) = 2x − 3
then fg(x) = g(x) − 12 then fg(x) = 3g(x) − 7 then fg(x) = 2g(x) − 3
Given Given Given
fg(x) = 3x − 4 fg(x) = 3x + 20 fg(x) = 10x + 5
g(x) − 12 = 3x − 4 3g(x) − 7 = 3x + 20 2g(x) − 3 = 10x + 5
g(x) = 3x − 4 + 12
= 3x + 8 3g(x) = 3x + 20 + 7 2g(x) = 10x + 5 + 3
= 3x + 27 = 10x + 8
g(x) = 3(x + 9) g(x) = 2(5x + 4)
3 2
= x + 9 = 5x + 4
10
Exercise 14 Solve each of the following.
Selesaikan setiap yang berikut.
TP 3 Mengaplikasikan kefahaman tentang fungsi untuk melaksanakan tugasan mudah.
Example 14 1 f : x → 14 − 3x 2 f : x → 2x − 7
g : x → 2 − 4x g : x → 3x
f : x → 2x + k fg : x → hx + k gf : x → ax + b
g:x→x−4
fg : x → mx + 6 Find the value of h and of k. Find the value of a and of b.
Cari nilai h dan nilai k. Cari nilai a dan nilai b.
Find the value of k and of m.
Cari nilai k dan nilai m. fg(x) = f(2 − 4x) gf(x) = g(2x − 7)
= 14 − 3(2 − 4x) = 3(2x − 7)
Solution = 14 − 6 + 12x = 6x − 21 PAK-21
fg(x) = f(x − 4) = 8 + 12x
= 2(x − 4) + k = 12x + 8 Compare QR CODE
= 2x − 8 + k gf(x) = ax + b
Compare gf(x) = 6x − 21
Compare/Bandingkan fg(x) = hx + k ∴ a = 6
fg(x) = 2x – 8 + k fg(x) = 12x + 8
∴ h = 12 b = −21
fg(x) = mx + 6
− 8 + k = 6 , m=2 k=8
k = 14
∴ m = 2, k = 14
3 f : x → 8x + k 4 f : x → 6 − 4x 5 f : x → 2x + h
g:x→x–4 g : x → ax + b g : x → 3x − 9
fg : x → mx − 18 gf : x → 2 − 12x gf : x → kx + 12
Find the value of k and of m. Find the value of a and of b. Find the value of h and of k.
Cari nilai k dan nilai m. Cari nilai a dan nilai b. Cari nilai h dan nilai k.
fg(x) = f(x − 4) gf(x) = g(6 − 4x) gf(x) = g(2x + h)
= 8(x − 4) + k = a(6 − 4x) + b = 3(2x + h) − 9
= 8x − 32 + k = 6a − 4ax + b = 6x + 3h − 9
= 6a + b − 4ax
Compare Compare
fg(x) = mx − 18 Compare gf(x) = kx + 12
fg(x) = 8x − 32 + k gf(x) = 2 − 12x gf(x) = 6x + 3h − 9
m = 8; −32 + k = −18 gf(x) = 6a + b − 4ax k = 6; 3h − 9 = 12
k = −18 + 32 −4a = −12; 6a + b = 2 3h = 12 + 9
= 14 a = 3 6(3) + b = 2 3h = 21
∴ m = 8, k = 14 18 + b = 2 h = 7
b = 2 − 18 ∴ h = 7, k = 6
= −16
∴ a = 3, b = −16
1.3 Inverse Functions / Fungsi Songsang
Exercise 15 Find the inverse function for each of the following functions.
Cari fungsi songsang bagi setiap fungsi berikut.
TP 3 Mengaplikasikan kefahaman tentang fungsi untuk melaksankan tugasan mudah.
Example 15
f(x) = 2x + 3 , x ≠ 0 From/Daripada , QR CODE
x
f –1(y) = 3 Scan or visit https://
Solution y−2 www.purplemath.com/
3 modules/invrsfcn3.htm
( )x = f –1 − for additional notes on
2x + 3 ∴ f –1(x) = x 2 , x ≠ 2 the inverse functions.
x
2x + 3
Let/Biarkan y = x
Common Error
xy = 2x + 3
3
xy − 2x = 3 f –1(x) = x – 2 is incomplete./adalah tidak lengkap.
The condition x ≠ 2 must be written in the answer.
x(y − 2) = 3 Syarat x ≠ 2 mesti ditulis dalam jawapan.
3
x = y−2
11
1.41 f(x) = 3x + 2 2 f(x) = 3x − 5 3 f(x) = 6 − 4x
f(x) = 3x + 2 f(x) = 3x − 5 f(x) = 6 − 4x
x = f −1(3x + 2) x = f –1(3x − 5) x = f –1(6 − 4x)
Let y = 3x + 2 Let y = 3x − 5 Let y = 6 − 4x
y − 2 = 3x y + 5 = 3x 4x = 6 − y
y − 2 = x y+5 =x x = 6−y
3 3 4
Hence, y−2 Hence, y+5 Hence,
3 3
f −1(y) = f –1(y) = f –1(y) = 6−y
4
x−2 x+5 6−x
f –1(x) = 3 f –1(x) = 3 f –1(x) = 4
4 f(x) = 5 − x 5 f(x) = x − 6 ,x≠0 6 f(x) = 7 + 7x
3 x
f(x) = 7 + 7x
f(x) = 5 − x f(x) = x − 6 x = f –1 (7 + 7x)
3 x
( ) x − 6
( ) x = f –1 5 − x x = f –1 x Let y = 7 + 7x
3
x − 6
Let y = 5 − x Let y = x y − 7 = 7x
3
xy = x − 6 y − 7 = x
7
3y = 5 − x 6 = x − xy
x = 5 − 3y 6 = x(1 − y) Hence,
Hence, 6 = x f –1(y) = y − 7
f –1(y) = 5 − 3y − 7
f –1(x) = 5 − 3x 1 y x − 7
7
Hence, f –1(x) =
f –1(y) = 1 6 y
−
6
f –1(x) = 1 − x , x≠ 1
Exercise 16 Find the values of p and q for each of the following. HOTS Applying
Cari nilai p dan q bagi setiap yang berikut.
TP 3 Mengaplikasikan kefahaman tentang fungsi untuk melaksanakan tugasan mudah.
Example 16
f(x) = 4x + q
f –1(x) = px + 2
Solution Compare/Bandingkan
f(x) = 4x + q
x = f –1(4x + q) f –1(x) = x − q
4 4
Let/Biarkan y = 4x + q
f –1(x) = px + 2
y−q
Then/Maka x = 4 ∴ 1 = p and/dan − q = 2
4 4
f –1(y) = y−q 1
4 p= 4 q = −8
x−q
f –1(x) = 4
= x − q
4 4
12
1 f(x) = 2x − 14 2 f(x) = 6x + q 3 f(x) = px + 6
f –1(x) = px + q f –1(x) = px − 12 f –1(x) = 4x + q
f(x) = 2x − 14 f(x) = 6x + q f(x) = px + 6
x = f –1(2x − 14) x = f –1(6x + q) x = f –1(px + 6)
Let y = 2x − 14 Let y = 6x + q Let y = px + 6
Then x = y + 14 Then x = y−q Then x = y−6
2 6 p
y + 14 y− q
f –1(y) = 2 f –1(y) = 6 f –1(y) = y− 6
p
x + 14 x− q x− 6
f –1(x) = 2 f –1(x) = 6 f –1(x) = p
= x +7 = x − q = x − 6
2 6 6 p p
Compare with Compare with Compare with
f –1(x) = px + q f –1(x) = px − 12 f –1(x) = 4x + q
∴ p = 1 ∴p= 1 ∴ 1 = 4
2 6 p
and and p = 1
q = 7 4
q
− 6 = −12 and
q = 72 − 6 = q
p
( ) − 6 = q
1
4
q = −6 4
= −24
4 f(x) = px + q 5 f(x) = px − 18 6 f(x) = 2p − 3x
f –1(x) = 2x − 6 f –1(x) = 3x + q q
f –1(x) =1 − 3 x
f(x) = px + q f(x) = px − 18 f(x) = 2p − 3x
x = f –1(px + q) x = f –1(px − 18) x = f –1(2p − 3x)
Let y = px + q Let y = px − 18 Let y = 2p − 3x
Then x = y−q Then x = y + 18 Then x = 2p − y
p p 3
f –1(y) = y− q y + 18 2p − y
p p 3
f –1(y) = f –1(y) =
f –1(x) = x− q x + 18 2p − x
p p 3
x q f –1(x) = f –1(x) =
p p
= − = x + 18 Compare with
p p
Compare with Compare with f –1(x) =1 − q x
f –1(x) = 3x + q 3
f –1(x) = 2x − 6 = 3 −3qx
∴ 1 = 2 ∴ 1 = 3
p p
p = 1 1 ∴ 2p = 3
2 3
p = 3
2
and and p =
− q = –6 18 = q and
p p –q = –1
q q = 1
p = 6 18
1
q = 6p ( ) = q
( ) 1 3
q = 6 2 q = 54
= 3
13
Exercise 17 Solve each of the following. HOTS Evaluating
Selesaikan setiap yang berikut.
TP 3 Mengaplikasikan kefahaman tentang fungsi untuk melaksanakan tugasan mudah.
TP 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang fungsi dalam konteks penyelesaian masalah rutin yang mudah.
1 Given h(x) = − 6 , find 2 It is given that g(x) = 7 − 3x.
x Diberi bahawa g(x) = 7 – 3x.
Diberi h(x) = − 6 , cari (a) Find/Cari
x (i) g2(x) and/dan g–1(x).
(ii) (g2)–1.
(a) h2(x), (b) Determine the relationship between (g2)–1
(b) h8(x),
(c) h17(x). and (g–1)2.
6 Tentukan hubungan antara (g2)–1 dan (g–1)2.
( ) (a) h2(x) = − ( ) (a) (i)
− 6 g2(x) = 7 − 3(7 − 3x) 7− 7−x
x 3
x = 97x−−2114+ 9x (b) (g–1)2 =
6 = 3
=6×
Let y = g(x) ( )21 − 7 + x
=x 3
y = 7 − 3x = 3
( ) (b) h3(x) = 6 6 3x = 7 − y
− x =− x
7 − y x + 14
6 x = 3 = 9
( ) h4(x) = − =x
6 7 − x ∴ (g2)–1 = (g–1)2
− x g–1(x) = 3
h8(x) = h4(h4(x)) (ii) Let y =g2(x)
=x y = 9x − 14
9x = y + 14
(c) h17(x) = h16(h(x)) x = y + 14
9
( ) = − 6 x + 14
x 9
∴ (g2)–1 =
6
= − x
3 Mr Lim is a fried noodles hawker. The daily profit that he can obtain, in RM, is given by w : x → 3x − 118 ,
where x is the number of packets of fried noodles sold in a day. 4
Encik Lim ialah seorang penjaja mi goreng. Keuntungan harian yang dapat diperolehnya, dalam RM, diberi oleh
w : x → 3x –4118, dengan keadaan x ialah bilangan peket mi goreng yang dijual dalam sehari.
(a) Calculate the average daily profit obtained by Mr Lim if he sold 1 197 packets of fried noodles in a
week.
Hitung keuntungan harian yang diperoleh Encik Lim jika dia menjual 1 197 bungkusan mi goreng dalam seminggu.
(b) Find the minimum number of packets of fried noodles that must be sold in a day so that Mr Lim does
not experience any loss.
Cari bilangan minimum peket mi goreng yang perlu dijual dalam sehari supaya Encik Lim tidak mengalami sebarang
kerugian.
(a) 1 week : 1 197 packets of fried noodles
1 day : 171 packets of fried noodles
w(x) = 3x − 118
4
= 3(171) − 118
4
= 98.75
∴ The daily profit = RM98.75
(b) 3x − 118 >0
4
3x − 118 > 0
3x > 118
x > 39.33
x = 40
The minimum number of packets of fried noodles that must be sold is 40 packets.
14
Review 1
Paper 1 Questions
1 Given h : x → 2mx − n, g : x → 6x − 5 and 8 It is given that f(x) = 6x and g(x) = 3h + kx. Express
SPM hg : x → 12mx + n. Express m in terms of n. SPM h in terms of k such that gf(2) = 15. h
Diberi h : x → 2mx − n, g : x → 6x − 5 dan hg : x → 12mx + n. Diberi bahawa f(x) = 6x dan g(x) = 3h + kx. Ungkapkan
CL `O16NE CL `O15NE
P1Q12 Ungkapkan m dalam sebutan n. P1Q2 dalam sebutan k dengan keadaan gf(2) = 15.
[3 marks/markah]
[3 marks/markah]
2 Given the function f : x → 4x − 3, find 9 The inverse function f is defined by
Diberi fungsi f : x → 4x − 3, cari
SPM Fungsi songsang f ditakrifkan kepada
(a) f(2), CLONE
(b) the value of k when f –1(k) = 5. f –1: x → 4 6 x , x ≠ 4.
nilai bagi k apabila f –1(k) = 5. `11 –
[3 marks/markah] P1Q3
Find/Cari
(a) f(x),
3 Diagram 1 shows the graph of the function (b) the value of x such that f(x) = –12.
CSL `OP17MNE fR:ajxah→1 |3 − 2x| for the domain −2 <x < 6. nilai bagi x dengan keadaan f(x) = –12.
menunjukkan graf bagi fungsi f : x→ |3 − 2x|
[4 marks/markah]
P1Q9 untuk domain −2 < x < 6. 10 Given the function f(x) = –ax + b, where a and b
are constants. Find the values of a and b such that
f(x) f –1(7) = 4 and f –1(–3) = 9.
9 Diberi fungsi f(x) = –ax + b, dengan keadaan a dan b ialah pemalar.
(–2, 7) Cari nilai a dan b dengan keadaan f –1(7) = 4 dan f –1(–3) = 9.
[4 marks/markah]
O 6x 11 Diagram 2 shows the function f : x → x2 + x + 4.
Rajah 2 menunjukkan fungsi f : x → x2 + x + 4.
Diagram 1/ Rajah 1 f
x x2 + x + 4
State/ Nyatakan
(a) the object of 9, (b) the image of 4, 3
16
objek bagi 9, imej bagi 4,
h
(c) the domain of 0 < f(x) < 7.
domain bagi 0 < f(x) < 7. [3 marks/markah] Diagram 2/ Rajah 2
4 Given the function h : x → 2x − 10, find Find the value of h.
Cari nilai h.
S PM Diberi fungsi h : x → 2x − 10, cari [3 marks/markah]
CLONE
`17 (a) h−1(x), 12 Diagram 3 shows the relation between set P, set Q
( )P1Q10 5m
(b) the value of m such that h2 2 = 20. SPM and set R.
Rajah 3 menunjukkan hubungan antara set P, set Q dan set R.
( ) 5m CL` O18NE
nilai bagi m dengan keadaan h2 2 = 20. P1Q22
[4 marks/markah] P
5 The function f : x → x – 7 and g : x → x 3, x ≠ 3 . Q fg : x → x2 + 6x + 7
SPM Find the value of gf(9). 4x − 4 R
CLONE
P` 11Q03 x 3
Fungsi f : x → x – 7 dan g : x → 4x – 3 , x ≠ 4 . Cari nilai
bagi gf(9). Diagram 3/ Rajah 3
[3 marks/markah] It is given that set P maps to set Q by the function
6 Given the functions g(x) = 6x – 9 and h(x) = 3x. x+3 and maps to set R by fg: x → x2 + 6x + 7.
2
SPM Find the value of gh(4). Diberi bahawa set P dipetakan kepada set Q oleh fungsi
CLONE
`11 Diberi fungsi g(x) = 6x – 9 dan h(x) = 3x.
Cari nilai bagi gh(4). x+3 dan dipetakan kepada set R oleh fg: x → x2 + 6x + 7.
P1Q2 2
[2 marks/markah]
(a) Write down the function which maps set P to
7 It is given that g(x) = x+4 , x ≠ 5 . Find the value set Q by using the function notation.
2x – 5 2
3 Tuliskan fungsi yang memetakan set P kepada set Q
11
of g–1(–3). dengan menggunakan tatatanda fungsi.
Diberi bahawa g(x) = x+4 , x ≠ 5 . Cari nilai 3 g–1(–3). (b) Find the function which maps set Q to set R.
2x – 5 2 11
Cari fungsi yang memetakan set Q kepada set R.
[3 marks/markah] [4 marks/markah]
15
Paper 2 Questions
1 Diagram 1 shows the function g maps set P to set Q and the function h maps set Q to set R.
SPM Rajah 1 menunjukkan fungsi g memetakan set P kepada set Q dan fungsi h memetakan set Q kepada set R.
