Page 33 - Hybrid PBD 2022 Form 4 Additional Mathematics

Matematik Tambahan Tingkatan 4 Bab 2 Fungsi Kuadratik

(c) 2x – 5x – 2 = 0 (d) 2x + 8x – 5 = 0
2
2
5
5
x – x – 1 = 0 x + 4x – = 0
2
2
2 2
5
5
   
x – x + – 5 2 – – 5 2 – 1 = 0 x + 4x + 2 – 2 – = 0
2
2
2
2
2 4 4 2
 x – 5  2 = 41 (x + 2) = 13
2
2
4
16
41
5
13
x – = ±  x + 2 = ± 
4 16 2
x = 2.851, –0.351 x = 0.550, –4.550
2
(e) 3x – 5x = 7 (f) 2x – 3x = –9
2
5
2
x – x = 7 x – x = 3
2
2
3 3 3
5
2
   
   
x – x + – 5 2 – – 5 2 = 7 x – x + – 1 2 – – 1 2 = 3
2
2
3 6 6 3 3 3 3
7
 x – 5  2 = + 25  x – 1  2 = 3 + 1
6
9
3
36
3
28
1
109
5
x – = ±  x – = ± 
6 36 3 9
x = 2.573, –0.907 x = 2.097, –1.431
2. Selesaikan setiap persamaan kuadratik berikut dengan menggunakan rumus. SP 2.1.1 TP3
Solve each of the following quadratic equations by using formula.
(a) x + 3x – 5 = 0 (b) 4x – 7x – 2 = 0
2
2
2
5x – 3x – 1 = 0
3 ± x = –3 ±9 – 4(1)(–5) x = 7 ±49 – 4(4)(–2)
9 – 4(5)(–1)
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2(4)
2(1)
2(5)
29
29
3 ± Tip Penting = –3 ± = 7 ±81
= 2 8
10 –b ± = 1.193, –4.193 = 2, –0.25
ac
– 4
b
2
= 0.839, –0.239 x = 2a
(c) 5 – x – 2x = 0 (d) 2x + 4x – 3 = 0 (e) 2x – 7x + 4 = 0
2
2
2
x = 1 ±1 – 4(–2)(5) x = –4 ±16 – 4(2)(–3) x = 7 ±49 – 4(2)(4)
2(–2) 2(2) 2(2)
17
= 1 ±41 = –4 ±40 = 7 ±
–4 4 4
= –1.851, 1.351 = 0.581, –2.581 = 2.781, 0.719
3. Selesaikan setiap yang berikut. SP 2.1.1 TP4
Solve each of the following.
(a) Rajah di sebelah menunjukkan segi empat tepat PQRS. Diberi luas PQRS P (x + 3) cm Q
ialah 20 cm , cari nilai x.
2
The diagram on the right shows a rectangle PQRS. Given that the area of PQRS is 20 cm , find the (2x – 1) cm
2
value of x.
(2x – 1)(x + 3) = 20 5 209 S R
2
2x + 5x – 23 = 0 x + = ± 
16
4
5
x + x – 23 = 0 x = 2.364, –4.864
2
2 2
5
5
5
2
–
x + x +     2 – 23 = 0 Maka, / Thus, x = 2.364
2
2 4 4 2
 x + 5  2 = 209
4
16
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