Page 22 - Modul A+1 MM Tingkatan 4

3. Selesaikan persamaan kuadratik berikut.
Solve the following quadratic equations. TP 3
(a) (x + 2)(x + 1) = 3(3x – 1) (b) x + 1 = (3x – 7) 2
x + 3x + 2 = 9x – 3 x + 1 = 9x – 42x + 49
2
2
x – 6x + 5 = 0 9x – 43x + 48 = 0
2
2
(x – 1)(x – 5) = 0 (9x – 16)(x – 3) = 0
x – 1 = 0 atau/or x – 5 = 0 9x – 16 = 0 atau/or x – 3 = 0
x = 1 atau/or x = 5 16
x = atau/or x = 3
9
4. (a) Selesaikan.
Solve. TP 3
(i) 12t – 31t – 15 = 0 (ii) 5w + 27w + 8 = 0
2
2
(12t + 5)(t – 3) = 0 (5w + 7)(w + 4) = 0
12t + 5 = 0 atau/or t – 3 = 0 5w + 7 = 0 atau/or w + 4 = 0
t = – 5 atau/or t = 3 7
12 w = – atau/or w = –4
5

(b) Dalam rajah yang diberi, PQRS ialah sebuah segi empat tepat. Sebuah bulatan dan dua Q R
buah semi bulatan yang kongruen terterap dalam segi empat tepat itu. Jejari setiap semi
bulatan itu ialah 1 cm. 2x cm
In the diagram given, PQRS is a rectangle. A circle and two congruent semicircles are inscribed in
the rectangle. The radius of each semicircle is 1 cm. P S
Ungkapkan
Express TP 5 (ii) luas kawasan berlorek, dalam bentuk ax + bx + c.
Lembaran PBD ©PAN ASIA PUBLICATIONS
(i) panjang PS, dalam sebutan x,
2
the length of PS, in terms of x,
the area of the shaded region, in the form ax + bx + c.

2
Luas kawasan berlorek

PS = 1 + 2x + 1
Area of the shaded region

= (2 + 2x) cm

= (2 + 2x)(2x) – p(1) – px
2
2
2
= [(4 – p)x + 4x – p] cm
= 4x + 4x – p – px 2 2
2
5. Rajah yang diberi menunjukkan suatu pepejal yang terdiri daripada gabungan sebuah kuboid
dan sebuah piramid tegak. Tinggi pepejal itu ialah (x + 8) cm.
Diagram given shows a solid consisting of the combination of a cuboid and a right pyramid. The height 3 cm
of the solid is (x + 8) cm.
12 cm
(4x – 5) cm
(a) Bentukkan satu ungkapan kuadratik, dalam bentuk (b) Jika isi padu pepejal itu ialah 476 cm , hitung nilai x.

3
ax + bx + c, bagi isi padu pepejal itu. If the volume of the solid is 476 cm , calculate the value of x.
3
2
Form a quadratic equation, in the form ax + bx + c, for the TP 4
2
volume of the solid. TP 2
16x + 204x – 280 = 476
2
Isi padu pepejal 16x + 204x – 756 = 0
2
Volume of solid 4x + 51x – 189 = 0
2
1 (4x + 63)(x – 3) = 0
= (4x – 5)(12)(3) + (4x – 5)(12)(x + 5)
3
= 144x – 180 + 4(4x + 20x – 5x – 25) x = – 63 atau/or x = 3
2
4
= 144x – 180 + 16x + 60x – 100 5
2
= (16x + 204x – 280) cm 3 x  , x = 3
2
4
MG-10