Latihan soal

1. R.A
Latihan Soal Persamaan dan Pertidaksamaan
1. Dengan menggunakan cara memfaktorkan tentukanlah himpunan penyelesaian persamaan
kuadrat berikut
a. 𝑥2
+ 12𝑥 + 35 = 0
b. 𝑥2
− 13𝑥 + 42 = 0
c. 𝑥2
+ 5𝑥 − 24 = 0
d. 𝑥2
− 3𝑥 − 54 = 0
Jawab :
a. 𝑥2
+ 12𝑥 + 35 = 0
( 𝑥 + 7)( 𝑥 + 5) = 0
( 𝑥 + 7) = 0 𝑎𝑡𝑎𝑢 ( 𝑥 + 5) = 0
𝑥 = −7 𝑎𝑡𝑎𝑢 𝑥 = −5
𝐻𝑝 ∶ {−7, −5}
b. 𝑥2
− 13𝑥 + 42 = 0
( 𝑥 − 7)( 𝑥 − 6) = 0
( 𝑥 − 7) = 0 𝑎𝑡𝑎𝑢 ( 𝑥 − 6) = 0
𝑥 = 7 𝑎𝑡𝑎𝑢 𝑥 = 6
𝐻𝑝 ∶ {7,6}
c. 𝑥2
+ 5𝑥 − 24 = 0
( 𝑥 − 3)( 𝑥 + 8) = 0
( 𝑥 − 3) 𝑎𝑡𝑎𝑢 ( 𝑥 + 8) = 0
𝑥 = 3 𝑎𝑡𝑎𝑢 𝑥 = −8
𝐻𝑝 ∶ {3, −8}
d. 𝑥2
− 3𝑥 − 54 = 0
( 𝑥 − 9)( 𝑥 + 6) = 0
( 𝑥 − 9) = 0 𝑎𝑡𝑎𝑢 ( 𝑥 + 6) = 0
𝑥 = 9 𝑎𝑡𝑎𝑢 𝑥 = −6
𝐻𝑝 ∶ {9, −6}
2. Dengan menggunakan cara melengkapkan kuadrat sempurna tentukanlah himpunan
penyelesaian dari persamaan kuadara berikut
a. 𝑥2
+ 12𝑥 + 35 = 0
b. 𝑥2
− 13𝑥 + 42 = 0
c. 𝑥2
+ 12𝑥 + 35 = 0
d. 𝑥2
− 13𝑥 + 42 = 0

2. R.A
Jawab :
a. 𝑥2
+ 12𝑥 + 35 = 0
𝑥2
+ 12𝑥 = −35
𝑥2
+ 12𝑥 + 36 = −35 + 36 { 𝑘𝑒𝑑𝑢𝑎 𝑟𝑢𝑎𝑠 𝑑𝑖𝑡𝑎𝑚𝑏𝑎ℎ (1/2 𝑥 12)2
= 36 }
( 𝑥 + 6)2
= 1
𝑥 + 6 = ±√1
𝑥 = −6 ± √1
𝑥 = −6 + 1 = −5 𝑎𝑡𝑎𝑢 𝑥 = −6 − 1 = −7
b. 𝑥2
− 13𝑥 + 42 = 0
(𝑥 -
13
2
)2
− (
−13
2
)2
= −42
(𝑥 -
13
2
)2
− (
169
4
)2
= −42
(𝑥 -
13
2
)2
= −42 +
169
4
(𝑥 -
13
2
)2
=
1
4