CLONE
`14
P2Q3 g h R
P Q
x 2x + 5 10x – 3
Diagram 1/ Rajah 1
(a) Express in terms of x, the function
Ungkapkan dalam sebutan x, fungsi
(i) which maps set Q to set P,
yang memetakan set Q kepada set P,
(ii) h(x).
(b) Find the value of x such that gh(x) = 5x + 9. [5 marks/markah]
Cari nilai x dengan keadaan gh(x) = 5x + 9. [2 marks/markah]
2 It is given that f : x → 2x – 4 and g : x → 2 – 3x. [5 marks/markah]
SPM Diberi bahawa f : x → 2x – 4 dan g : x → 2 – 3x. [3 marks/markah]
CLONE
`18 (a) Find/Cari
P2Q2 (i) g(6),
(ii) the value of p if 3f()p=+123g)(=6),12 g(6),
nilai p jika f(p +
(iii) gf(x).
(b) Hence, sketch the graph of y = |gf(x)| for –1 ≤ x ≤ 5.
Seterusnya, lakarkan graf bagi y = |gf(x)| untuk –1 ≤ x ≤ 5.
H OTS Zone
1 Given fg : x → 6x + 5 and g : x → 2x + 1. Find f(x).
Diberi fg : x → 6x + 5 dan g : x → 2x + 1. Cari f(x).
2 Given fg(x) = 4x + 3 and g : x → x 3 1 , x ≠ 1. Find f(x).
–
Diberi fg(x) = 4x + 3 dan g : x → 3 , x ≠ 1. Cari f(x).
x–1
3 Given f : x → 3x + 7 and fg : x → 6x2 – 3x – 5, find g(x).
Diberi f : x → 3x + 7 dan fg : x → 6x2 – 3x – 5, cari g(x).
4 Given f : x → 2x – 4 and fg : x → x 6 2 , x ≠ 2, find g(x).
–
Diberi f : x → 2x – 4 dan fg : x → 6 , x ≠ 2, cari g(x).
x–2
5 Desa Kindergarten started operating in 2015 and the number of students for the first 12 years is given by
f : t → 16 + 7t, such that t is the number of years after 2015. HOTS Applying
Tadika Desa mula beroperasi pada tahun 2015 dan bilangan murid untuk 12 tahun yang pertama diberi oleh f : t → 16 + 7t, dengan
keadaan t ialah bilangan tahun selepas 2015.
(a) Find the number of students after 5 years.
Cari bilangan murid selepas 5 tahun.
(b) In which year will the number of students be 79?
Pada tahun berapakah bilangan murid akan menjadi 79 orang?
16
Chapter Learning Area: Algebra
2 Quadratic Functions
Fungsi Kuadratik
2.1 Quadratic Equations and Inequalities / Persamaan dan Ketaksamaan Kuadratik
Smart Tip
1 The general form of a quadratic equation is (b) Has an equal sign “=” and can be expressed in
ax2 + bx + c = 0, where a, b, c are constants and a ≠ 0. the form ax2 + bx + c = 0.
Bentuk am persamaan kuadratik ialah ax2 + bx + c = 0, dengan
keadaan a, b, c ialah pemalar dan a ≠ 0. Mempunyai tanda “=” dan boleh dinyatakan dalam bentuk
ax2 + bx + c = 0.
2 Characteristics of a quadratic equation:
Ciri-ciri suatu persamaan kuadratik: (c) The highest power of the variable is 2.
Kuasa tertinggi bagi pemboleh ubah ialah 2.
(a) Involves only one variable.
Melibatkan hanya satu pemboleh ubah.
Exercise 1 Solve the following quadratic equations by completing the square. Give the answer correct
to three decimal places.
Selesaikan persamaan kuadratik berikut dengan penyempurnaan kuasa dua. Beri jawapan betul kepada tiga
tempat perpuluhan.
Example 1 1 x2 − 4x − 5 = 0 2 x2 − 6x − 16 = 0
x2 + 4x − 3 = 0 x2 − 4x − 5 = 0 x2 − 6x − 16 = 0
Solution x2 − 4x = 5 x2 − 6x = 16
( ) ( )x2 − 6x + – 62 2 = 16 + – 26 2
– 42 2 – 24 2
=5+
x2 + 4x − 3 = 0 ( ) ( )x2 − 4x +
x2 + 4x = 3 ( )x − 24 2 = 5 + 4 ( )x− 6 2 = 16 + 9
2
24 2 = 3 + 42
( ) ( )x2 + 4x + 2 (x − 2)2 = 9 (x − 3)2 = 25
( )x4
+ 2 2 = 3 + 4 x − 2 = ± 9 x − 3 = ± 25
(x + 2)2 = 7 x =2±3 x =3±5
x + 2 = ± 7
x = –2 ± 7 x = 2 − 3 or x = 2 + 3 x = 3 − 5 or x = 3 + 5
x = –2 − 7 or/atau x = –2 + 7
x = –1 or x = 5 x = –2 or x = 8
x = –4.646 or x = 0.646
3 x2 + 3x − 7 = 0 4 x2 + 5x + 2 = 0 5 x2 − 7x + 3 = 0
x2 + 3x − 7 = 0 x2 + 5x + 2 = 0 x2 − 7x + 3 = 0
2 2x2 + 3x = 7 3 2 32 x2 + 5x = –2 52 x2 − 7x = –3
2 2 5 2 x2 − 7x + – 72 2 = –3 + – 72 2
x2 + 3x + = 7 + x2 + 5x + 2 2 = –2 +
x+ 3 2 = 7 + 9 x+ 5 2 = –2 + 25 x− 7 2 = –3 + 49
2 4 2 4 2 4
3 2 37
x + = 4 x + 5 2 = 17 x − 7 2 = 37
2 4 4
2 2
x + 3 = ± � 37 x + 5 = ± � 17 x − 7 = ± � 37
2 4 2 4 2 4
x+ 3 = ± 37 x + 5 = ± 17 x − 7 = ± 37
2 2 2
2 2 2
37
x = – 23 ± x = – 52 ± 17 x = 72 ± 37
2
37 2 2
37 17 17 37 37
x = – 23 − 2 or x = – 23 + x = – 52 − or x = – 52 + x = 7 − or x = 7 +
2 2 2 2 2 2 2
x = –4.541 or x = 1.541 x = –4.562 or x = –0.438 x = 0.459 or x = 6.541
17
Exercise 2 Solve the following quadratic equations by completing the square. Give the answers correct
to three decimal places.
Selesaikan persamaan kuadratik berikut dengan penyempurnaan kuasa dua. Beri jawapan betul kepada tiga
tempat perpuluhan.
Example 2 1 3x2 − 5x − 7 = 0 2 3x2 − x − 5 = 0
3x2 − 2x − 4 = 0 3x2 − 5x − 7 = 0 (÷3) 3x2 − x − 5 = 0 (÷3)
Solution x2 − 35 x − 7 = 0 x2 − 31 x − 5 = 0
3 3
3x2 − 2x − 4 = 0 (÷3) 13 x 5
x2 − 53 x = 7 x2 − = 3
32 x 4 3
x2 − − 3 = 0 x2 − 35 x + – 56 2 7 – 56 2 x2 − 31 x + – 16 2 5 – 61 2
3 3
23 x 4 = + = +
3
x2 − = x 5 2 7 25 x− 1 2 = 5 + 1
6 3 36 6 3 36
x2 − 32 x + – 26 2 = 34 + – 62 2 − = +
x 5 2 109 x− 1 2 = 61
6 36 6 36
x− 26 2 4 1 − =
3 9
= + 5 � 109 1 � 61
6 36 6 36
x 1 2 13 x − = ± x − = ±
9
− 3 = 5 109 1 61
6 6
1 ± 313 x − = ± 6 x − = ± 6
3
x − = 5 109 1 61
6 6
1 13 1 13 x = ± 6 x = ± 6
3 3
x = − 3 or x = + 3 5 109 5 109 1 61 1 61
6 6 6 6
x = –0.869 or x = 1.535 x= − 6 or x = + 6 x = − 6 or + 6
Smart Tip x = –0.907 or x = 2.573 x = –1.135 or x = 1.468
If a ≠ 1, the coefficient of x2
must be reduced to 1.
Jika a ≠ 1, pekali bagi x2 perlulah
diturunkan kepada 1.
3x2 ÷ 3 x2
3 3x2 + 2x − 6 = 0 4 5x2 − 3x − 4 = 0 5 7x2 − 5x − 9 = 0
3x2 + 2x − 6 = 0 (÷3) 5x2 − 3x − 4 = 0 (÷5) 7x2 − 5x − 9 = 0 (÷7)
x2 + 32 x − 2 = 0 x2 − 35 x − 4 = 0 x2 − 75 x − 9 = 0
x2 + 32 x = 2 5 7
35 x 4
x2 − = 5 x2 − 75 x = 9
7
x2 + 23 x + 13 13 2 x2 − 35 x + – 130 4 – 130 2
2 2 = 5 + x2 − 57 x + – 154 2 9 – 154 2
7
=2+ = +
x + 1 2 = 2 + 1 x 3 2 4 9 x 5 2 9 25
3 9 10 5 100 14 7 196
− = + − = +
x + 1 2 = 19 x− 3 2 = 89 x 5 2 277
9 10 100 14 196
3 − =
x + 1 = ± � 19 x − 3 = ± � 89 x − 5 = ± � 277
3 9 10 100 14 196
x + 1 = ± 19 x − 3 = ± 89 x − 5 = ± 277
3 10 14
3 10 14
x = – 1 ± 19 x = 3 ± 89 x = 5 ± 277
3 10 14
3 10 14
x = – 1 − 19 or x = – 1 + 19 x = 3 − 89 or x = 3 + 89 x = 5 − 277 or x = 5 + 277
3 3 10 10 14 14
3 3 10 10 14 14
x = –1.786 or x = 1.120 x = –0.643 or x = 1.243 x = –0.832 or x =1.546
18
Exercise 3 Solve the following quadratic equations by using formula. Give the answers correct to three
decimal places.
Selesaikan persamaan kuadratik berikut dengan menggunakan rumus. Beri jawapan betul kepada tiga tempat
perpuluhan.
Example 3 1 x2 − 3x = 5 2 3x2 − 5x − 7 = 0
3x2 − 4x − 6 = 0 x2 − 3x = 5 3x2 − 5x − 7 = 0
x2 − 3x − 5 = 0 a = 3, b = –5, c = –7
Solution a = 1, b = –3, c = –5 x= −(–5) ± (–5)2 − 4(3)(–7)
3x2 − 4x − 6 = 0 2(3)
x= −(–3) ± (–3)2 − 4(1)(–5)
a = 3, b = –4, c = –6 2(1) 5 ± 25 + 84
= 6
3 ± 9 + 20 3 ± 29
−(–4) ± (–4)2 − 4(3)(–6) x = 2 = 2
x= 2(3) 5 ± 109
= 6
3 − 29 3 + 29
4 ± 16 + 72 x = 2 or x = 2 5 − 109 5 + 109
= 6 x = 6 or x = 6
4 ± 88 x = –1.193 or x = 4.193 x = –0.907 or x = 2.573
6
=
x = 4 − 88 or/atau x = 4 + 88 3 x2 + 4x = 6 4 2x2 − 3x = 8
6 6
x2 + 4x = 6 2x2 − 3x = 8
x = –0.897 or/atau 2.230 x2 + 4x − 6 = 0 2x2 − 3x − 8 = 0
a = 1, b = 4, c = –6 a = 2, b = –3, c = –8
−(–3) ± (–3)2 − 4(2)(–8)
Smart Tip x= –4 ± 42 − 4(1)(–6) x= 2(2)
2(1)
3 ± 9 + 64
x = –4 ± 16 + 24 x = 4
If/Jika ax2 + bx + c = 0 2(1)
then/maka x = –b ± b2 – 4ac –4 ± 40 x = 3 ± 73
2 4
2a x =
x = –2 + 10 or x = –2 − 10 x = 3 − 73 or x = 3 + 73
4 4
x = 1.162 or –5.162 x = –1.386 or x = 2.886
5 The diagram below shows a cuboid-shaped tissue 6 The diagram below shows a trapezium.
box. Rajah di bawah menunjukkan sebuah trapezium.
Rajah di bawah menunjukkan sebuah kotak tisu berbentuk (x + 3) cm
kuboid.
(x + 4) cm (2x) cm
4 cm
x cm (2x + 1) cm
(2x + 1) cm Find the value of x.
It is given that the volume of the box is 92 cm3. Cari nilai x.
Find the value of x. x+3
Diberi bahawa isi padu kotak itu ialah 92 cm3. Cari nilai x.
x+4 x+4 2x
Volume = 4x(2x + 1) = 92 x+3 x–2
2x2 + x = 23
2x2 + x − 23 = 0 Using Phythagoras’ theorem,
–b ± b2 − 4ac (x + 4)2 + (x − 2)2 = (2x)2
x = 2a x2 + 8x + 16 + x2 − 4x + 4 = 4x2
2x2 − 4x − 20 = 0
= –(1) ± (1)2 − 4(2)(–23) x2 − 2x − 10 = 0
b2
2(2) x = –b ± − 4ac
2a
= –3.650 or 3.150 −(–2) ± (–2)2 − 4(1)(–10)
The negative value for length is not valid.
∴ x = 3.150 = 2(1)
= –2.317 or 4.317
∴ x = 4.317
19
Exercise 4 Find the quadratic equation from the given roots.
Cari persamaan kuadratik daripada punca-punca yang diberi.
Example 4 Smart Tip
Roots/Punca: 3, – 21 A quadratic equation with roots given can be expressed as
Suatu persamaan kuadratik dengan punca−punca yang diberi boleh diungkap sebagai
Solution x2 − (S.O.R)x + (P.O.R) = 0 where/dengan keadaan P.O.R = product of roots
S.O.R = sum of roots hasil darab punca
Sum of roots = 3 + – 21 = 52 hasil tambah punca
Hasil tambah punca
Product of roots = (3) – 21 = – 23 1 Roots: 3, 2 2 Roots: 5, –3
Punca Punca
Hasil darab punca
Sum of roots = 3 + 2 = 5 Sum of roots = 5 + (–3) = 2
The quadratic equation: Product of roots = (3)(2) = 6 Product of roots = (5)(–3) = –15
Persamaan kuadratik: The quadratic equation The quadratic equation
required is required is
x2 − 5 x + – 23 = 0 x2 − (5)x + (6) = 0 x2 − (2)x + (–15) = 0
2 x2 − 5x + 6 = 0 x2 − 2x − 15 = 0
2x2 − 5x − 3 = 0
Alternative Method Alternative Method Alternative Method
x = 3, x = – 21 x = 3, x = 2 x = 5, x = –3
x − 3 = 0, 2x = –1 x − 3 = 0, x − 2 = 0 x − 5 = 0, x + 3 = 0
x − 3 = 0, 2x + 1 = 0 (x − 3)(x − 2) = 0 (x – 5)(x + 3) = 0
(x – 3)(2x + 1) = 0 x2 − 2x − 3x + 6 = 0 x2 + 3x − 5x − 15 = 0
2x2 + x – 6x – 3 = 0 x2 − 5x + 6 = 0 x2 − 2x − 15 = 0
2x2 − 5x – 3 = 0
3 Root: 5 only/sahaja 4 Roots: 3, – 23 5 Roots: 23, – 13
Punca 7 Punca Punca
Sum of roots Sum of roots = 3 + – 23 = 7 Sumof = 3 + – 31 = 7
3 roots 2 6
= 75 + 57 = 10
7 Product of roots
Product of roots
Product of roots = (3)– 32 = –2
= 23 – 31 = – 12
= 75 75 = 25
49
The quadratic equation
The quadratic equation required is The quadratic equation
required is
required is x2 − 37 x + (–2) = 0
x2 − 170x + 2495 = 0 3x2 − 7x − 6 = 0 x2 − 67 x + – 12 = 0
x2 − 4790x + 4295 = 0 6x2 − 7x − 3 = 0
49x2 − 70x + 25 = 0 Alternative Method Alternative Method
Alternative Method x = 3, x = – 32 x = 3 , x = – 13
x − 3 = 0, 3x + 2 = 0 2
5 (x − 3)(3x + 2) = 0 2x − 3 = 0, 3x + 1 = 0
x = 7 3x2 + 2x − 9x − 6 = 0
3x2 − 7x − 6 = 0 (2x − 3)(3x + 1) = 0
7x − 5 = 0
6x2 + 2x − 9x − 3 = 0
(7x − 5)2 = 0
6x2 − 7x − 3 = 0
(7x − 5)(7x − 5) = 0
49x2 − 35x − 35x + 25 = 0
49x2 − 70x + 25 = 0
20
Exercise 5 Solve each of the following.
Selesaikan setiap yang berikut.
Example 5
The roots of the quadratic equation 2x2 + px + 6q = 12 are 3 and –5. Find the values of p and q.
