Page 32 - Hybrid PBD 2022 Form 4 Additional Mathematics

Matematik Tambahan Tingkatan 4 Bab 2 Fungsi Kuadratik
BAB

2 Fungsi Kuadratik

Quadratic Functions

PBD
PBD 2.1 Persamaan dan Ketaksamaan Kuadratik Buku Teks
PBD
Quadratic Equations and Inequalities ms. 36 – 44
FOKUS TOPIK
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1. Bentuk am persamaan kuadratik 4. Ketaksamaan kuadratik boleh diselesaikan menggunakan
General form of a quadratic equation kaedah
ax + bx + c = 0 Quadratic inequality can be solved using methods of
2
dengan keadaan a, b dan c ialah pemalar dan a ≠ 0. (a) lakaran graf
where a, b and c are constants and a ≠ 0. graph sketching
2. Penyelesaian atau punca-punca bagi persamaan (b) garis nombor

number line
kuadratik boleh diperoleh dengan menggunakan (c) jadual
The solutions or roots of quadratic equation can be obtained by table
using
(a) penyempurnaan kuasa dua / completing the square 5. Bagi persamaan kuadratik dalam bentuk (x − a)(x − b) = 0,
(b) kaedah rumus / formula method dengan a  b,
–b ± For quadratic equation in the form of (x − a)(x − b) = 0, where a  b,
2
– 4
ac
b
x = (a) jika / if
2a (x − a)(x − b)  0,
3. Persamaan kuadratik dengan punca-punca α dan β maka / then
Quadratic equation with roots α and β
x  a atau / or x  b,
2
x − (hasil tambah punca)x + (hasil darab punca) = 0 (b) jika / if
x − (sum of roots)x + (product of roots) = 0
2
dengan keadaan / where (x − a)(x − b)  0,
hasil tambah punca / sum of roots = α + β = – b maka / then
a
c a  x  b.
hasil darab punca / product of roots = αβ =
a
1. Selesaikan setiap persamaan kuadratik berikut dengan menggunakan kaedah penyempurnaan kuasa dua.
Solve each of the following quadratic equations by using completing the square method. SP 2.1.1 TP3
x – 6x + 9 = 0 Tambah dan tolak sebutan  pekali x 2 .
2

2
–6
–6
2
2
x – 6x +     2 + 9 = 0 Add and subtract term  coefficient of x 2  . Tip Penting
–
2
2
2
2
(x + a) = x + 2ax + a 2 Dalam penyempurnaan kuasa dua,
2
pekali x mesti bernilai 1.
2
x – 6x + (–3) – 9 + 9 = 0 In completing the square, the coefficient
2
2
(x – 3) = 0 of x must be 1.
2
2
x – 3 = 0
x = 3
(a) x + 6x – 2 = 0 (b) –x – 4x + 3 = 0
2
2
2
6
2
6
x + 6x +     2 – 2 = 0 4 2 x + 4x – 3 = 0
2
–
2
2
2
4
–
2
2
(x + 3) – 9 – 2 = 0 x + 4x +     – 3 = 0
2
2
2
(x + 3) = 11 (x + 2) – 4 – 3 = 0
2
2
x + 3 = ±11 (x + 2) = 7
7
x = 0.317, –6.317 x + 2 = ±
x = 0.646, –4.646
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