(𝑥 −
13
2
) = ±√
1
4
𝑥 =
13
2
+ √
1
4
atau 𝑥 =
13
2
− √
1
4
c. 𝑥2
+ 12𝑥 + 35 = 0
𝑥2
+ 12𝑥 = −35
𝑥2
+ 12𝑥 + 36 = −35 + 36 { 𝑘𝑒𝑑𝑢𝑎 𝑟𝑢𝑎𝑠 𝑑𝑖𝑡𝑎𝑚𝑏𝑎ℎ (1/2 𝑥 12)2
= 36 }
( 𝑥 + 6)2
= 1
𝑥 + 6 = ±√1
𝑥 = −6 ± √1
𝑥 = −6 + 1 = −5 𝑎𝑡𝑎𝑢 𝑥 = −6 − 1 = −7
d. Sama dengan soal b
3. Dengan menggunakan cara rumus ABC tentukanlah himpunan penyelesaian persamaan kuadrat
berikut
a. 𝑥2
+ 13𝑥 + 36 = 0
b. 𝑥2
− 3𝑥 − 28 = 0
c. 𝑥2
+ 2𝑥 + 10 = 0
d. 𝑥2
− 8𝑥 + 20 = 0
Jawab :
a. 𝑥2
+ 13𝑥 + 36 = 0
𝑎 = 1, 𝑏 = 13 𝑑𝑎𝑛 𝑐 = 36
𝑥1,2 =
−𝑏 ± √𝑏2 − 4𝑎𝑐
2𝑎
=
−(13)± √132 − 4.1.36
2.1
=
−13 ± √169− 144
2
=
−13 ± √25
2
=
−13 ± 5
2

3. R.A
𝑚𝑎𝑘𝑎 𝑥1 =
13 + 5
2
= 9 𝑎𝑡𝑎𝑢 𝑥2 =
13 − 5
2
= 4
b. 𝑥2
− 3𝑥 − 28 = 0
𝑎 = 1, 𝑏 = −3 𝑑𝑎𝑛 𝑐 = −28
𝑥1,2 =
−𝑏 ± √𝑏2 − 4𝑎𝑐
2𝑎
=
−(−3) ± √−32 − 4.1. (−28)
2.1
=
3 ± √9 + 112
2
=
3 ± √121
2
=
3 ± 11
2
𝑚𝑎𝑘𝑎 𝑥1 =
3 + 11
2
= 7 𝑎𝑡𝑎𝑢 𝑥2 =
3 − 11
2
= −4
c. 𝑥2
+ 2𝑥 + 10 = 0
𝑎 = 1, 𝑏 = 2 𝑑𝑎𝑛 𝑐 = 10
𝑥1,2 =
−𝑏 ± √𝑏2 − 4𝑎𝑐
2𝑎
=
−2 ± √22 − (4.1.10
2.1
=
−2 ± √4 − 40
2
=
−2 ± √−36
2
=
−2 ± 6𝑖
2
= −1 ± 3𝑖
𝑚𝑎𝑘𝑎 𝑥1 = −1 + 3𝑖 𝑎𝑡𝑎𝑢 𝑥2 = −1 − 3𝑖
d. 𝑥2
− 8𝑥 + 20 = 0
𝑎 = 1, 𝑏 = −8 𝑑𝑎𝑛 𝑐 = 20
𝑥1,2 =
−𝑏 ± √𝑏2 − 4𝑎𝑐
2𝑎
=
−(−8)± √82 − (4.1.20
2.1
=
8 ± √64 − 80
2
=
8 ± √−16
2
=
8 ± 4𝑖
2
= 4 ± 2𝑖
𝑚𝑎𝑘𝑎 𝑥1 = 4 + 2𝑖 = 7 𝑎𝑡𝑎𝑢 𝑥2 = 4 − 2𝑖