Punca-punca bagi persamaan kuadratik 2x2 + px + 6q = 12 ialah 3 dan –5. Cari nilai p dan q.
Solution Sum of roots Product of roots
Sum of roots Hasil tambah punca Hasil darab punca
Hasil tambah punca – p2 = –2
∴p=4 3q − 6 = –15
3 + (–5) = 3 − 5 = –2 3q = –9
Smart Tip ∴q = –3
Product of roots
Hasil darab punca To solve this question, the Alternative Method
equation must be expressed as
(3)(–5) = –15 x2 – (S.O.R)x + (P.O.R) = 0.
Therefore, the coefficient of x2
2x2 + px + 6q = 12 must be reduced to 1. x = 3, x = –5
2x2 + px + 6q − 12 = 0 Untuk menyelesaikan soalan ini,
(÷2) persamaan perlu diungkap sebagai (x – 3)(x + 5) = 0
x2 – (S.O.R)x + (P.O.R) = 0. Maka,
2p x 6q − 12 pekali bagi x2 mesti diturunkan kepada 1. x2 + 5x − 3x − 15 = 0
2
x2 + + = 0 x2 + 2x − 15 = 0 (32)
x2 + p2 x + 3q − 6 = 0 2x2 + 4x − 30 = 0
Compare/Bandingkan
2x2 + px + 6q − 12 = 0
Compare with/Bandingkan dengan p = 4, 6q − 12 = –30
x2 − (S.O.R) x + (P.O.R) = 0
6q = –18
q = –3
1 The roots of the quadratic equation 2x2 − 8x + p = 5 2 The roots of the quadratic equation 2x2 + px − 16 = 0
are 1 and 3. Find the value of p. are 2 and –4. Find the value of p.
Punca-punca bagi persamaan kuadratik 2x2 − 8x + p = 5 Punca-punca bagi persamaan kuadratik 2x2 + px − 16 = 0
ialah 1 dan 3. Cari nilai p. ialah 2 dan –4. Cari nilai p.
Given roots 1 and 3 Given roots 2 and –4
Sum of roots = 1 + 3 = 4 Sum of roots = 2 + (–4) = 2 − 4 = –2
Product of roots = (1)(3) = 3 Product of roots = (2)(–4) = –8
2x2 − 8x + p = 5 Product of roots 2x2 + px − 16 = 0 (÷2) Sum of roots
2x2 − 8x + p − 5 = 0 x2 + p2 x − 8 = 0 – p2 = –2
(÷2) p − 5 = 3 ∴p=4
2
x2 − 4x + p − 5 = 0
2 p – 5 = 6
Sum of roots = 4 p = 5 + 6
= 11
3 The roots of the quadratic equation 3x2 + px + 6 = q 4 Given p and 5 are the roots of 2x2 + 4x + q = 0.
are –2 and 6. Find the values of p and q. Find the values of p and q.
Punca-punca bagi persamaan kuadratik 3x2 + px + 6 = q Diberi p dan 5 ialah punca bagi 2x2 + 4x + q = 0. Cari nilai
ialah –2 dan 6. Cari nilai p dan q. p dan q.
Given roots –2 and 6 Given roots p and 5
Sum of roots = –2 + 6 = 4 Sum of roots = p + 5
Product of roots = (–2)(6) = –12 Product of roots = 5p
3x2 + px + 6 = q 2x2 + 4x + q = 0 (42)
3x2 + px + 6 − q = 0
(43) x2 + 2x + q =0
2
x2 + p3 x + 6 − q = 0
3
Sum of roots Product of roots
p + 5 = –2
Sum of roots Product of roots p = –2 − 5 5p = q
– 3p = 4 p = –7 2
6 − q q
∴ p = –12 3 = –12 5(–7) = 2
6 – q = –36 –35 = q
2
q = 42
q = –70
21
Exercise 6 Solve.
Selesaikan.
Example 6
(a) a and β are the roots of the quadratic equation (b) Given that 3 and β are the roots of
2x2 + 4x − 9 = 0. Form a new quadratic equation 4
with roots a + 2 and β + 2.
20x2 − 7x − 6 = 0, find the value of β.
dan ialah punca bagi persamaan kuadratik 2x2 + 4x − 9 = 0.
Diberi bahawa 3 dan β ialah punca bagi persamaan
Bentukkan satu persamaan kuadratik yang baharu dengan 4
punca + 2 dan + 2.
kuadratik 20x2 − 7x − 6 = 0, cari nilai β.
PAK-21 Solution Solution
PAK-21 2x2 + 4x − 9 = 0 20x2 − 7x − 6 = 0
a = 2, b = 4, c = –9
Roots: a, β a = 20, b = –7, c = –6
QR CODE Punca: a, β Roots: 43 , β Sum of roots
a + β = – ba = – 24 = –2 3 + β = – ba Hasil tambah punca-punca
4
α + β = – b
c – 92 β = – (2–70) 3 a
aβ = a = − 4
New roots: a + 2, β + 2 = 7 − 15
Punca baharu: a + 2, β + 2 20
ACTIVITY Sum of new roots = 2–08
Hasil tambah punca-punca baharu
a + 2 + β + 2 = (a + β) + 2 + 2 = – 25
= (–2) + 4
=2 Alternative Method
Product of new roots 3 (β) = c Product of roots
Hasil darab punca-punca baharu 4 a
Hasil darab punca-punca
(a + 2)(b + 2) –6 4
= aβ + 2a + 2β + 4 β = 20 3 3 αβ = c
= aβ + 2(a + β) + 4 a
β = – 25
= 1– 292 + 2(–2) + 4
Or/Atau
= – 29
20x2 – 7x − 6 = 0
New quadratic equation (4x − 3)(5x + 2) = 0
Persamaan kuadratik yang baharu 4x − 3 = 0 or/atau 5x + 2 = 0
4x = 3 5x = −2
x2 − (S.O.R)x + (P.O.R) = 0 x = 3 x = – 52
4
2x2 − (2)x + – 29 = 0
Roots/Punca: 43 , β
x2 − 2x − 9 = 0
2 ∴ β = – 52
2x2 − 4x − 9 = 0
ACTIVITY PAK-21 Trade a problem QR CODE
Steps/Langkah-langkah: Scan or visit
1 Students are required to provide a question of forming quadratic equations from given https://www.
purplemath.
roots. com/modules/
Murid-murid dikehendaki menyediakan satu soalan mengenai membentuk persamaan quadform.htm
for additional
kuadratik daripada punca yang diberi. notes on how to
2 Students are divided into groups of 4. Trade the question with other members. solve quadratic
Murid-murid dibahagi kepada kumpulan yang terdiri daripada 4 orang murid. Tukar equations.
soalan dengan ahli yang lain.
3 Teacher hold a discussion of results of the problem solving for each group.
Guru membuat perbincangan bagi hasil penyelesaian masalah bagi setiap kumpulan.
22
1 Given α and β are the roots of 2 Given α and β are the roots of 3 If α and β are the roots of
the quadratic equation the quadratic equation the quadratic equation
x2 − 7x + 14 = 0, form a new 3x2 − x − 15 = 0, form a new 4x2 + 3x + 12 = 0, form a new
quadratic equation with roots quadratic equation with roots quadratic equation with
α and β3 . 5 − α and 5 − β. roots 4 and 4 .
3 α β
Diberi α dan β ialah punca bagi
Diberi α dan β ialah punca bagi persamaan kuadratik 3x2 − x − 15 = 0, Jika α dan β ialah punca bagi
persamaan kuadratik x2 − 7x + 14 = 0, bentukkan persamaan kuadratik yang persamaan kuadratik 4x2 + 3x + 12 = 0,
baharu dengan punca 5 − α dan 5 − β. bentukkan persamaan kuadratik yang
bentukkan persamaan kuadratik yang
4 4
αβ 3x2 − x − 15 = 0 baharu dengan punca α dan β .
baharu dengan punca 3 dan 3 .
a = 3, b = –1, c = –15
x2 − 7x + 14 = 0 Roots: α, β 4x2 + 3x + 12 = 0
a = 1, b = –7, c = 14 a = 4, b = 3, c = 12
α + β = – ba = – (–31) = 1
Roots: α, β 3 Roots: α, β
α + β = – ba = – (–17) = 7 αβ = c = –15 = –5 α + β = – ba = – 43
a 3
αβ = c = 14 = 14 New roots: 5 − α, 5 − β αβ = c = 12 = 3
a 1 a 4
4 4
αβ Sum of the new roots New roots: α , β
New roots: 3 , 3 (5 − α) + (5 − β)
Sum of the new roots = 10 − α – β Sum of the new roots
α + β = α +3 β = 10 − (α + β) 4 + β4 = (4β + 4α)
3 3 α αβ
= 10 − 1
= 37 = 3 = 4(βα+β α)
29
Product of the new roots 3 4– 43
α3 β3 = αβ Product of the new roots = 3
9 (5 − α)(5 − β)
= 25 − 5β − 5α + αβ = –33
= 14 = 25 − 5(β + α) + αβ
9
New quadratic equation = 25 − 5 1 − 5 = –1
x2 − (S.O.R)x + (P.O.R) = 0 3
55 Product of the new roots
3 α4 β4 16
x2 − 7 x + 194 = 0 = = αβ
3
16
9x2 − 21x + 14 = 0 New quadratic equation = 3
x2 − (S.O.R)x + (P.O.R) = 0
x2 − 239 x + 535 = 0 New quadratic equation
x2 − (S.O.R)x + (P.O.R) = 0
3x2 − 29x + 55 = 0
16
x2 − (–1)x + 3 = 0
3x2 + 3x + 16 = 0
4 α and –1 are the roots of 2x2 – 6x + k = 0, find the values of α and k.
α dan –1 ialah punca bagi persamaan kuadratik 2x2 − 6x + k = 0, cari nilai bagi α dan k.
2x2 − 6x + k = 0 α(–1) = c
a = 2, b = –6, c = k a
k
Roots: α, –1 –α = 2 , (α = 4)
α + (–1) = – ba –4 = k
2
α − 1 = – (–6) –8 = k
2
α − 1 = 3
α = 4
23
5 One of the roots of the quadratic equation 6 One of the roots of the quadratic equation
2x2 − mx + m = 0, m ≠ 0, is twice the other root. 3x2 − 5x + p = 0 is three times the value of the
other root. Find the value of p.
Find the value of m and determine the values of
the roots. HOTS Applying Satu daripada punca bagi persamaan kuadratik 3x2 − 5x + p
Satu daripada punca bagi persamaan kuadratik 2x2 − mx + = 0 ialah tiga kali punca yang satu lagi. Cari nilai p.
m = 0, m ≠ 0, ialah dua kali punca yang satu lagi. Cari nilai 3x2 − 5x + p = 0
m dan tentukan nilai bagi punca-punca itu.
2x2 – mx + m = 0 Substitute 1 into 2 x2 − 35x + p = 0
a = 2, b = –m, c = m 3
m6 2 = m
4 Roots: α and 3α
Assume the roots are m2 = m Sum of roots = α + 3α = 4α
α and 2α, 36 4
4 α = 5
Sum of roots: m2 – 9m = 0 3
α + 2α = – b m(m – 9) = 0 α = 5
a 12
m = 0 or m = 9,
3α = – (–2m)
m≠0∴m=9 Product of roots = α(3α) = 3α2
α = m 1 Substitute m = 9 into p = 3α2
6 3
2x2 − mx + m = 0
p = 9α2
Product of roots: 2x2 − 9x + 9 = 0
5
α(2α) = ac (2x − 3)(x − 3) = 0 Substitute α = 12 into p = 9α2
2α2 = m2 2x − 3 = 0 and x − 3 = 0 5 25
12 16
x = 3 x = 3 p = 9 22 =
2
α2 = m4
2
Exercise 7 Find the range of values of x.
Cari julat nilai x.
Smart Tip
(a) (x − m)(x − n) < 0 or/atau (x − m)(x − n) ≤ 0 (b) (x − m)(x − n) > 0 or/atau (x − m)(x − n) ≥ 0
m<x<n m ≤ x ≤ n x < m or/atau x > n x ≤ m, or/atau x ≥ n
m nx m nx
Example 7 1 x2 < 6 + x 2 x2 + 2x > 15
x2 < 6 + x x2 + 2x > 15
x2 + 7x > 8 x2 − x − 6 < 0 x2 + 2x − 15 > 0
(x + 2)(x − 3) < 0 (x + 5)(x − 3) > 0
Solution x
x2 + 7x > 8 Roots: x = –2, 3 Roots: x = –5, 3
x2 + 7x − 8 > 0 –8
(x − 1)(x + 8) > 0 1 –5 3 x
Roots/Punca: x = 1, –8 –2 3x x < –5 or x > 3
x < –8 or/atau x > 1
–2 < x < 3
3 2x2 − 6x 8 4 –x2 + 2x + 8 0 5 2 > 3x − x2
2x2 − 6x 8 –x2 + 2x + 8 0 2 > 3x − x2
2x2 − 6x − 8 0 x2 − 2x − 8 0 x2 − 3x + 2 > 0
2(x2 − 3x − 4) 0 (x − 4)(x + 2) 0 (x − 1)(x − 2) > 0
2(x + 1)(x − 4) 0
Roots: x = 4, –2 Roots: x = 1, 2
Roots: x = –1, 4
–2 4 x
–1 4x x<1 1 2x
–2 x 4
–1 x 4 or x > 2
24
Exercise 8 Solve.
Selesaikan. HOTS Applying
TP 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang fungsi kuadratik dalam konteks penyelesaian masalah rutin yang mudah.
1 The graph of a quadratic function g(x) = px2 – 6x + q, where p and q are constants, has a minimum point.
Graf bagi fungsi kuadratik g(x) = px2 – 6x + q, dengan keadaan p dan q ialah pemalar, mempunyai satu titik minimum.
(a) Given p is an integer such that –2 < p < 2, state the value of p.
Diberi p ialah satu integer dengan keadaan –2 < p < 2, nyatakan nilai bagi p.
(b) Using the answer in (a), find the value of q when the graph touches the x-axis at one point.
Dengan menggunakan jawapan di (a), cari nilai bagi q apabila graf itu menyentuh paksi-x pada satu titik.
(c) Sketch the graph of g(x).
Lakarkan graf bagi g(x).
(a) p = –1 dan 1 (c) g(x) = x2 − 6x + 9
g(x) has a minimum point, g(x) > 0 = (x − 3)2
∴p=1
g(x) = x2 − 6x + q g(x)
(b) Since the graph touches the x-axis at
one point, then 9 x
b2 − 4ac = 0 03
(–6)2 − 4(1)(q) = 0
36 − 4q = 0
–4q = –36
q=9
2.2 Types of Roots of Quadratic Equations / Jenis-jenis Punca Persamaan Kuadratik
Smart Tip
1 Given a quadratic equation ax2 + bx + c = 0./Diberi persamaan kuadratik ax2 + bx + c = 0. a>0 x a<0 x
(a) If b2 – 4ac > 0, then the equation has two different real roots. x x
Jika b2 – 4ac > 0, maka persamaan mempunyai dua punca nyata yang berbeza. x
(b) If b2 – 4ac = 0, then the equation has two equal real roots.
Jika b2 – 4ac = 0, maka persamaan mempunyai dua punca nyata yang sama.
a>0 a<0
(c) If b2 – 4ac < 0, then the equation has no real roots. x
Jika b2 – 4ac < 0, maka persamaan tidak mempunyai punca nyata.
a>0 a<0
2 The term ‘discriminant’ stands for b2 – 4ac./ Istilah ‘pembezalayan’ merujuk kepada b2 – 4ac.
Exercise 9 Determine the types of roots for each of the following quadratic equations.
Tentukan jenis punca bagi setiap persamaan kuadratik berikut.
Example 8 1 x2 − 2x − 3 = 0 2 4x2 − 4x + 1 = 0
3x2 – 2x + 5 = 0 a = 1, b = –2, c = –3 a = 4, b = –4, c = 1
b2 – 4ac = (–2)2 – 4(1)(–3) b 2 – 4ac = (–4)2 – 4(4)(1)
Solution = 4 + 12
a = 3, b = –2, c = 5 = 16 = 16 – 16
b2 – 4ac = (–2)2 – 4(3)(5) 16 > 0 = 0
So, the quadratic equation So, the quadratic equation
= 4 – 60 x2 – 2x – 3 = 0 has two different 4x2 – 4x + 1 = 0 has two equal
= –56 real roots. real roots.
–56 < 0
So, the quadratic equation
3x2 – 2x + 5 = 0 has no real roots.
Maka, persamaan kuadratik
3x2 – 2x + 5 = 0 tidak mempunyai
punca nyata.
