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Soal Matematika Bab 2 Persamaan dan Pertidaksamaan
- 1. 1. Dengan menggunakan cara memfaktorkan tentukanlah himpunan
penyelesaian dari persamaan kuadratberikut:
a. π₯2
+ 12π₯ + 35 = 0
b. π₯2
β 13π₯ + 42 = 0
c. π₯2
+ 5π₯ β 24 = 0
d. π₯2
β 3π₯ β 54 = 0
2. Dengan menggunakan cara melengkapkan kuadrat sempurna
tentukanlah himpunan penyelesaian dari persamaan kuadratberikut:
a. π₯2
+ 12π₯ + 35 = 0
b. π₯2
β 13π₯ + 42 = 0
c. π₯2
+ 12π₯ + 35 = 0
d. π₯2
β 13π₯ + 42 = 0
3. Dengan menggunakan cara rumus ABCtentukanlah himpunan
penyelesaian dari persamaan kuadratberikut:
a. π₯2
+ 13π₯ + 36 = 0
b. π₯2
β 3π₯ β 28 = 0
c. π₯2
+ 2π₯ + 10 = 0
d. π₯2
β 8π₯ + 20 = 0
4. Tentukanlah himpunan penyelesaian dari pertidaksamaan berikut
a. π₯2
+ 14π₯ + 45 < 0
b. π₯2
β 15π₯ + 54 β€ 0
c. π₯2
β 3π₯ β 10 > 0
d. π₯2
+ 5π₯ β 14 β₯ 0
5. Tentukanlah penyelesaian daripersamaan mutlak berikut:
a. |x + 3| = 5
b. |x β 4| = 7
c. |2x + 8| = 9
d. |3x β 4| = 5
- 2. 6. Tentukanlah himpunan penyelesaian dari pertidaksamaan mutlak
berikut
a. |2π₯ + 3| < 10
b. |5π₯ β 4| β€ 10
c. |2π₯ + 3| > | π₯ β 4|
d. |3π₯ β 2| β₯ |2π₯ β 1|
Jawaban
1. a. π₯2
+ 12π₯ + 35 = 0 β ( π₯ + 7)( π₯ + 5)
π₯1 = β7 & π₯2 = β5 β π»π = { β7,β5}
b. π₯2
β 13π₯ + 42 = 0 β ( π₯ β 7)( π₯ β 6)
π₯1 = 7 & π₯2 = 6 β π»π= { 7,6}
c. π₯2
+ 5π₯ β 24 = 0 β ( π₯ + 8)( π₯ β 3)
π₯1 = β8 & π₯2 = π β π―π = { 3, β8}
d. π₯2
β 3π₯ β 54 = 0 β ( π₯ β 9)( π₯ + 6)
π₯1 = 9 & π₯2 = β6 β π»π = { 9, β6}
- 3. 3. a. π₯2
+ 13π₯ + 36 = 0
=> π₯1,2 =
β13 Β± β169β (4.1.36)
2.1
=> π₯1,2 =
β13 Β± β169 β 144
2
=> π₯1,2 =
β13 Β± 5
2
b. π₯2
β 3π₯ β 28 = 0
=> π₯1,2 =
3 Β± β9 β (4.1.β28)
2.1
=> π₯1,2 =
3 Β± β9 + 112
2
=> π₯1,2 =
3 Β± 11
2
c. π₯2
+ 2π₯ + 10 = 0
=> π₯1,2 =
β2 Β± β4 β (4.1.10
2.1
=> π₯1,2 =
β2 Β± β4 β 40
2
=> π₯1,2 =
β2 Β± 6π
2
= β1 Β± 3π
d. π₯2
β 8π₯ + 20 = 0
=> π₯1,2 =
8 Β± β64 β (4.1.20)
2.1
=> π₯1,2 =
8 Β± β64 β 80
2
=> π₯1,2 =