4. R.A
4. Tentukanlah himpunan penyelesaian dari pertidaksamaan berikut
a. 𝑥2
+ 14𝑥 + 45 < 0
b. 𝑥2
− 15𝑥 + 54 ≤ 0
c. 𝑥2
− 3𝑥 − 10 > 0
d. 𝑥2
+ 5𝑥 − 14 ≥ 0
Jawab :
a. 𝑥2
+ 14𝑥 + 45 < 0
𝑁𝑖𝑙𝑎𝑖 𝑛𝑜𝑙 𝑑𝑎𝑟𝑖 𝑏𝑎𝑔𝑖𝑎𝑛 𝑘𝑖𝑟𝑖 𝑝𝑒𝑟𝑡𝑖𝑑𝑎𝑘𝑠𝑎𝑚𝑎𝑎𝑛
𝑥2
+ 14𝑥 + 45 = 0
( 𝑥 + 9)( 𝑥 + 5) = 0
𝑥 + 9 = 0 𝑎𝑡𝑎𝑢 𝑥 + 5 = 0
𝑥1 = −9 𝑎𝑡𝑎𝑢 𝑥2 = −5
−9 − 5
HP = { 𝑥I − 9 < 𝑥 < −5 }
b. 𝑥2
− 15𝑥 + 54 ≤ 0
𝑁𝑖𝑙𝑎𝑖 𝑛𝑜𝑙 𝑑𝑎𝑟𝑖 𝑏𝑎𝑔𝑖𝑎𝑛 𝑘𝑖𝑟𝑖 𝑝𝑒𝑟𝑡𝑖𝑑𝑎𝑘𝑠𝑎𝑚𝑎𝑎𝑛
𝑥2
− 15𝑥 + 54 = 0
( 𝑥 − 9)( 𝑥 − 6) = 0
𝑥 − 9 = 0 𝑎𝑡𝑎𝑢 𝑥 − 6 = 0
𝑥1 = 9 𝑎𝑡𝑎𝑢 𝑥2 = 6
6 9
HP = { 𝑥I 6 ≤ 𝑥 ≤ 9}
c. 𝑥2
− 3𝑥 − 10 > 0
𝑁𝑖𝑙𝑎𝑖 𝑛𝑜𝑙 𝑑𝑎𝑟𝑖 𝑏𝑎𝑔𝑖𝑎𝑛 𝑘𝑖𝑟𝑖 𝑝𝑒𝑟𝑡𝑖𝑑𝑎𝑘𝑠𝑎𝑚𝑎𝑎𝑛
𝑥2
− 3𝑥 − 10 = 0
( 𝑥 − 5)( 𝑥 + 2) = 0
𝑥 − 5 = 0 𝑎𝑡𝑎𝑢 𝑥 + 2 = 0
𝑥1 = 5 𝑎𝑡𝑎𝑢 𝑥2 = −2
−2 5
HP = { 𝑥I x < −2 𝑎𝑡𝑎𝑢 𝑥 > 9}
d. 𝑥2
+ 5𝑥 − 14 ≥ 0
𝑁𝑖𝑙𝑎𝑖 𝑛𝑜𝑙 𝑑𝑎𝑟𝑖 𝑏𝑎𝑔𝑖𝑎𝑛 𝑘𝑖𝑟𝑖 𝑝𝑒𝑟𝑡𝑖𝑑𝑎𝑘𝑠𝑎𝑚𝑎𝑎𝑛
𝑥2
+ 5𝑥 − 14 = 0
( 𝑥 + 7)( 𝑥 − 2) = 0
𝑥 + 7 = 0 𝑎𝑡𝑎𝑢 𝑥 − 2 = 0
𝑥1 = −7 𝑎𝑡𝑎𝑢 𝑥2 = 2
−7 2
HP = { 𝑥I x ≤ −7 𝑎𝑡𝑎𝑢 𝑥 ≥ 2}

5. R.A
5. Tentukanlah penyelesaian dari persamaan mutlak berikut :
a. | 𝑥 + 3| = 5
b. | 𝑥 − 4| = 7
c. |2𝑥 + 8| = 9
d. |3𝑥 − 4| = 5
Jawab :
a. | 𝑥 + 3| = 5
( 𝑥 + 3)2
= 52
𝑥2
+ 6𝑥 + 9 = 25
𝑥2
+ 6𝑥 − 16 = 0
( 𝑥 + 8)( 𝑥 − 2) = 0
𝑥 + 8 = 0 𝑎𝑡𝑎𝑢 𝑥 − 2 = 0
𝑥1 = −8 𝑑𝑎𝑛 𝑥2 = 2
b. | 𝑥 − 4| = 7
( 𝑥 − 4)2
= 72
𝑥2
− 8𝑥 + 16 = 49