25
3 2x2 + 3x – 5 = 0 4 2x2 – x + 6 = 0 5 9x2 – 12x + 4 = 0
a = 2, b = 3, c = –5 a = 2, b = –1, c = 6 a = 9, b = –12, c = 4
b2 – 4ac = (3)2 – 4(2)(–5) b2 – 4ac = (–1)2 – 4(2)(6) b2 – 4ac = (–12)2 – 4(9)(4)
= 9 + 40 = 1 – 48
= 49 = –47 = 144 – 144
49 > 0 –47 < 0 = 0
So, the quadratic equation So, the quadratic equation So, the quadratic equation
2x2 + 3x – 5 = 0 has two 2x2 – x + 6 = 0 has no real 9x2 – 12x + 4 = 0 has two equal
different real roots. roots. real roots.
Exercise 10 Find the range of values of k if the quadratic equation has two different real roots.
Cari julat bagi nilai k jika persamaan kuadratik mempunyai dua punca nyata yang berbeza.
Example 9 1 x2 – 5x + 3 = k 2 3x2 + 2x + k = 5
2x2 + 6x + 5 = k x2 − 5x + 3 = k 3x2 + 2x + k = 5
x2 − 5x + 3− k = 0 3x2 + 2x + k – 5 = 0
Solution ∴ a = 1, b = –5, c = 3 – k ∴ a = 3, b = 2, c = k – 5
Given/Diberi
2x2 + 6x + 5 = k With two different real roots, With two different real roots,
2x2 + 6x + 5 − k = 0 b2 − 4ac > 0
(–5)2 − 4(1)(3 − k) > 0 b2 − 4ac > 0
∴ a = 2, b = 6, c = 5 – k 25 – 12 + 4k > 0
13 + 4k > 0 (2)2 − 4(3)(k − 5) > 0
4k > –13
With two different real roots, 4 − 12k + 60 > 0
k > – 143
Dengan dua punca nyata yang berbeza, –12k + 64 > 0
b2 − 4ac > 0 –12k > –64
(6)2 − 4(2)(5 − k) > 0 12k < 64
36 − 40 + 8k > 0 k < 64
12
–4 + 8k > 0
8k > 4 k < 16
3
k > 1
2
3 kx2 + 6x = –1 4 kx2 + 2x = 1 – x2 5 2x2 + 4x + k = 1
kx2 + 6x = –1 kx2 + 2x = 1 − x2 2x2 + 4x + k = 1
kx2 + 6x + 1 = 0 kx2 + x2 + 2x − 1 = 0 2x2 + 4x + k − 1 = 0
∴ a = k, b = 6, c = 1 (k + 1)x2 + 2x − 1 = 0 ∴a = 2, b = 4, c = k – 1
∴ a = k + 1, b = 2, c = –1
With two different real roots, With two different real roots,
b2 − 4ac > 0 With two different real roots, b2 − 4ac > 0
(6)2 − 4(k)(1) > 0 b2 − 4ac > 0 (4)2 − 4(2)(k − 1) > 0
(2)2 − 4(k + 1)(–1) > 0 16 − 8k + 8 > 0
–4k > –36 4 + 4k + 4 > 0 24 − 8k > 0
4k < 36 8 + 4k > 0 –8k > –24
k < 9
4k > –8 8k < 24
k > –2 k < 3
26
Exercise 11 Find the value of k if the given quadratic equation has two equal real roots.
Cari nilai bagi k jika persamaan kuadratik yang diberi mempunyai dua punca nyata yang sama.
Example 10 1 x2 − 6x + 1 = k 2 3x2 + 4x + k = 2
2x2 + 3x + 6 = k x2 − 6x + 1 = k 3x2 + 4x + k = 2
x2 − 6x + 1 − k = 0 3x2 + 4x + k − 2 = 0
Solution ∴ a = 1, b = –6, c = 1 − k ∴ a = 3, b = 4, c = k − 2
Given/Diberi
With two equal real roots, With two equal real roots,
2x2 + 3x + 6 = k b2 − 4ac = 0
( –6)2 − 4(1)(1 − k) = 0 b2 − 4ac = 0
2x2 + 3x + 6 − k = 0 36 − 4 + 4k = 0
32 + 4k = 0 (4)2 − 4(3)(k − 2) = 0
∴ a = 2, b = 3, c = 6 − k
4k = –32
With two equal real roots, k = –8 16 − 12k + 24 = 0
Dengan dua punca nyata yang sama, 40 − 12k = 0
b2 − 4ac = 0 –12k = –40
(3)2 − 4(2)(6 − k) = 0 10
3
9 − 48 + 8k = 0 k =
–39 + 8k = 0
8k = 39
k = 39
8
3 kx2 − 2x = 6 4 x2 + kx + 1 = –x 5 x2 + 2x = kx2 + 3
kx2 − 2x = 6 x2 + kx + 1 = –x x2 + 2x = kx2 + 3
kx2 − 2x − 6 = 0 x2 + kx + x + 1 = 0 x2 − kx2 + 2x – 3 = 0
∴ a = k, b = –2, c = –6 x2 + (k + 1)x + 1 = 0 (1 − k)x2 + 2x – 3 = 0
∴ a = 1, b = k + 1, c = 1 ∴ a = 1 − k, b = 2, c = –3
With two equal real roots, With two equal real roots, With two equal real roots,
b2 − 4ac = 0
b2 − 4ac = 0 (k + 1)2 − 4(1)(1) = 0 b2 − 4ac = 0
(k + 1)2 − 4 = 0
(–2)2 – 4(k)(–6) = 0 (k + 1 + 2)(k + 1 − 2) = 0 (2)2 − 4(1 − k)(–3) = 0
(k + 3)(k − 1) = 0
4 + 24k = 0 k + 3 = 0 or k − 1 = 0 4 + 12 − 12k = 0
k = –3 k = 1
24k = –4 16 − 12k = 0
k = – 1 –12k = –16
6
4
k = 3
Exercise 12 Find the range of values of k if the given quadratic equation has no real roots.
Cari julat bagi bagi nilai k jika persamaan kuadratik yang diberi tidak mempunyai punca nyata.
Example 11 1 x2 – 5x + 3 = k 2 2x2 + 4x + k = 1
3x2 + 2x + 1 = k x2 − 5x + 3 = k 2x2 + 4x + k = 1
x2 − 5x + 3 − k = 0 2x2 + 4x + k − 1 = 0
Solution ∴ a = 1, b = –5, c = 3 − k ∴ a = 2, b = 4, c = k − 1
Given/Diberi
3x2 + 2x + 1 = k Has no real roots, Has no real roots,
3x2 + 2x + 1 − k = 0 b2 − 4ac < 0 b2 − 4ac < 0
∴ a = 3, b = 2, c = 1 − k (–5)2 − 4(1)(3 − k) < 0 (4)2 − 4(2)(k − 1) < 0
25 − 12 + 4k < 0 16 – 8k + 8 < 0
Has no real roots 13 + 4k < 0 –8k < –24
4k < –13 8k > 24
Tidak mempunyai punca nyata, k < – 143 k > 3
b2 − 4ac < 0
(2)2 − 4(3)(1 − k) < 0
4 − 12 + 12k < 0
–8 + 12k < 0
12k < 8
k < 2
3
27
3 kx2 + 3x = 7 4 kx2 + x = x2 – 2 5 2kx2 + 3x = x2 + 2
kx2 + 3x = 7 kx2 + x = x2 – 2 2kx2 + 3x = x2 + 2
kx2 + 3x − 7 = 0 kx2 − x2 + x + 2 = 0 2kx2 − x2 + 3x − 2 = 0
∴ a = k, b = 3, c = –7 (k − 1)x2 + x + 2 = 0 (2k − 1)x2 + 3x − 2 = 0
∴ a = k – 1, b = 1, c = 2 ∴ a = 2k − 1, b = 3, c = –2
Has no real roots,
b2 − 4ac < 0 Has no real roots, Has no real roots,
(3)2 − 4(k)(–7) < 0 b2 − 4ac < 0
9 + 28k < 0 b2 − 4ac < 0 ( 3)2 − 4(2k − 1)(–2) < 0
28k < –9 9 + 16k − 8 < 0
k < – 298 (1)2 − 4(k − 1)(2) < 0 1 + 16k < 0
16k < –1
1 − 8k + 8 < 0 k < – 116
–8k < –9
8k > 9
9
k > 8
2.3 Quadratic Functions / Fungsi Kuadratik
Smart Tip
1 If a is positive (a > 0), the graph of the function has a minimum point.
Jika a ialah positif (a > 0), graf fungsi itu mempunyai satu titik minimum.
2 If a is negative (a < 0), the graph of the function has a maximum point.
Jika a ialah negatif (a < 0), graf fungsi itu mempunyai satu titik maksimum.
Exercise 13 Determine whether the graph of each of the following quadratic functions has a maximum
point or minimum point.
Tentukan sama ada graf bagi setiap fungsi kuadratik berikut mempunyai titik maksimum atau titik minimum.
TP 2 Mempamerkan kefahaman tentang fungsi kuadratik.
Example 12 1 f(x) = 5x − 6 + 2x2
f(x) = 2x – 4x2 – 3 Since a = 2 > 0, the graph of
the function is a parabola with
Solution a minimum point.
Since a = –4 < 0, the graph of the function is a parabola with a
maximum point.
Oleh sebab a = –4 < 0, graf fungsi ialah satu parabola dengan satu titik maksimum.
Common Error
f(x) = 2x – 4x2 – 3 a = 2
Incorrect because 2 is the value for b. Students must be aware of the
arrangement of f(x) = ax2 + bx + c in the equation given.
Salah kerana 2 ialah nilai bagi b. Pelajar perlu berhati-hati dengan susunan
f(x) = ax2 + bx + c dalam persamaan yang diberi.
2 f(x) = –3x2 − 2x − 6 3 f(x) = –5x2 − 6x + 2 4 f(x) = 6x + 3x2 − 4
Since a = –3 < 0, the graph of Since a = –5 < 0, the graph of Since a = 3 > 0, the graph of
the function is a parabola with the function is a parabola with the function is a parabola with
a maximum point. a maximum point. a minimum point.
28
Exercise 14 Sketch the graph for each of the following quadratic functions.
Lakarkan graf bagi setiap fungsi kuadratik berikut.
TP 3 Mengaplikasikan kefahaman tentang fungsi kuadratik untuk melaksanakan tugasan mudah.
Example 13 1 Given/Diberi 2 Given/Diberi
f(x) = x2 + 2x – 3 f(x) = x2 – 2x + 3
Given/Diberi
f(x) = x2 + 2x + 3 f(x) f(x)
Solution
f(x)
3 0x 3
x –3 0
0 x
3 Given/Diberi 4 Given/Diberi 5 Given/Diberi
f(x) = x2 – 2x – 3 f(x) = –x2 + 2x + 3 f(x) = –x2 + 2x – 3
f(x) f(x) f(x)
0x
0 3
–3 x –3
x
0
Smart Tip x x
1 (a) If b2 − 4ac > 0, then the function has two different real roots. a>0 a<0
Jika b2 − 4ac > 0, maka fungsi mempunyai dua punca nyata yang berbeza.
x
(b) Graph f(x) intersects the x-axis at two distinct points.
Graf f(x) menyilang paksi-x pada dua titik yang berlainan. a>0 x a<0
2 (a) If b2 − 4ac = 0, then the function has two equal real roots. x
Jika b2 − 4ac = 0, maka fungsi mempunyai dua punca nyata yang sama.
(b) Graph f(x) intersects the x-axis at one point only, where x-axis is the a>0 x a<0
tangent to the curve.
Graf f(x) menyilang paksi-x pada satu titik sahaja, dengan keadaan paksi-x ialah tangen
kepada lengkung itu.
3 (a) If b2 − 4ac < 0, then the function has no real roots.
Jika b2 − 4ac < 0, maka fungsi tidak mempunyai punca nyata.
(b) Graph f(x) does not intersect the x-axis.
Graf f(x) tidak menyilang paksi-x.
29
Exercise 15 Find the range of values of k if the curve of each of the quadratic functions below intersects
with the x-axis at two different points.
Cari julat nilai k jika lengkung bagi setiap fungsi kuadratik di bawah menyilang paksi-x pada dua titik yang berlainan.
TP 3 Mengaplikasikan kefahaman tentang fungsi kuadratik untuk melaksanakan tugasan mudah.
Example 14 1 f(x) = x2 + 6x + k
f(x) = 3x2 − 2x + k − 5 a = 1, b = 6, c = k
Solution b2 – 4ac > 0
(6)2 − 4(1)(k) > 0
a = 3, b = –2, c = k − 5 36 − 4k > 0
–4k > –36
If the curve intersects with the x–axis at two different points, then f(x) 4k < 36
k<9
has two different real roots.
Jika lengkung menyilang paksi-x pada dua titik yang berlainan, maka f(x)
mempunyai dua punca nyata yang berbeza.
b2 − 4ac > 0
(–2)2 – 4(3)(k − 5) > 0 Common Error
4 – 12k + 60 > 0 –12k > –64 When multiplying or dividing an
–12k + 64 > 0
–64 inequality by a negative number, reverse
–12k > –64 k > –12 the direction of the inequality notation.
16 Apabila mendarab atau membahagi suatu
k < 3 16 ketaksamaan dengan suatu nombor negatif,
k > 3
songsangkan arah tatatanda ketaksamaan itu.
2 f(x) = –x2 + 2x – k + 4 3 f(x) = 4x2 + kx + 4 4 f(x) = (k − 3)x2 – 2x – 4
a = –1, b = 2, c = –k + 4 a = 4, b = k, c = 4 a = k – 3, b = –2, c = –4
b2 – 4ac > 0 b2 − 4ac > 0 b2 − 4ac > 0
( 2)2 − 4(–1)(–k + 4) > 0 k2 − 4(4)(4) > 0
4 + 4(–k + 4) > 0 k2 − 64 > 0 (–2)2 − 4(k − 3)(–4) > 0
4 − 4k + 16 > 0 (k + 8)(k − 8) > 0
–4k + 20 > 0 k < –8 or 4 + 16k − 48 > 0
–4k > –20
4k < 20 16k > 44
k < 5
k>8 k > 11
4
Exercise 16 Find the value (or values) of m if the curve of each of the quadratic functions below intersects
with the x-axis at one point only.
Cari nilai (atau nilai-nilai) bagi m jika lengkung bagi setiap fungsi kuadratik di bawah menyilang paksi-x pada
satu titik sahaja.
TP 3 Mengaplikasikan kefahaman tentang fungsi kuadratik untuk melaksanakan tugasan mudah.
Example 15 1 f(x) = –2x2 + x + m
f(x) = mx2 + 3x + 6 b2 − 4ac = 0 a = –2, b = 1, c = m
Intersects with x-axis at one
Solution (3)2 − 4(m)(6) = 0 point only,
a = m, b = 3, c = 6 b2 – 4ac = 0
If the curve intersects with the x–axis at one 9 − 24m = 0 (1)2 – 4(–2)(m) = 0
point only, then f(x) has two equal roots. 1 + 8m = 0
Jika lengkung menyilang paksi-x pada satu titik –24m = –9
sahaja, maka f(x) mempunyai dua punca yang sama. 3 8m = –1
m = 8 m = – 81
2 f(x) = 4x2 − 6x − m − 2 3 f(x) = 1 − 8x − mx2 4 f(x) = x2 − mx + m + 8
a = 4, b = –6, c = –m − 2 a = –m, b = –8, c = 1 a = 1, b = –m, c = m + 8
Intersects with x-axis at one Intersects with x-axis at one Intersects with x-axis at one
point only, point only, point only,
b2 − 4ac = 0 b2 − 4ac = 0 b2 – 4ac = 0
(–6)2 − 4(4)(–m − 2) = 0 (–8)2 − 4(–m)(1) = 0 (–m)2 − 4(1)(m + 8) = 0
36 + 16m + 32 = 0 64 + 4m = 0 m2 − 4m − 32 = 0
16m = –68 4m = –64 (m + 4)(m − 8) = 0
m = – 147 m = –16 m + 4 = 0 or m − 8 = 0
m = –4 m = 8
30
Exercise 17 Find the range of values of n if the curve of each of the quadratic function below does not
intersect with the x-axis.
Cari julat nilai n jika lengkung bagi fungsi kuadratik di bawah tidak menyilang paksi-x.
TP 3 Mengaplikasikan kefahaman tentang fungsi kuadratik untuk melaksanakan tugasan mudah.
