Page 19 - Modul A+1 Matematik Tambahan Tingkatan 4

2.1 Persamaan dan Ketaksamaan Kuadratik Buku Teks
Quadratic Equations and Inequalities m.s. 36-44

1. Selesaikan persamaan kuadratik yang berikut dengan menggunakan kaedah penyempurnaan kuasa dua. Berikan jawapan
betul kepada dua tempat perpuluhan.
Solve the following quadratic equations by using completing the square method. Give answers correct to two decimal places. TP 2

Contoh/Example
(i) x – 6x + 3 = 0 (ii) –2x – 4x + 1 = 0
2
2
Tip SPM Tip SPM BAB 2
1
–6
2
–6
2
x – 6x +   2 = –3 +   2 1. Pindahkan sebutan x + 2x – = 0 Bahagikan kedua-dua
2
belah persamaan dengan
pemalar ke sebelah
2
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2
1
2
2
2
x – 6x + (−3) = –3 + (−3) 2 kanan. x + 2x +   2 = +   2 –2 supaya pekali bagi x
2
2
2
2
2
2
(x – 3) = –3 + 9 Move the constant adalah 1.
2
Divide both sides of the
term to the right hand
1
(x – 3) = 6 side. (x + 1) = + 1 equation by –2 so that the
2
2
x – 3 = ±AB 6 2. Tambahkan sebutan 2 coefficient of x is 1.
2
x = 3 ± AB 6 pekali x 2 di (x + 1) = 3
2
x = 5.45 atau/or x = 0.55  2  2
3
kedua-dua belah x + 1 = ± 
Tip SPM persamaan. 2
3
Add the term x = –1 ± 
Perhatikan bahawa  coefficient of x  2 to 2
2
Note that both sides of the x = –2.22 atau/or x = 0.22
(x + a) = x + 2ax + a 2 equation.
2
2
(a) x + 4x + 1 = 0 (b) x + 7x + 3 = 0
2
2
4
4
7
7
x + 4x +   2 = –1 +   2 x + 7x +   2 = –3 +   2
2
2
2
2
2
2
(x + 2) = –1 + 4  x + 7  2 = –3 + 49 = 37
2
(x + 2) = 3 2 4 4
2
7
x = –2 ± AB 3 x = – ± A 37
x = – 0.27 atau x = –3.73 2 4
x = –0.46 atau x = –6.54
(c) –x + 6x + 4 = 0 (d) 2x – 5x + 2 = 0
2
2
5 –5 2 –5 2
x – 6x – 4 = 0 x – x +   = –1 +  
2
2
4
4
2
–6
–6
x – 6x +   2 = 4 +   2  x – 5  2 = –1 + 25 = 9
2
2
2
(x – 3) = 4 + 9 = 13 4 5 16 9 16
2
13
x = 3 ± ABBB x = ± A 16
4
x = 6.61 atau x = –0.61 x = 0.5 atau x = 2
2
2
(e) –2x + 6x + 1 = 0 (f) 4x – 8x + 2 = 0
1 1
2
2
x – 3x – = 0 x – 2x + = 0
2
2
–2
–3
1
–3
1
–2
x – 3x +   2 = +   2 x – 2x +   2 = – +   2
2
2
2
2
2
2
2
2
9
1
 x – 3  2 = + = 11 (x – 1) = 1
2
2
2
2
4
4
3 11 1
x = ± A 4 x = 1 ± A 2
2
x = –0.16 atau x = 3.16 x = 1.71 atau x = 0.29
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