𝑥2
− 8𝑥 − 33 = 0
( 𝑥 − 11)( 𝑥 + 3) = 0
𝑥 − 11 = 0 𝑎𝑡𝑎𝑢 𝑥 + 3 = 0
𝑥1 = −11 𝑑𝑎𝑛 𝑥2 = −3
c. |2𝑥 + 8| = 9
(2𝑥 + 8)2
= 92
4𝑥2
+ 32𝑥 + 64 = 81
4𝑥2
+ 32𝑥 − 17 = 0
𝑥1,2 =
−𝑏 ± √𝑏2 − 4𝑎𝑐
2𝑎
=
−(32) ± √322 − (4.4. −17
2.4
=
−32± √1024+272
2.4
=
−32 ± 36
8
=
−8±9
2
𝐽𝑎𝑑𝑖 𝑥1 =
−8 + 9
2
=
1
2
𝑑𝑎𝑛 𝑥2 =
−8 − 9
2
=
−17
2
d. |3𝑥 − 4| = 5
(3𝑥 − 4)2
= 52
9𝑥2
− 24𝑥 + 16 = 25
9𝑥2
− 24𝑥 − 9 = 0 ∶ 3
3𝑥2
− 8𝑥 − 3 = 0
(3𝑥 + 1)( 𝑥 − 3) = 0
3𝑥 + 1 = 0 𝑎𝑡𝑎𝑢 𝑥 − 3 = 0

6. R.A
𝑥 = −
1
3
𝑑𝑎𝑛 𝑥 = 3
6. Tentukan himpunan penyelesaian dari pertidaksamaan mutlak berikut :
a. |2𝑥 + 3| < 0
b. |5𝑥 + 4 | ≤ 0
c. |2𝑥 + 3| > | 𝑥 − 4 |
d. |3𝑥 − 2| ≥ |2𝑥 − 1|
Jawab :
a. |2𝑥 + 3| < 0
0 < 2𝑥 + 3 < 0
0 − 3 < 2𝑥 < 0 + 3
−3 < 2𝑥 < 3
−
3
2
< 𝑥 <
3
2
HP = {𝑥I x < −
3
2
𝑎𝑡𝑎𝑢 𝑥 >
3
2
}
b. |5𝑥 + 4 | ≤ 0
0 ≤ 5𝑥 + 4 ≤ 0
0 − 4 ≤ 5𝑥 ≤ 0 + 4
−4 ≤ 5𝑥 ≤ 4
−
4
5
≤ 𝑥 ≤
4
5
HP = {𝑥I x ≤ −
4
5
𝑎𝑡𝑎𝑢 𝑥 ≥
4
5
}
c. |2𝑥 + 3| > | 𝑥 − 4 |
[(2𝑥 + 3) +(𝑥 − 4)][(2𝑥 + 3) − (𝑥 − 4)]
(3𝑥 − 1)(𝑥 + 7) > 0
3𝑥 − 1 = 0 𝑎𝑡𝑎𝑢 𝑥 + 7 = 0
3𝑥 = 1 𝑎𝑡𝑎𝑢 𝑥 = −7
𝑥 =
1
3
𝑎𝑡𝑎𝑢 𝑥 = −7
HP = {𝑥I x < −7 𝑎𝑡𝑎𝑢 𝑥 >
1
3
}
d. |3𝑥 − 2| ≥ |2𝑥 − 1|
[(3𝑥 − 2) +(2𝑥 − 1)][(3𝑥 − 2) − (2𝑥 − 1)]
(5𝑥 − 3)(𝑥 − 1) ≥ 0
5𝑥 − 3 = 0 𝑎𝑡𝑎𝑢 𝑥 − 1 = 0
𝑥 =
3
5
𝑎𝑡𝑎𝑢 𝑥 = 1
HP = {𝑥I 1 ≤ x ≥
3
5
}