Example 16 1 f(x) = –2x2 – 4x − n + 1
f(x) = 2x2 − 6x + 1 − n b2 − 4ac < 0 a = –2, b = –4, c = –n + 1
(–6)2 − 4(2)(1 − n) < 0 No intersection with x-axis,
Solution b2 − 4ac < 0
a = 2, b = –6, c = 1 − n 3 6 − 8 + 8n < 0 (–4)2 − 4(–2)(–n + 1) < 0
If the curve does not intersect with the 16 – 8n + 8 < 0
x–axis, then f(x) does not have real roots. 8n < –28 –8n < –24
Jika lengkung itu tidak menyilang paksi-x, maka n < – 27 8n > 24
f(x) tidak mempunyai punca yang nyata. n > 3
2 f(x) = nx2 – 2x – 8 3 f(x) = nx2 − 12x + 6 4 f(x) = 6 − 2x + (n − 1)x2
a = n, b = –2, c = –8 a = n, b = –12, c = 6 a = n – 1, b = –2, c = 6
No intersection with x-axis,
No intersection with x-axis, b2 − 4ac < 0 No intersection with x-axis,
(–12)2 − 4(n)(6) < 0
b2 − 4ac < 0 144 − 24n < 0 b2 − 4ac < 0
( –2)2 − 4(n)(–8) < 0 –24n < –144 (–2)2 − 4(n − 1)(6) < 0
24n > 144
4 + 32n < 0 4 − 24n + 24 < 0
n > 6
32n < –4 –24n < –28
n < – 1 24n > 28
8
n > 7
6
Exercise 18 By using the method of completing the square, express each of the following quadratic functions
in the form of f(x) = a(x – h)2 + k. Hence, state the coordinates of the vertex of the function.
Dengan menggunakan kaedah penyempurnaan kuasa dua, ungkapkan setiap fungsi kuadratik berikut dalam
bentuk f(x) = a(x – h)2 + k. Seterusnya, nyatakan koordinat verteks bagi fungsi itu.
TP 3 Mengaplikasikan kefahaman tentang fungsi kuadratik untuk melaksanakan tugasan mudah.
Example 17 1 f(x) = x2 – 6x – 8 2 f(x) = 3x2 – 12x + 15
f(x) = 2x2 – 7x – 4 f(x) = x2 – 6x – 8 f(x) = 3x2 – 12x + 15
1 2 = x2 – 6x + – 6 2 = 3[x2 – 4x + 5]
2
Solution – 3 1 2 4 2
f(x) = 2x2 – 7x – 4 2
1– 6 22 = 3 x2 – 4x + – –
2
7 – 8 2 1– 4 42
2 2 +5
4= 2
x2 – x – 2 = (x – 3)2 – 9 – 8
1 2= 2 x2 – 7 7 2 = (x – 3)2 – 17 = 3[(x – 2)2 – 4 + 5]
2 4
x + – – a = 1 > 0, f(x) has a minimum = 3[(x – 2)2 + 1]
1– 7 22 24 value. = 3(x – 2)2 + 3
4
– Minimum vertex = (3 , –17) a = 3 > 0, f(x) has a minimum
2 47 49 value.
16
= 21x – 4 2 – – 2 Minimum vertex = (2 , 3)
= 21x – 7 22 – 81 4 3 f(x) = –x2 – 8x + 13 4 f(x) = –3x2 – 24x + 27
4 16
f(x) = –x2 – 8x + 13 f(x) = –3x2 – 24x + 27
1 2= 2 x – 7 2 – 81 = –[x2 + 8x – 13] = –3[x2 + 8x – 9]
4 8
3 1 2 8 2 3 1 2 8 2
= – x2 + 8x + 2 = –3 x2 + 8x + 2
– –
a = 2 > 0, f(x) has a minimum 1 28 22 – 134 1 82 22 – 94
value.
a = 2 > 0, f(x) mempunyai nilai = –[(x + 4)2 – 16 – 13] = –3[(x + 4)2 – 16 – 9]
minimum. = –[(x + 4)2 – 29] = –3[(x + 4)2 – 25]
1 2Minimum vertex = 7 , – 81 = –(x + 4)2 + 29 = –3(x + 4)2 + 75
4 8
Verteks minimum a = –1 < 0, f(x) has a maximum a = –3 < 0, f(x) has a maximum
value. value.
Maximum vertex = (–4 , 29) Maximum vertex = (–4, 75)
31
Exercise 19 Make generalisation about the shape and the position of the graph when the values of a, h
and k are changed.
Buat generalisasi terhadap bentuk dan kedudukan graf apabila nilai-nilai a, h dan k berubah.
TP 3 Mengaplikasikan kefahaman tentang fungsi kuadratik untuk melaksanakan tugasan mudah.
The diagram shows a graph Example 18 f(x)
f(x) = (x + 1)2 + 2, where a = 1, 11
h = –1 and k = 2. f(x) = 9(x + 1)2 + 2
Rajah menunjukkan graf (–1, 2) 2
f(x) = (x + 1)2 + 2, dengan keadaan Solution –1 0
a = 1, h = –1 dan k = 2. When the value of a changes from 1 to 9, the width
of the graph decreases. The axis of symmetry and
f(x) the minimum value of the graph do not change.
When x = 0,
3 f(x) = 9(0 + 1)2 + 2 = 11 x
(–1, 2) Apabila nilai a berubah daripada 1 kepada 9, kelebaran graf
x semakin berkurang. Paksi simetri dan nilai minimum graf tidak
0 berubah.
Apabila x = 0,
f(x) = 9(0 + 1)2 + 2 = 11
1 f(x) = (x – 1)2 + 2 2 f(x) = (x + 1)2 – 2
When the value of h changes from –1 to 1, the When the value of k changes from 2 to –2,
graph with the same shape moves 2 units to the graph with the same shape moves 4 units
the right. The axis of symmetry is x = 1 and the downward. The minimum value becomes –2 and
minimum value does not change, that is 2. axis of symmetry does not change, that is x = –1.
When x = 0, When x = 0,
f(x) = (0 – 1)2 + 2 = 3 f(x) = (0 + 1)2 – 2 = –1
f(x) f(x)
3 –1 0 x
2 (1, 2) –1
x
01 –2
(–1, –2)
Exercise 20 Sketch the graph for each of the following quadratic functions. HOTS Analysing
Lakarkan graf bagi setiap fungsi kuadratik berikut.
TP 3 Mengaplikasikan kefahaman tentang fungsi kuadratik untuk melaksanakan tugasan mudah.
Example 19
f(x) = –x2 + 4x + 12 The maximum point = (2, 16)
Titik maksimum = (2, 16)
Solution To find the x-intercept, substitute f(x) = 0.
a = –1 < 0, graph has the shape of ∩ Untuk mencari pintasan-x, gantikan f(x) = 0.
a = –1 < 0, graf mempunyai bentuk ∩
f(x) = –x2 + 4x + 12 –x2 + 4x + 12 = 0
= –(x2 − 4x − 12) (–x − 2)(x − 6) = 0 Common Error
= – x2 − 4x + – 4 2 − – 4 2 − 12 –x − 2 = 0 or/atau x − 6 = 0 f(x) = –x2 + 4x + 12
= x2 – 4x – 12
2 2 x = –2 x = 6
= –[x2 − 4x + (–2)2 − (–2)2 − 12] To find the f(x)-intercept, substitute x = 0.
Untuk mencari pintasan-f(x), gantikan x = 0.
= –[(x − 2)2 – 16] = –(x − 2)2 + 16 f(x) = –(0)2 + 4(0) + 12 = 12
Alternative Method f(x)
The maximum point can also be obtained by 16 (2, 16)
finding the image of 2, for example:
Titik maksimum boleh juga diperoleh dengan mencari 12
imej bagi 2, contohnya:
f(2) = –(2)2 + 4(2) + 12 = 16 –1 2 6 x
∴ Maximum point/Titik maksimum
= (2, 16)
32
1 f(x) = –x2 + 6x + 16 2 f(x) = x2 − 4x − 5
a = –1 < 0, graph has the shape of ∩ a = 1 > 0, graph has the shape of ∪
f(x) = –x2 + 6x + 16 f(x) = x2 – 4x – 5
= –(x2 – 6x – 16)
– 42 2 – 42 2
[= – x2 – 6 2 – 6 2 16 = x2 − 4x +
– −5
2 2
– 6x + – – = x2 – 4x + (–2)2 – (–2)2 − 5
= –[x2 – 6x + (–3)2 – (–3)2 – 16] = (x – 2)2 – 9
= –[(x – 3)2 − 25] The minimum point = (2, –9)
= –(x – 3)2 + 25 When f(x) = 0,
x2 – 4x – 5 = 0
The maximum point = (3, 25) (x + 1)(x – 5) = 0
x + 1 = 0 or x – 5 = 0
When f(x) = 0, x = –1 x = 5 f(x)
–x2 + 6x + 16 = 0 When x = 0,
f(x) = (0)2 – 4(0) − 5
(–x – 2)(x – 8) = 0 f(x) –1 0 2 5 x
= –5
–x – 2 = 0 or x − 8 = 0 25 (3, 25)
16
x = –2 x = 8
When x = 0, –5
f(x) = –(0)2 + 6(0) + 16 –9 (2, –9)
= 16
x
–2 0 3 8
3 f(x) = –x2 − 7x + 8 4 f(x) = 2x2 − 4x − 16
a = –1 < 0, graph has the shape of ∩ a = 2 > 0, graph has the shape of ∪
f(x) = –x2 − 7x + 8 f(x) = 2x2 – 4x – 16
= –(x2 + 7x − 8) = 2[x2 – 2x – 8]
= – x2 + 7x + 722 – 7 2 – 8 = 2 x2 – 2x + – 2 2 – – 2 2 – 8
2 2 2
= –x + 7 2 − 841 = 2[x2 – 2x + (–1)2 − (–1)2 – 8]
= 2[(x – 1)2 – 9]
2 = 2(x – 1)2 – 18
The minimum point = (1, –18)
= – x + 7 2 + 81
2 4
The maximum point = – 27, 81 When f(x) = 0,
4
When f(x) = 0, 2x2 + 4x − 16 = 0
–x2 – 7x + 8 = 0 f(x) 2(x2 − 2x − 8) = 0 f(x)
(–x + 1)(x + 8) = 0 ( )–—27, —841 7 —841 2(x + 2)(x − 4) = 0
2
–x + 1 = 0 or x + 8 = 0 8 x + 2 = 0 or x − 4 = 0
x
x = 1 x = –8 x = –2 x=4 –2 0 1 4
When x = 0, –8 –—7 0 1 When x = 0, –16
f(x) = –(0)2 – 7(0) + 8 2 x f(x) = 2(0)2 + 4(0) − 16 –18 (1, –18)
= 8 = –16
5 f(x) = –3x2 + 12x + 15
a = –3 < 0, graph has the shape of ∩ When f(x) = 0,
f(x) = –3x2 + 12x + 15
= –3(x2 − 4x − 5) –3x2 + 12x + 15 = 0 f(x)
27
–3(x2 – 4x – 5) = 0 15 (2, 27)
= –3 x2 – 4x + – 4 2 – – 4 2 – 5 –3(x + 1)(x – 5) = 0
2 2
x + 1 = 0 or x – 5 = 0
= –3[x2 – 4x + (–2)2 – (–2)2 – 5] x = –1 x = 5
= –3[(x – 2)2 − 9] When x = 0, –1 0 2 5 x
f(x) = –3(0)2 + 12(0) + 15
= –3(x – 2)2 + 27
= 15
The maximum point = (2, 27)
33
Exercise 21 Solve.
Selesaikan.
TP 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang fungsi kuadratik dalam konteks penyelesaian masalah rutin yang mudah.
TP 5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang fungsi kuadratik dalam konteks penyelesaian masalah rutin yang kompleks.
Example 20
An object is thrown vertically upwards from a Solution
position. The height, h in metre, of the object at (a) h(t) = –4t2 + 8t + 12
time t seconds is given by the function When/Apabila t = 1 ,
2
h(t) = –4t2 + 8t + 12.
1 1
Suatu objek dilambung ke atas dari suatu kedudukan. Tinggi, h h(t) = –41 2 22 + 81 2 2 + 12
dalam meter, objek itu pada masa t saat diberi oleh fungsi
h(t) = –4t2 + 8t + 12. = –1 + 4 + 12 = 15 m
(a) Find the height of the object when t = 1 second. (b) h(t) = –4t2 + 8t + 12
2
1 = –4[t2 – 2t – 3]
2
Cari tinggi objek itu apabila t = saat. = –4[t2 – 2t + (–1)2 – (–1)2 – 3]
(b) In how many seconds will the object reach the = –4[(t – 1)2 – 4]
maximum height? = –4(t – 1)2 + 16
Dalam masa berapa saatkah objek itu akan mencapai tinggi The object reaches the maximum height in
maksimum? 1 second.
(c) What is the maximum height of the object? Objek itu mencapai tinggi maksimum dalam masa 1 saat.
Berapakah tinggi maksimum objek itu? (c) Maximum point/Titik maksimum = (1, 16)
The maximum height is 16 m.
Tinggi maksimum ialah 16 m.
1 The number of students who graduated from a college is given by the function f(x) = x2 – 6x + 84. State the
first year that at least 100 students graduated.
Bilangan pelajar yang tamat pengajian di sebuah kolej diberi oleh fungsi f(x) = x2 – 6x + 84. Nyatakan tahun pertama yang
sekurang-kurangnya 100 orang pelajar tamat pengajian.
f(x) ≥ 100 –2 8 x
x2 – 6x + 84 ≥ 100
x2 – 6x – 16 ≥ 0
( x + 2)(x – 8) ≥ 0
From the graph, x ≤ –2, x ≥ 8
So, the first year that at least 100 students
graduated is in the 8th year.
2 A ball is thrown vertically upwards from a position. The height, h in metre, of the ball at time t seconds is
given by the function h(t) = –6t2 + 24t + 30.
Sebiji bola dilambung ke atas dari suatu kedudukan. Tinggi, h dalam meter, bola itu pada masa t saat diberi oleh fungsi
h(t) = –6t2 + 24t + 30.
(a) What is the height of the ball when t = 1 second?
Berapakah tinggi bola itu apabila t = 1 saat?
(b) In how many seconds will the object reach the maximum height?
Dalam masa berapa saatkah objek itu akan mencapai tinggi maksimum?
(c) What is the maximum height of the object?
Berapakah tinggi maksimum objek itu?
(a) When t = 1, (c) Maximum point = (2, 54)
h(t) = –6t2 + 24t + 30 The maximum height is 54 m.
= –6(1)2 + 24(1) + 30
= –6 + 24 + 30
= 48 m
(b) h(t) = –6t2 + 24t + 30
= –6[t2 – 4t – 5]
= –6[t2 – 4t + (–2)2 – (–2)2 – 5]
= –6[(t – 2)2 – 9]
= –6(t – 2)2 + 54
The ball reaches the maximum height in 2 seconds.
34
3 The diagram shows a parabolic gateway that is represented by the graph y = – 1 x2 + 8. T
50
The distance between the two ends of the curve, PQ is 70 m and the height of PQ from
the horizontal ground, EF is 12 m. Find the maximum height of the gateway from the
horizontal ground.
Rajah menunjukkan sebuah pintu gerbang berbentuk parabola yang diwakili oleh graf y = – 1 x2 + 8. P Q
50 F
Jarak di antara dua hujung lengkung, PQ ialah 70 m dan tinggi PQ dari permukaan tanah mengufuk,
EF ialah 12 m. Cari tinggi maksimum pintu gerbang itu dari permukaan tanah mengufuk. E
y = – 1 x2 + 8
50
When x = 0, y = – 1 (0)2 + 8 = 8
50
When x = 70 = 35, y = – 1 (35)2 + 8 = –16.5
2 50
y
T
0 8m
P x
16.5 m
Q
(35, –16.5)
12 m
E 70 m F
The maximum height = 8 m + 16.5 m + 12 m = 36.5 m
4 In an experiment, an object is launched at an acute angle from a horizontal plane. The height of the object,
y, is represented by the graph y = – 1 x2 + 2x, where x is the horizontal distance from the point of the
launching. 36
Dalam satu eksperimen, sebuah objek dilancarkan pada suatu sudut tirus dari satah mengufuk. Tinggi objek itu, y, diwakili oleh
graf y = – 1 x2 + 2x, dengan keadaan x ialah jarak mengufuk dari titik pelancaran.
36
(a) What is the horizontal distance of the object from the point of launching when it reaches the
maximum point?
Berapakah jarak mengufuk objek itu dari titik pelancaran apabila ia mencapai titik maksimum?
(b) Determine the maximum height of the object.
Tentukan tinggi maksimum objek itu.
(c) How far has the object travelled horizontally before it falls to the horizontal plane again?
Berapa jauhkah objek itu telah bergerak secara mengufuk sebelum objek itu jatuh ke satah mengufuk semula?
(a) y = – 1 x2 + 2x (b) 36 m
36
(c) When y = 0
= – 1 [x2 – 72x]
36 – 1 x2 + 2x = 0
36
1 3x2 1– 72 22 1– 72 224 x1– 22
= – 36 – 72x + 2 – 2 1 x + = 0
36
3 4 1 x = 0 or – 1 x + 2 = 0
= – 36 x2 – 72x + (–36)2 – 362 36
– 1 x = –2
= – 1 [(x – 36)2 – 362] 36
36
= – 1 (x – 36)2 + 36 x = 72
36
The object travelled 72 m horizontally before
Maximum point = (36, 36) it falls to the horizontal plane again.