8 Β± 4π
2
= 4 Β± 2π
π₯1 =
β8
2
= β4
π₯2 =
β18
2
= β9
π»π = {β4,β9}
π₯1 =
14
2
= 7
π₯2 =
β8
2
= β4
π»π = {7,β4}
π₯1 = β1 + 3π
π₯2 = β1 β 3π
π»π = {β1 + 3π, β1 β 3π}
π₯1 = 4 + 2π
π₯2 = 4 β 2π
π»π = {4 + 2π, 4 β 2π}
- 4. 4. a. π₯2
+ 14π₯ + 45 < 0
β ( π₯ + 9)( π₯ + 5) < 0
β π₯ > β9 π π₯ < β5
b. π₯2
β 15π₯ + 54 β€ 0
β ( π₯ β 9)( π₯ + 6) β€ 0
β π₯ β₯ 6 π π₯ β€ 9
c. π₯2
β 3π₯ β 10 > 0
β ( π₯ β 5)( π₯ + 2) > 0
β π₯ < β2 π π₯ > 5
d. π₯2
+ 5π₯ β 14 β₯ 0
β ( π₯ + 7)( π₯ β 2) β₯ 0
β π₯ β€ β7 π π₯ β₯ 2
π»π = { π₯|β9 < π₯ < β5}
π»π = { π₯|6 < π₯ < 9}
π»π = { π₯| π₯ < 2 ππ‘ππ’ π₯ > 5}
π»π = { π₯| π₯ β€ β7 ππ‘ππ’ π₯ β₯ 2}
- 5. π₯1 =
β8 + 9
2
=
1
2
π₯2 =
β8 β 9
2
=
β17
2
5. a. | π₯ + 3| = 5 β ( π₯ + 3)2
= 52
π₯2
+ 6π₯ + 9 = 25
π₯2
+ 6π₯ β 16 = 0
( π₯ + 8)( π₯ β 2) = 0
π₯1 = β8 πππ π₯2 = 2
b. | π₯ β 4| = 7 β ( π₯ β 4)2
= 72
π₯2
β 8π₯ + 16 = 49
π₯2
β 8π₯ β 33 = 0
( π₯ β 11)( π₯ + 3) = 0
π₯1 = 11 πππ π₯2 = β3
c. |2π₯ + 8| = 9 β (2π₯ + 8)2
= 92
4π₯2
+ 32π₯ + 64 = 81
4π₯2
+ 32π₯ β 17 = 0
=> π₯1,2 =
β32Β± β1024+272
4.2
=> π₯1,2 =
β32 Β± 36
8
=> π₯1,2 =
β8 Β± 9
2
d. |3π₯ β 4| = 5 β (3π₯ β 4)2
= 52
9π₯2
β 24π₯ + 16 = 25
9π₯2
β 24π₯ β 9 = 0 βΆ 3
3π₯2
β 8π₯ β 3 = 0
(3π₯ + 1)( π₯ β 3) = 0
π₯1 = β
1
3
πππ π₯2 = 3
- 6. π»π = { π₯|
β13
2
< π₯ <
7
2
, π₯ β π
}
π»π = { π₯|
β6
5
< π₯ <
14
5
, π₯ β π
}
π₯1,2 =
β10 Β± 11
3
π₯1 =
1
3
πππ π₯2 =
β10 β 11
3
= β7
π»π = { π₯|π₯ β€
3
5
π π₯ β₯ 1}
π»π = { π₯|π₯ β€
3
5
π π₯ β₯ 1}
6. a. |2π₯ + 3| < 10
β10 < 2π₯ + 3 < 10
β10β 3
2
< π₯ <
10 β 3
2
β13
2
< π₯ <
7
2
b. |5π₯ β 4| β€ 10
β10 β€ 5π₯ β 4 β€ 10
β10+ 4
5
< π₯ <
10 + 4
5
β6
5
< π₯ <
14
5
c. |2π₯ + 3| > |π₯ β 4|
β(2π₯ + 3)2 > β(π₯ β 4)2
4π₯2
+ 12π₯ + 9 > π₯2
β 8π₯ + 16
3π₯2
+ 20π₯ β 7 > 0
=> π₯1,2 =
β20Β± β400β(4.3.β7)
2.3
=> π₯1,2 =
β20 Β± β400 + 84
6
=> π₯1,2 =
β20 Β± β484
6
c. |3π₯ β 2| β₯ |2π₯ β 1|
β(3π₯ β 2)2 β₯ β(2π₯ β 1)2
9π₯2
β 12π₯ + 4 β₯ 4π₯2
β 4π₯ + 1
5π₯2
β 8π₯ + 3 β₯ 0
(5π₯ β 3)( π₯ β 1) β₯ 0
π₯ β€
3
5
π π₯ β₯ 1