When the object reaches the maximum
point, the horizontal distance is 36 m.
35
Review 2
Paper 1 Questions
1 Given the quadratic equation x2 + kx + 4 = 3x has 6 Hassan has a rectangular board with a
two equal real roots, find the values of k.
Diberi persamaan kuadratik x2 + kx + 4 = 3x mempunyai SPM measurement of 6x metre in length and 3x metre
dua punca nyata yang sama, cari nilai-nilai k. CLONE
`18 in width. He cuts part of the board to form a
[3 marks/markah] P1Q19 square with sides of x metre. Find the range of
values of x if the remaining area of the board is at
2 Given –4 is one of the roots of the quadratic least (x2 + 9) metre2 .
SPM equation 3(x + h)2 = 75, where h is a constant. Find Hassan mempunyai sekeping papan berbentuk segi empat
CLONE the values of h. tepat yang mempunyai ukuran 6x meter panjang dan
`15
P1Q5 Diberi –4 ialah salah satu punca bagi persamaan kuadratik 3x meter lebar. Dia memotong sebahagian daripada papan
3(x + h)2 = 75, dengan keadaan h ialah satu pemalar. itu kepada bentuk segi empat sama yang bersisi x meter. Cari
Cari nilai-nilai h. julat nilai x jika luas papan yang tinggal adalah sekurang-
[2 marks/markah] kurangnya (x2 + 9) meter2.
[3 marks/markah]
3 (a) It is given that one of the roots of the
SPM quadratic equation x2 + (h + 4)x − h2 = 0, 7 Find the range of values of p if the quadratic
CLONE where h is a constant, is negative of the other.
Find the value of the product of roots. equation (2 − p)x2 − (2p + 1)x = p has no roots.
`17
P1Q13 Cari julat nilai p jika persamaan kuadratik
Diberi bahawa satu daripada punca-punca bagi (2 − p)x2 − (2p + 1)x = p tidak mempunyai punca.
persamaan kuadratik x2 + (h + 4)x − h2 = 0, dengan [3 marks/markah]
keadaan h ialah pemalar, adalah negatif kepada yang
satu lagi. Cari nilai bagi hasil darab punca. 8 The quadratic equation ax2 + bx + c = 0 has roots
2
[2 marks/markah] 3 and – 1 1 . Find the values of a, b and c.
2 ax2 + bx + c = 0 mempunyai punca
(b) It is given that the quadratic equation 2
Persamaan kuadratik 3
px2 − 9qx + 4p = 0, where p and q are
dan –1 1 . Cari nilai a, b and c.
constants, has two equal roots. Find p : q. 2 [3 marks/markah]
Diberi bahawa persamaan kuadratik px2 − 9qx + 4p = 0,
dengan keadaan p dan q ialah pemalar, mempunyai dua
punca yang sama. Cari p : q. 9 (a) Express 3x(x − 1) = (3 − x)(x + 1) in general form.
[2 marks/markah] Ungkapkan 3x(x − 1) = (3 − x)(x + 1) dalam bentuk am.
4 Diagram 1 shows the graph y = a(x – p)2 + q, (b) Hence, solve the quadratic equation, giving
SPM where a, p and q are constants. The straight line the answers correct to three decimal places.
CLONE y = –12 is the tangent to the curve at point U.
Seterusnya, selesaikan persamaan kuadratik, berikan
`18
P1Q18 Rajah 1 menunjukkan graf y = a(x – p)2 + q, dengan keadaan jawapan betul kepada tiga tempat perpuluhan.
a, p dan q ialah pemalar. Garis lurus y = –12 ialah tangen [4 marks/markah]
kepada lengkung pada titik U.
y 10 The quadratic equation x2 – kx + 4 = 3x has two
equal roots. Find the possible values of k.
Persamaan kuadratik x2 – kx + 4 = 3x mempunyai dua
–2 0 8 x punca yang sama. Cari nilai-nilai k yang mungkin.
[3 marks/markah]
y = –12 11 Given the quadratic equation 3x2 – 6px + p = 0
U
satisfies only one value of x. Find the value of p.
Diagram 1/ Rajah 1 Diberi persamaan kuadratik 3x2 – 6px + p = 0 hanya
memuaskan satu nilai bagi x. Cari nilai p.
(a) State the coordinates of U. [2 marks/markah]
Nyatakan koordinat U.
(b) Find the value of a.
Cari nilai a. 12 It is given that α and β are the roots of quadratic
[3 marks/markah] SPM equation 2x2 − 5x + 6 = 0. Form a quadratic
CLONE
equation which has the roots α and β2.
`16 2
P1Q17
5 The quadratic equation x2 + x = 3kx − k2, has two Diberi bahawa α dan β adalah punca-punca bagi persamaan
different real roots. Find the range of values of k.
kuadratik 2x2− 5x + 6 = 0. Bentuk persamaan kuadratik
Persamaan kuadratik x2 + x = 3kx − k2, mempunyai dua αβ
punca nyata yang berbeza. Cari julat bagi nilai k. yang mempunyai punca-punca 2 and 2.
[3 marks/markah]
[3 marks/markah]
36
Paper 2 Questions
1 (a) The quadratic equation x2 – 8x + 12 = 0 has roots p and q, where p > q. Find
Persamaan kuadratik x2 – 8x + 12 = 0 mempunyai punca-punca p dan q, dengan keadaan p > q. Cari
(i) the values of p and q,
nilai bagi p dan q,
(ii) the range of values of x if x2 – 8x + 12 > 0.
julat nilai x jika x2 – 8x + 12 > 0.
[5 marks/markah]
(b) Using the values of p and q from (a)(i), form a quadratic equation with roots p + 3 and 2q + 3.
Menggunakan nilai p dan q dari (a)(i), bentukkan suatu persamaan kuadratik dengan punca p + 3 dan 2q + 3.
[2 marks/markah]
2 Given α and β are the roots of the quadratic equation x(x – 6) = 2m – 8, where m is a constant.
SPM Diberi α dan β ialah punca bagi persamaan kuadratik x(x – 6) = 2m – 8, dengan keadaan m ialah pemalar.
CLONE
`15 (a) Find the range of values of m if α ≠ β.
P2Q5 Cari julat nilai m jika α ≠ β.
[3 marks/markah]
(b) Given α and β are the roots of another quadratic equation 2x2 + px – 16 = 0, where p is a constant. Find
2 2
the values of m and p.
αβ
Diberi 2 dan 2 ialah punca-punca bagi persamaan kuadratik lain 2x2 + px – 16 = 0, dengan keadaan p ialah pemalar. Cari
nilai bagi m dan p.
[4 marks/markah]
H OTS Zone
1 Azmi cut squares with sides measuring 5 cm from each corner of a square-shaped cardboard as shown in
Diagram 1(a). He then folded the cardboard to be an open box as shown in Diagram 1(b).
Azmi memotong segi empat sama bersisi 5 cm daripada setiap bucu sekeping kadbod yang berbentuk segi empat sama seperti dalam
Rajah 1(a). Dia kemudian melipat kadbod tersebut menjadi sebuah kotak terbuka seperti dalam Rajah 1(b).
5 cm 5 cm
5 cm
Diagram 1(a)/ Rajah 1(a) Diagram 1(b)/ Rajah 1(b)
(a) If the volume of the box is 1 280 cm3, form a quadratic equation representing the volume. HO TS Creating
Jika isi padu kotak itu ialah 1 280 cm3, bentuk satu persamaan kuadratik mewakili isi padu tersebut.
(b) Hence, find the length of the sides of the original cardboard. HOTS Applying
Seterusnya, cari panjang sisi kadbod asal.
37
Chapter Learning Area: Algebra
3 Systems of Equations
Sistem Persamaan
3.1 Systems of Linear Equations in Three Variables / Sistem Persamaan Linear dalam Tiga Pemboleh Ubah
Smart Tip
General form of system of linear equations in three variables is
Bentuk am bagi sistem persamaan linear dalam tiga pemboleh ubah ialah
ax + by + cz = d, where/dengan keadaan a, b and/dan c ≠ 0
Exercise 1 Solve the following problems. 1 Based on a certain arrangement, three workers
Selesaikan masalah yang berikut. are required to work for 120 hours in a week. The
first worker should work 6 hours more than the
TP 1 Mempamerkan pengetahuan asas tentang sistem persamaan. second worker and the second worker should
work 12 hours more than the third worker. Write
Example 1 down three equations that satisfy the above
conditions.
A collection of 220 coins consists of 10 cents,
20 cents and 50 cents coins. The total number Berdasarkan suatu penetapan, tiga orang pekerja
of 10 cents coins and 20 cents coins exceeds the dikehendaki bekerja selama 120 jam dalam seminggu.
number of 50 cents coins by 100. If the total value Pekerja pertama mesti bekerja 6 jam lebih daripada pekerja
of the collection is RM64.00, write down three kedua dan pekerja kedua mesti bekerja 12 jam lebih daripada
equations that satisfy the above conditions. pekerja ketiga. Tuliskan tiga persamaan yang memuaskan
Suatu koleksi 220 keping duit syiling terdiri daripada duit syarat-syarat di atas.
syiling 10 sen, 20 sen dan 50 sen. Jumlah bilangan duit
syiling 10 sen dan 20 sen melebihi bilangan duit syiling Let x = number of working hours of the first
50 sen sebanyak 100 keping. Jika koleksi duit syiling worker
itu bernilai RM64.00, tuliskan tiga persamaan yang
memuaskan syarat-syarat di atas. y = number of working hours of the second
worker
Solution
Let/Katakan z = number of working hours of the third
x = number of 10 cents coins worker
x = bilangan duit syiling bernilai 10 sen
y = number of 20 cents coins Hence,
y = bilangan duit syiling bernilai 20 sen x + y + z = 120
z = number of 50 cents coins x–y=6
z = bilangan duit syiling bernilai 50 sen y – z = 12
Hence/Oleh itu,
x + y + z = 220
x + y – z = 100
0.1x + 0.2y + 0.5z = 64
2 Johan buys 4 kg durians, 3 kg jackfruits and 2 kg watermelons for RM95. Isyak buys 3 kg durians, 2 kg
jackfruits and 4 kg watermelons for RM75. The price of 1 kg durians is RM5 more than the price of 1 kg
jackfruits. Write down three equations that satisfy the above conditions.
Johan membeli 4 kg durian, 3 kg nangka dan 2 kg tembikai dengan harga RM95. Isyak membeli 3 kg durian, 2 kg nangka dan
4 kg tembikai dengan harga RM75. Harga 1 kg durian ialah RM5 lebih daripada harga 1 kg nangka. Tuliskan tiga persamaan
yang memuaskan syarat-syarat di atas.
Let x = the price of 1 kg durians
y = the price of 1 kg jackfruits
z = the price of 1 kg watermelons
Hence,
4x + 3y + 2z = 95
3x + 2y + 4z = 75
x–y=5
38
3 A quiz competition consists of section A, section B and section C. The marks of the questions in section A,
section B and section C are different. The table below shows the results of three participants, the number
of questions answered correctly in each section and the total marks obtained. Write down three equations
that satisfy the conditions in the given table.
Suatu pertandingan kuiz terdiri daripada bahagian A, bahagian B dan bahagian C. Markah bagi soalan dalam bahagian A,
bahagian B dan bahagian C adalah tidak sama. Jadual di bawah menunjukkan keputusan bagi tiga orang peserta, bilangan soalan
yang dijawab betul dalam setiap bahagian dan jumlah markah yang diperoleh. Tuliskan tiga persamaan yang memuaskan syarat-
syarat dalam jadual yang diberi.
Participant Section A Section B Section C Total marks
Peserta Bahagian A Bahagian B Bahagian C Jumlah markah
Jamil 12 7 10 59
Saiful 10 11 6 56
Hasan 16 9 4 64
Let x = the marks for each question in section A
y = the marks for each question in section B
z = the marks for each question in section C
Hence,
12x + 7y + 10z = 59
10x + 11y + 6z = 56
16x + 9y + 4z = 64
4 A stationery shop offers three sale packages for pencils, pens and erasers.
Sebuah kedai alat tulis menawarkan tiga pakej jualan untuk pensel, pen dan pemadam.
RM7.20 RM6.30 RM6.90
Write down three equations that satisfy the sale packages.
Tuliskan tiga persamaan yang memuaskan pakej jualan itu.
Let x = the price of a pencil
y = the price of a pen
z = the price of an eraser
Hence,
x + 3y + 2z = 7.20
4x + 2y + z = 6.30
2x + y + 3z = 6.90
39
Exercise 2 Solve each of the following.
Selesaikan setiap yang berikut.
TP 3 Mengaplikasikan kefahaman tentang sistem persamaan untuk melaksanakan tugasan mudah.
Example 2 1 x + 2y + z = 5
x + y + 2z = 8
x + 2y + z = 3 2x + y + z = 11
2x – y + 3z = 13
2x + 3y + 4z = 11 x + 2y + z = 5 ——— 1
x + y + 2z = 8 ——— 2
Solution 2x + y + z = 11 ——— 3
x + 2y + z = 3 ——— 1
2x – y + 3z = 13 ——— 2 Eliminate x from 1 and 2 ,
2x + 3y + 4z = 11 ——— 3 1 : x + 2y + z = 5
2 : x + y + 2z = 8 (–)
Eliminate z from 1 and 2 , y – z = –3 ——— 4
Hapuskan z daripada 1 dan 2 ,
3 × 1 : 3x + 6y + 3z = 9 Eliminate x from 2 and 3 ,
2 : 2x – y + 3z = 13 (–) 2 × 2 : 2x + 2y + 4z = 16
3 : 2x + y + z = 11 (–)
x + 7y = –4 ——— 4
y + 3z = 5 ——— 5
Eliminate z from 1 and 3 ,
Hapuskan z daripada 1 dan 3 , Solve 4 and 5 to find the values of y and z,
4 × 1 : 4x + 8y + 4z = 12 4 : y – z = –3
3 : 2x + 3y + 4z = 11 (–) 5 : y + 3z = 5 (–)
2x + 5y = 1 ——— 5 –4z = –8
z = 2
Solve 4 and 5 to find the values of x and y,
Selesaikan 4 dan 5 untuk mencari nilai x dan y, From 5 , y + 3(2) = 5
2 × 4 : 2x + 14y = –8 y = –1
5 : 2x + 5y = 1 (–)
From 1 , x + 2(–1) + (2) = 5
9y = –9 x – 2 + 2 = 5
y = –1 x = 5
From/Dari 5 , 2x + 5(–1) = 1 Hence, x = 5, y = –1, z = 2
2x – 5 = 1
2x = 6
x = 3
From/Dari 1 , (3) + 2(–1) + z = 3
3 – 2 + z = 3
z = 2
Hence/Oleh itu,
x = 3, y = –1, z = 2
Calculator Corner
A calculator can also be used to solve system of linear equations in three unknowns.
Kalkulator juga boleh digunakan untuk menyelesaikan persamaan linear dalam tiga pemboleh ubah.
Example:
x + 2y + z = 3
2x – y + 3z = 13
2x + 3y + 4z = 11
MODE MODE MODE
131=2=1=3=2=–1=
3 = 13 = 2 = 3 = 4 = 11 = = =
40
2 x + 4y + 2z = 1 3 x + y + 2z = 5
x – y – 3z = 6 x–y+z=3
x + 2y + z = 4 2x – y – z = 14
x + 4y + 2z = 1 ——— 1 x + y + 2z = 5 ——— 1
x – y – 3z = 6 ——— 2 x – y + z = 3 ——— 2
x + 2y + z = 4 ——— 3 2x – y – z = 14 ——— 3
Eliminate x from 1 and 2 , Eliminate y from 1 and 2 ,
1 : x + 4y + 2z = 1 1 : x + y + 2z = 5
2 : x – y – 3z = 6 (–) 2 : x – y + z = 3 (+)
5y + 5z = –5 2x + 3 z = 8 ——— 4
y + z = –1 ——— 4
Eliminate y from 2 and 3 ,
Eliminate x from 2 and 3 , 2 : x – y + z = 3
2 : x – y – 3z = 6 3 : 2x – y – z = 14 (–)
3 : x + 2y + z = 4
–3y – 4z = 2 ——— 5 –x + 2z = –11 ——— 5
Solve 4 and 5 to find the values of y and z, Solve 4 and 5 to find the values of x and z,
4 × 4 : 4y + 4z = –4 4 : 2x + 3z = 8
5 : –3y – 4z = 2 (+) 2 × 5 : –2x + 4z = –22 (+)
y = –2 7z = –14
z = –2
From 4 , (–2) + z = –1
z = 1 From 4 , 2x + 3(–2) = 8
2x – 6 = 8
From 1 , x + 4(–2) + 2(1) = 1 2x = 14
x – 8 + 2 = 1 x = 7
x = 7
From 1 , (7) + y + 2(–2) = 5
Hence, x = 7, y = –2, z = 1 y = 2
4 2x + y + z = 9 Hence, x =7, y = 2, z = –2
x–y+z=1
5 3x – y + 4z = 14
–x – 3y + 2z = 3 x + 6y – z = 2
2x + y + z = 9 ——— 1 2x – y + 3z = 10
x – y + z = 1 ——— 2
–x – 3y + 2z = 3 ——— 3 3x – y + 4z = 14 ——— 1
x + 6y – z = 2 ——— 2
Eliminate x from 1 and 2 , 2x – y + 3z = 10 ——— 3
1 : 2x + y + z = 9
2 × 2 : 2x – 2y + 2z = 2 (–) Eliminate x from 1 and 2 ,
3y – z = 7 ——— 4 1 : 3x – y + 4z = 14
Eliminate x from 2 and 3 , 3 × 2 : 3x + 18y – 3z = 6 (–)
2 : x – y + z = 1 –19y + 7z = 8 ——— 4
3 : –x – 3y + 2z = 3 (+)
–4y + 3z = 4 ——— 5 Eliminate x from 2 and 3 ,
Solve 4 and 5 to find the values of y and z, 2 × 2 : 2x + 12y – 2z = 4
3 × 4 : 9y – 3z = 21
5 : –4y + 3z = 4 (+) 3 : 2x – y + 3z = 10 (–)
13y – 5z = –6 ——— 5
5y = 25
y = 5 Solve 4 and 5 to find the values of y and z,
5 × 4 : –95y + 35z = 40
From 4 , 3(5) – z = 7 7 × 5 : 91y – 35z = -42 (+)
z = 8
–4y = –2
From 1 , 2x + (5) + (8) = 9 y = 0.5
2x = –4
x = –2 From 5 , 13(0.5) – 5z = –6
Hence, x = –2, y = 5, z = 8 5z = 12.5
z = 2.5
From 1 , 3x – (0.5) + 4(2.5) = 14
3x = 4.5
x = 1.5
Hence, x = 1.5, y = 0.5, z = 2.5
41
Smart Tip
System of linear equations in three variables are used to solve problems in daily life. The given problem is expressed
as a system of linear equations and then solved to determine the value of each variable. Sometimes, the system of
equations consists of three linear equations but not every linear equation involves three variables.
Sistem persamaan linear dalam tiga pemboleh ubah digunakan untuk menyelesaikan masalah dalam kehidupan seharian. Masalah yang diberi
diungkapkan sebagai satu sistem persamaan linear dan kemudian diselesaikan untuk menentukan nilai bagi setiap pemboleh ubah. Kadang-kadang,
sistem persamaan itu terdiri daripada tiga persamaan linear tetapi tidak semestinya setiap persamaan linear melibatkan tiga pemboleh ubah.
Exercise 3 Solve each of the following.
Selesaikan setiap yang berikut.
TP 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang sistem persamaan dalam konteks penyelesaian masalah rutin yang mudah.
Example 3
Siti buys 2 kg carrots, 1 kg cabbages and 3 kg potatoes for RM21.50, while Salmah buys 4 kg carrots, 1 kg
cabbages and 2 kg potatoes for RM27.50. Given the total price of 1 kg carrots and 1 kg cabbages exceeds the
price of 1 kg potatoes by RM5. Find the price for a kilogram of each type of vegetable, in RM.
Siti membeli 2 kg lobak merah, 1 kg kubis dan 3 kg ubi kentang dengan harga RM21.50, manakala Salmah membeli 4 kg lobak
merah, 1 kg kubis dan 2 kg ubi kentang dengan harga RM27.50. Diberi jumlah harga 1 kg lobak merah dan 1 kg kubis adalah
melebihi harga 1 kg ubi kentang sebanyak RM5. Cari harga untuk setiap kilogram bagi setiap jenis sayur, dalam RM.
Let the price of 1 kg carrots = x Solve 4 and 5 to find the values of x and z,
the price of 1 kg cabbages = y
the price of 1 kg potatoes = z Selesaikan 4 dan 5 untuk mencari nilai x dan z,
Katakan harga 1 kg lobak merah = x
harga 1 kg kubis = y 4 : –2x + z = –6
harga 1 kg ubi kentang = z
2x + y + 3z = 21.5 ——— 1 5 : x + z = 7.5 (–)
4x + y + 2z = 27.5 ——— 2
–3x = –13.5
x + y – z = 5 ——— 3
Eliminate y from 1 and 2 , x = 4.5
Hapuskan y daripada 1 dan 2 , From/Daripada 5 , (4.5) + z = 7.5
1 : 2x + y + 3z = 21.5 z = 3
2 : 4x + y + 2z = 27.5 (–)
From/Daripada 1 , 2(4.5) + y + 3(3) = 21.5
–2x + z = –6 ——— 4 y = 3.5
Eliminate y from 2 and 3 ,
Hence, the price of 1 kg carrots = RM4.50, the
Hapuskan y daripada 2 dan 3 , price of 1 kg cabbages = RM3.50 and the price of
2 : 4x + y + 2z = 27.5 1 kg potatoes = RM3.
3 : x + y – z = 5 (–) Maka, harga bagi 1 kg lobak merah = RM4.50, harga bagi
1 kg kubis = RM3.50 dan harga bagi 1 kg ubi kentang = RM3.
3x + 3z = 22.5
x + z = 7.5 ——— 5
1 Jamil, Minah and Chandran went to a shop to purchase some items to decorate their classroom. Jamil
bought two manila cards, three boxes of marker pens and four glue sticks for RM16.00. Minah bought six
manila cards, four boxes of marker pens and two glue sticks for RM23.00. Chandran bought two manila
cards, five boxes of marker pens and three glue sticks for RM21.00. Find the unit price of each item, in RM.
Jamil, Minah dan Chandran pergi ke sebuah kedai untuk membeli beberapa barang untuk menghias kelas mereka. Jamil membeli
dua keping kad manila, tiga kotak pen penanda dan empat botol gam dengan harga RM16.00. Minah membeli enam keping kad
manila, empat kotak pen penanda dan dua botol gam dengan harga RM23.00. Chandran membeli dua keping kad manila, lima
kotak pen penanda dan tiga botol gam dengan harga RM21.00. Cari harga seunit bagi setiap jenis barang tersebut, dalam RM.
Let the price of a manila card = x Solve 4 and 5 to find the values of x and y,
4 : 10x + 5y = 30
the price of a box of marker pens = y 5 × 5 : 10x + 55y = 180 (–)
the price of a glue stick = z –50y = –150
2x + 3y + 4z = 16 ——— 1 y = 3
6x + 4y + 2z = 23 ——— 2 From 4 , 10x + 5(3) = 30
2x + 5y + 3z = 21 ——— 3
Eliminate z from 1 and 2 , 10x = 15
x = 1.5
1 : 2x + 3y + 4z = 16 From 1 , 2(1.5) + 3(3) + 4z = 16
2 × 2 : 12x + 8y + 4z = 46 (–) 4z = 4
10x + 5y = 30 ——— 4 z = 1
Eliminate z from 1 and 3 , Hence, the price of a manila card = RM1.50, the
price of a box of marker pens = RM3 and the price
1 × 3 : 6x + 9y + 12z = 48 of a glue stick = RM1.
3 × 4 : 8x + 20y + 12z = 84 (–)
2x + 11y = 36 ——— 5
42
2 The table below shows the number of stamps bought by three students.
Jadual di bawah menunjukkan bilangan setem yang dibeli oleh tiga orang murid.
Student Number of stamps
Murid Bilangan setem
Ali Type P Type Q Type R Total price
Bala Jenis P Jenis Q Jenis R Jumlah harga
Chan
8 10 20 (RM)
30 15 25
20 10 50 23.60
35.00
50.00
Find the unit price of each type of stamp, in RM.
Cari harga seunit bagi setap jenis setem, dalam RM.
Let the price of a stamp type P = x Solve 4 and 5 to find the values of x and y,
the price of a stamp type Q = y
the price of a stamp type R = z 2 × 4 : 160x + 20y = 44
8x + 10y + 20z = 23.6 ——— 1 5 : 40x + 20y = 20 (–)
30x + 15y + 25z = 35 ——— 2
2 0x + 10y + 50z = 50 ——— 3 120x = 24
Eliminate z from 1 and 2 , x = 0.2
5 × 1 : 40x + 50y + 100z = 118
4 × 2 : 120x + 60y + 100z = 140 (–) From 4 , 80(0.2) + 10y = 22
16 + 10y = 22
–80x – 10y = –22 10y = 6
80x + 10y = 22 ——— 4 y = 0.6
Eliminate z from 2 and 3 , From 1 , 8(0.2) + 10(0.6) + 20z = 23.6
2 × 2 : 60x + 30y + 50z = 70 20z = 16
3 : 20x + 10y + 50z = 50 (–) z = 0.8
40x + 20y = 20 ——— 5 Hence, the price of a stamp type P = RM0.20, the
price of a stamp type Q = RM0.60 and the price of
a stamp type R = RM0.80.
3 The table below shows the number of sales for three types of handphones by three stores in a week.
Jadual di bawah menunjukkan bilangan jualan bagi tiga jenis telefon bimbit oleh tiga buah kedai dalam seminggu.
Store Handphone A Number of sales Total (RM)
Kedai Telefon bimbit A Bilangan jualan Jumlah (RM)
P 8 Handphone B Handphone C 18 400
Q Telefon bimbit B Telefon bimbit C
R 2 22 300
54
4 22 400
69
68
Calculate the unit price of each type of handphone, in RM.
Hitung harga seunit bagi setiap jenis telefon bimbit, dalam RM.
Let the price of 1 unit handphone A = x Solve 4 and 5 to find the values of y and z,
the price of 1 unit handphone B = y 4 : –19y – 32z = –70 800
the price of 1 unit handphone C = z 3.2 × 5 : 19.2y + 32z = 71 040 (+)
8x + 5y + 4z = 18 400 ——— 1 0.2y = 240
2x + 6y + 9z = 22 300 ——— 2 y = 1 200
4x + 6y + 8z = 22 400 ——— 3
From 5 , 6(1 200) + 10z = 22 200
Eliminate x from 1 and 2 , 10z = 15 000
1 : 8x + 5y + 4z = 18 400 z = 1 500
4 × 2 : 8x + 24y + 36z = 89 200 (–)
–19y – 32z = –70 800 ——— 4 From 1 , 8x + 5(1 200) + 4(1 500) = 18 400
8x = 6 400
Eliminate x from 2 and 3 , x = 800
2 × 2 : 4x + 12y + 18z = 44 600
3 : 4x + 6y + 8z = 22 400 (–) Hence the price of 1 unit handphone A = RM800,
the price of 1 unit handphone B = RM1 200 and the
6y + 10z = 22 200 ——— 5 price of 1 unit handphone C = RM1 500.
43
4 A souvenir shop offers three souvenir packages to customers as shown in the table below.
Sebuah kedai cenderamata menawarkan tiga pakej cenderamata kepada pelanggan seperti ditunjukkan dalam jadual di bawah.
RM92 RM101 RM111
Calculate the unit price of each of the item, in RM.
Hitung harga seunit bagi setiap barang tersebut, dalam RM.
Let the price of a cap = x Solve 4 and 5 to find the values of x and y,
the price of a pair of slippers = y 4 : x – y = –9
the price of a shirt = z 5 : x + 3y = 91 (−)
–4y = –100
2x + y + z = 92 ——— 1 y = 25
x + 2y + z = 101 ——— 2
x + y + 2z = 111 ——— 3 From 5 , x + 3(25) = 91
x = 16
Eliminate z from 1 and 2 ,
1 : 2x + y + z = 92 From 1 , 2(16) + (25) + z = 92
2 : x + 2y + z = 101 (−) z = 35
x – y = – 9 ——— 4 Hence, the price of a cap = RM16, the price of a
pair of slippers = RM25 and the price of a shirt
Eliminate z from 2 and 3 , = RM35.
2 × 2 : 2x + 4y + 2z = 202
3 : x + y + 2z = 111 (−)
x + 3y = 91 ——— 5
3.2 Simultaneous Equations Involving One Linear Equation and One Non-Linear Equation
Persamaan Serentak yang Melibatkan Satu Persamaan Linear dan Satu Persamaan Tak Linear
Exercise 4 State whether each of the following equations is a linear or a non-linear equation.
Nyatakan sama ada setiap persamaan yang berikut ialah persamaan linear atau persamaan tak linear.
TP 1 Mempamerkan pengetahuan asas tentang sistem persamaan.
Example 4 1 x2 − 6 = y 2 x + y = 16
2x + 3y = 5 Non-linear Linear
Solution
Linear
3 2 + 2y = x 4 y2 − 17 = x 5 30 − 5x = –7y
x Non-linear Linear
2 + 2xy = x2
Non-linear
44
Exercise 5 Solve each of the following simultaneous equations.
Selesaikan setiap persamaan serentak yang berikut.
TP 3 Mengaplikasikan kefahaman tentang sistem persamaan untuk melaksanakan tugasan mudah.
Smart Tip
Steps to solve simultaneous equations: /Langkah-langkah untuk menyelesaikan persamaan serentak:
i-THINK Flow Map
(1) Identify the linear (2) Make one of the variables (3) Substitute the variable into the non- PAK-21
equation. as the subject of the linear equation, giving a quadratic
Kenal pasti persamaan equation. equation in one variable.
linear. Jadikan salah satu pemboleh Gantikan pemboleh ubah itu dalam persamaan
ubah sebagai subjek persamaan. tak linear, menjadikan satu persamaan kuadratik
dengan satu pemboleh ubah.
(4) Solve the quadratic equation using factorisation (5) Substitute the solution in (4) one by one QR CODE
into the linear equation to obtain the other
or formula x/y = –b ± b2 − 4ac . related solutions.
2a Gantikan penyelesaian di (4) satu demi satu ke
dalam persamaan linear untuk mendapatkan
Selesaikan persamaan kuadratik itu menggunakan penyelesaian lain yang berkaitan.
pemfaktoran atau formula x/y = –b ± b2 − 4ac
2a .
Example 5 Step 4/Langkah 4
x + 3y = 5 (11y − 19)(y − 1) = 0
x2 + 2y2 = 6
11y − 19 = 0 or y − 1 = 0
Solution
Step 1/Langkah 1 y = 19 y = 1 QR CODE
x + 3y = 5 ——— 1 11
x2 + 2y2 = 6 ——— 2 Scan or visit
Step 5/Langkah 5 19 , https://goo.gl/
Step 2/Langkah 2 When/Apabila y = 11 m6Se9S for
From/Daripada 1 ,
x = 5 − 3y x = 5 − 311912 = – 121 additional
notes on the
Step 3/Langkah 3 When/Apabila y = 1,
Substitute into 2 , x = 5 − 3(1) = 2 methods
Gantikan ke dalam 2 , of solving
Hence,/Oleh itu, non-linear
(5 − 3y)2 + 2y2 = 6 equations.
25 − 30y + 9y2 + 2y2 = 6
11y2 − 30y + 19 = 0 ——— 3 x = – 121, y = 19
11
or/atau
x = 2, y = 1
Smart Tip When/Apabila x = – 121 , y = 19
11
Always substitute the answers back into the non-linear
equation for checking. – 1212 + 2 11912 = 6
Sentiasa ganti semula jawapan ke dalam persamaan tak linear untuk
semakan. When/Apabila x = 2, y = 1
(2)2 + 2(1)2 = 6
x2 + 2y2 = 6 (non-linear/tak linear)
The answers are correct.
Jawapan adalah betul.
45
1 2x + y = 5 2 x – y = 3 3 x + y = 5
xy = 4 xy – 2y = 2
xy = 1
4 2 x – y = 3 ——— 1 x + y = 5 ——— 1
xy = 4 ——— 2 xy – 2y = 2 ——— 2
2x + y = 5 ——— 1
From 1 ,
xy = 1 ——— 2 x=3+y From 1 ,
4 2 Substitute into 2 ,
(3 + y)y = 4 x=5–y
From 1 , 3y + y2 = 4
y2 + 3y – 4 = 0 Substitute into 2 ,
y = 5 − 2x (y – 1)(y + 4) = 0
y – 1 = 0 or y + 4 = 0 (5 – y)y – 2y = 2
Substitute into 2 ,
y = 1 y = –4 5y – y2 – 2y = 2
x(5 − 2x) = 1 When y = 1, x = 3 + (1) = 4 –y2 + 3y – 2 = 0
4 2 When y = –4, x = 3 + (–4) = –1
y2 − 3y + 2 = 0
Hence,
5x – 2x2 = 2 x = 4, y = 1 or x = –1, y = –4 (y – 1)(y – 2) = 0
2x2 – 5x + 2 = 0 y – 1 = 0 or y – 2 = 0
(2x – 1)(x – 2) = 0 y = 1 y = 2
2x – 1 = 0 or x – 2 = 0 When y = 1, x = 5 – (1) = 4
When y = 2, x = 5 – (2) = 3
x = 1 x = 2
2
When x = 1 , y = 5 – 2 1 =4 Hence,
2 2 x = 4, y = 1 or x = 3, y = 2
When x = 2, y = 5 – 2(2) = 1
Hence,
x = 1 , y = 4 or x = 2, y = 1
2
4 x − y = 2 5 3x + y = 5 6 2x – y = 3
x2 + xy + y = 13 x2 – y2 = –3 x2 – 3xy + y2 = 5
x – y = 2 ——— 1 3x + y = 5 ——— 1 2x – y = 3 ——— 1
x2 + xy + y = 13 ——— 2 x2 – y2 = –3 ——— 2 x2 – 3xy + y2 = 5 ——— 2
From 1 , From 1 , From 1 ,
x=2+y y = 5 − 3x
Substitute into 2 , Substitute into 2 , y = 2x − 3
(2 + y)2 + (2 + y)y + y = 13 x2 – (5 − 3x)2 = –3
4 + 4y + y2 + 2y + y2 + y = 13 x2 – (25 − 30x + 9x2) = –3 Substitute into 2 ,
2y2 + 7y − 9 = 0 x2 – 25 + 30x – 9x2 = –3
(2y + 9)(y − 1) = 0 –8x2 + 30x − 22 = 0 [÷ (–2)] x2 – 3x(2x – 3) + (2x – 3)2 = 5
2y + 9 = 0 or y − 1 = 0
x2 – 6x2 + 9x + 4x2 – 12x + 9 = 5
–x2 – 3x + 4 = 0 [× (–1)]
x2 + 3x – 4 = 0
4x2 – 15x + 11 = 0 (x – 1)(x + 4) = 0
y = – 92 y = 1 (4x – 11)(x − 1) = 0 x – 1 = 0 or x + 4 = 0
4x – 11 = 0 or x − 1 = 0 x = 1 x = –4
When y = – 92, x = 11 x = 1 When x = 1, y = 2(1) − 3 = –1
4
x = 2 + – 29 = – 52 141, When x = –4, y = 2(–4) – 3 =
5–
Whenx= –11
y =
When y = 1, x = 2 + (1) = 3 3 11 = – 143 Hence,
4 x = 1, y = –1 or x = –4, y = –11
Hence, When x = 1, y = 5 − 3(1) = 2
x = – 52, y = – 92 or x = 3, y = 1
Hence,
x = 141, y = – 143 or x = 1, y = 2
46
Exercise 6 Solve each of the following.
Selesaikan setiap yang berikut.
TP 3 Mengaplikasikan kefahaman tentang sistem persamaan untuk melaksanakan tugasan mudah.
Example 6 1 x − y = 2 2 x + y = 1
2x − y = 3 3 + 2y = 3 x + 3 = 3
x y
4 3
x + y = 5 x − y = 2 ——— 1 x + y = 1 ——— 1
Solution 3 + 2y = 3 ——— 2 x + 3 = 3 ——— 2
2x − y = 3 ——— 1 x y
4 + 3 = 5 ——— 2 From 2 , From 2 ,
x y 3 + 2xy = 3x ——— 3 xy + 3 = 3y ——— 3
From 1 ,
2 × (xy), From 1 ,
4y + 3x = 5xy ——— 3 y=x−2
Substitute into 3 , x=1−y
From/Daripada 1 , Substitute into 3 ,
3 + 2x(x − 2) = 3x (1 − y)y + 3 = 3y
y = 2x − 3 3 + 2x2 − 4x − 3x = 0 y − y2 + 3 − 3y = 0
Substitute into 3 ,
2x2 − 7x + 3 = 0 –y2 − 2y + 3 = 0
Gantikan ke dalam 3 , (2x − 1)(x − 3) = 0 y2 + 2y − 3 = 0
4(2x − 3) + 3x = 5x(2x – 3) 2x − 1 = 0 or x − 3 = 0 (y − 1)(y + 3) = 0
8x − 12 + 3x = 10x2 − 15x x = 1 x = 3 y − 1 = 0 or y + 3 = 0
2
0 = 10x2 − 15x − 8x + 12 − 3x 1 21 – 32 y = 1 y = –3
2
10x2 − 26x + 12 = 0 (÷2) When x = , y = − 2 = When y = 1, x = 1 − (1) = 0
5x2 − 13x + 6 = 0 When x = 3, y = (3) − 2 = 1 When x = –3, x = 1 − (–3) = 4
(5x − 3)(x − 2) = 0 Hence,
5x − 3 = 0 or/atau x − 2 = 0 Hence, x = 0, y = 1
x = 3 x = 2 1 – 23 or
5 2
x = , y = x = 4, y = –3
When/Apabila x = 3 , or
5
3 – 59 x = 3, y = 1
y = 2 5 − 3 =
When/Apabila x = 2,
y = 2(2) − 3 = 1
Hence/Oleh itu,
x = 3 , y = – 59
5
or/atau
x = 2, y = 1
3 x + 3y = 5
2 + 1 = 2
x y
x + 3y = 5 ——— 1 From 1 , When y = 5 , x = 5 − 3 5 = 5
6 6 2
2 + 1 = 2 ——— 2 x = 5 − 3y
x y
Substitute into 3 , When y = 1, x = 5 − 3(1) = 2
2 × (xy), 2y + (5 − 3y) = 2(5 − 3y)y Hence,
2y + x = 2xy ——— 3
2y + 5 − 3y = 10y − 6y2 x = 5 , y = 5
2 6
6y2 − 11y + 5 = 0
(6y − 5)(y − 1) = 0 or
6 y − 5 = 0 or y − 1 = 0 x = 2, y = 1
y = 5 y = 1
6
47
4 x + 2y = 7 5 2x + y = 2 6 x + 3y = 1
2 − 3 = 1 3 − 2 = 1 x(x − y)
x y 2x y 8 =0
x + 2y = 7 ——— 1 2x + y = 2 ——— 1 x + 3y = 1 ——— 1
2 − 3 = 1 ——— 2 3 − 2 = 1 ——— 2 x(x − y)
x y 2x y 8 =0
2 × (xy), 2 × (2xy), x2 − xy = 0 ——— 2
2y − 3x = xy ——— 3 3y − 4x = 2xy ——— 3
From 1 ,
From 1 , From 1 , x = 1 − 3y
x = 7 − 2y y = 2 − 2x Substitute into 2 ,
Substitute into 3 , Substitute into 3 ,
2y − 3(7 − 2y) = (7 − 2y)y 3(2 − 2x) − 4x = 2x(2 − 2x) (1 − 3y)2 − (1 − 3y)y = 0
2y − 21 + 6y = 7y − 2y2 6 − 6x − 4x = 4x − 4x2
2y − 21 + 6y − 7y + 2y2 = 0 6 − 6x − 4x − 4x + 4x2 = 0 1 − 6y + 9y2 − y + 3y2 = 0
2y2 + y − 21 = 0 4x2 − 14x + 6 = 0
2x2 − 7x + 3 = 0 12y2 − 7y + 1 = 0
(3y − 1)(4y − 1) = 0
3y − 1 = 0 or 4y − 1 = 0
(2y + 7)(y − 3) = 0 y = 1 y = 1
3 4
2y + 7 = 0 or y – 3 = 0 (2x − 1)(x − 3) = 0
2x − 1 = 0 or x − 3 = 0 When 1 1
y = – 27 y = 3 y = 3 , x = 1 − 3 3 =0
When y = – 27,
PAK-21 x = 1 x = 3 When y = 1 , x = 1 − 3 1 = 1
2 4 4 4
ACTIVITY x = 7 − 2– 72 = 14 When x = 1 , y = 2 − 2 1 =1 Therefore,
2 2
When y = 3, x = 7 − 2(3) = 1 When x = 3, y = 2 − 2(3) = –4 x = 0, y = 1
Hence, 3
or
Hence, – 72 1
x = 2 , y = 1
x= 14, y = x = 1 , y = 1
or 4 4
or
x = 1, y = 3 x = 3, y = –4
ACTIVITY PAK-21 Gallery Walk
Steps/Langkah-langkah:
1 Teacher is required to provide questions on Systems of Equations on the coloured cards, where each card
has 3 questions.
Guru dikehendaki menyediakan beberapa soalan mengenai Sistem Persamaan pada kad berwarna, dengan
setiap kad perlu mempunyai 3 soalan.
2 Students perform this activity in groups of 3 students. A coloured card is randomly selected for each group.
Murid-murid melakukan aktiviti ini secara berkumpulan yang terdiri daripada 3 orang murid. Satu kad
berwarna dipilih secara rawak bagi setiap kumpulan.
3 Each group is required to answer all questions on the selected card. Write each answer on a mahjung paper.
Setiap kumpulan dikehendaki menjawab semua soalan yang terdapat pada kad yang dipilih. Tulis setiap
jawapan pada kertas mahjung.
4 The group work of each group is posted on the class’s notice board. Students are required to stand next to
their group work.
Hasil kerja setiap kumpulan ditampal pada papan kenyataan kelas. Murid-murid dikehendaki berdiri di
sebelah hasil kerja masing-masing.
5 A group is required to move to each group to evaluate the work of the group. Once completed, other groups
need to do the same step.
Satu kumpulan perlu bergerak ke setiap kumpulan bagi menilai hasil kerja kumpulan tersebut. Setelah
selesai, kumpulan lain perlu melakukan langkah yang sama.
6 Teacher hold a discussion with students to enhance the students’ understanding on the Systems of Equations.
Guru mengadakan perbincangan dengan murid bagi menambahkan kefahaman murid bagi tajuk Sistem
Persamaan Serentak.
48
Exercise 7 Solve each of the following simultaneous equations. Give the answer correct to 3 decimal
places.
Selesaikan setiap persamaan serentak yang berikut. Beri jawapan betul kepada 3 tempat perpuluhan.
TP 3 Mengaplikasikan kefahaman tentang sistem persamaan untuk melaksanakan tugasan mudah.
Example 7
2x + 3y = 1 a = 5, b = –6, c = –7 When/Apabila y = 1.927,
x2 − y2 = 2
y = –b ± b2 − 4ac x = 1 − 3(1.927) = –2.391
2a 2
Solution
2x + 3y = 1 ——— 1 y = –(–6) ± (–6)2 − 4(5)(–7) Hence/Oleh itu,
x2 − y2 = 2 ——— 2 2(5)
x = 1.591, y = –0.727
From/Daripada 1 , = 6 ± 176 or/atau
2x = 1 − 3y 10 x = –2.391, y = 1.927
1 − 3y = 6 − 176 or/atau 6 + 176 Smart Tip
2 10 10
x =
Substitute into 2 , = –0.7266 or/atau 1.9266 Formula y or x = –b ± b2 − 4ac is applied
2a
Gantikan ke dalam 2 , = –0.727 or/atau 1.927
when the quadratic equation cannot be
1 − 3y 2 y2 = 2 When/Apabila y = –0.727, solved by factorisation.
2
− 1 – 3(–0.727) –b ± b2 – 4ac
2 Formula y atau x = 2a digunakan apabila
1 – 6y + 9y2 x = = 1.591 persamaan kuadratik tidak boleh diselesaikan
4 menggunakan pemfaktoran.
− y2 = 2 — (×4)
1 − 6y + 9y2 − 4y2 = 8
5y2 − 6y − 7 = 0
1 2x + y = 3 2 x + 2y = 1 3 x + 2y = x2 − y2 = 3
3x + y2 = 6 x2 − y = 3 x + 2y = 3 ——— 1
x2 − y2 = 3 ——— 2
2x + y = 3 ——— 1 x + 2y = 1 ——— 1
3x + y2 = 6 ——— 2 x2 − y = 3 ——— 2
From 1 , From 1 ,
y = 3 − 2x
Substitute into 2 , From 1 , x = 3 − 2y
3x + (3 − 2x)2 = 6 Substitute into 2 ,
3x + 9 − 12x + 4x2 = 6 x = 1 − 2y (3 – 2y)2 − y2 = 3
4x2 − 9x + 3 = 0 Substitute into 2 , 9 − 12y + 4y2 − y2 = 3
(1 − 2y)2 − y = 3
1 − 4y + 4y2 − y − 3 = 0 3y2 − 12y + 6 = 0
–(–9) ± (–9)2 − 4(4)(3) 4y2 − 5y − 2 = 0 y2 − 4y + 2 = 0
2(4)
x = y = –(–5) ± (–5)2 − 4(4)(–2) y = –(–4) ± (–4)2 − 4(1)(2)
2(4) 2(1)
9 ± 33
= 8 5 ± 57 4 ± 8
= 8 = 2
= 9 − 33 or 9 + 33 5 − 57 5 + 57 4 – 8 4 + 8
8 8 8 8 2 2
= or = or
= 0.4069 or 1.8430 = –0.319 or 1.569 = 0.586 or 3.414
= 0.407 or 1.843
When y = 0.586,
When y = –0.319,
When x = 0.407, x = 1 − 2(–0.319) = 1.638 x = 3 − 2(0.586) = 1.828
y = 3 − 2(0.407) = 2.186 When y = 1.569,
When x = 1.843, x = 1 − 2(1.569) = –2.138 When y = 3.414,
y = 3 − 2(1.843) = –0.686
x = 3 − 2(3.414) = –3.828
Hence, Hence, Hence,
x = 0.407 , y = 2.186 x = 1.638, y = –0.319 x = 1.828, y = 0.586
or or or
x = 1.843, y = –0.686 x = –2.138, y = 1.569 x = –3.828, y = 3.414
49
4 4x + 3y = 1 5 2x − 3y = 1
–x2 + y2 = 2 4xy − x2 = 6
4x + 3y = 1 ——— 1 2x − 3y = 1 ——— 1
–x2 + y2 = 2 ——— 2 4xy − x2 = 6 ——— 2
From 1 ,
From 1 ,
y = 1 − 4x 3y = 2x –1
3
2x − 1
Substitute into 2 , y = 3
–x2 + 1 − 4x 2 =2 Substitute into 2 ,
3
2x − 1
3
–x2 + 1 − 8x + 16x2 = 2 — (×9) 4x − x2 = 6 —— (×3)
9
4x(2x − 1) − 3x2 = 18
–9x2 + 1 − 8x + 16x2 = 18
8x2 − 4x − 3x2 = 18
7x2 – 8x − 17 = 0
5x2 − 4x − 18 = 0
–(–8) ± (–8)2 − 4(7)(–17)
x = 2(7) x = –(–4) ± (–4)2 − 4(5)(–18)
2(5)
8 ± 540
= 14 = 4 ± 376
10
540 8 + 540
= 8 − 14 or 14 = 4 − 376 4 + 376
10 10
or
= –1.088 or 2.231 = –1.539 or x = 2.339
When x = –1.088, When x = –1.539,
y = 1 − 4(–1.088) = 1.784 y = 2(–1.539) − 1 = –1.359
3 3
When y = 2.231, When x = 2.339,
y = 1 − 4(2.231) = –2.641 y= 2(2.339) − 1 = 1.226
3 3
Hence, Hence,
x = –1.088, y = 1.784
or x = –1.539, y = –1.359
x = 2.231, y = –2.641
or
x = 2.339, y = 1.226
Exercise 8 Solve each of the following problems. HOTS Applying
Selesaikan setiap masalah yang berikut.
TP 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang sistem persamaan dalam konteks penyelesaian masalah rutin yang mudah.
Example 8
The diagram shows a right-angled triangle. From/Daripada 1 , x = 33 − 2y
Rajah menunjukkan sebuah segi tiga bersudut tegak. 3
P Substitute x into/Gantikan x ke dalam 2 ,
33 − 2y y + 2y = 54
3
x+2 2x + 1
QR 33y − 2y2 + 6y = 162
2y –2y2 + 39y − 162 = 0
2y2 − 39y + 162 = 0
Given the perimeter is 36 cm and the area is 54 cm2. (y − 6)(2y − 27) = 0
Find the values of x and y.
Diberi perimeter ialah 36 cm dan luas ialah 54 cm2. y − 6 = 0 or 2y − 27 = 0
Cari nilai x dan y.
y = 6 y = 27
2
Solution
When/Apabila y = 6,
Perimeter = 36 cm
x = 33 − 2(6) = 7
(x + 2) + (2x + 1) + 2y = 36 3
3x + 2y = 33 ——— 1 When/Apabila y = 227,
Area/Luas = 54 cm2 33 − 2227
1 × (x + 2)(2y) = 54 2 x = 3 = 6 =2
2 xy + 2y = 54 ——— 3
50
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Whole Book Answers f4
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