matematika bab 2
TRANSCRIPT
BAB II PERSAMAAN DAN PERTIDAKSAMAAN KUADRAT
1. Dengan menggunakan cara memfaktorkan tentukanlah himpunan penyelesaian dari
persamaan kuadrat berikut :
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 kuadrat berikut :
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 ABC tentukanlah himpunan penyelesaian dari
persamaan kuadrat berikut :
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 dari persamaan mutlak berikut :
a. |x + 3| = 5
b. |x β 4| = 7
c. |2x + 8| = 9
d. |3x β 4| = 5
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
π₯ + 7 π₯ + 5 = 0
π₯1 = β7 ππ‘ππ’ π₯2 = β5
BAB II PERSAMAAN DAN PERTIDAKSAMAAN KUADRAT
b. π₯2 β 13π₯ + 42 = 0
π₯ β 7 π₯ β 6 = 0
π₯1 = 7 ππ‘ππ’ π₯2 = 6
c. π₯2 + 5π₯ β 24 = 0
π₯ + 8 π₯ β 3 = 0
π₯1 = β8 ππ‘ππ’ π₯2 = 3
d. π₯2 β 3π₯ β 54 = 0
π₯ β 9 π₯ + 6 = 0
2. a. π₯2 + 12π₯ + 35 = 0
π₯2 + 12π₯ = π₯2 β ππ₯ = (π₯ + π
2)2 β (
π
2)2 , π = 12
(π₯ + 12
2)2 β (
12
2)2 = β35
(π₯ + 6)2 β 36 = β35
(π₯ + 6)2 = 1
(π₯ + 6) = β1
π₯ = β6 Β± β1
π₯ = β6 + β1 atau π₯ = β6 β β1
b. π₯2 β 13π₯ + 42 = 0
(π₯ - 13
2)2 β (
β13
2)2 = β42
(π₯ - 13
2)2 β (
169
4)2 = β42
(π₯ - 13
2)2 = β42 +
169
4
(π₯ - 13
2)2 =
1
4
(π₯ β13
2) = Β±β
1
4
π₯ =13
2+ β
1
4 atau π₯ =
13
2β β
1
4
3. a. π₯2 + 13π₯ + 36 = 0
=> π₯1,2 =β13 Β± 169 β (4.1.36)
2.1
=> π₯1,2 =β13 Β± β169 β 144
2
=> π₯1,2 =β13 Β± 5
2
π₯1 = β13 + 5
2= β4
π₯2 = β13 β 5
2= β9
π₯1 = 9 ππ‘ππ’ π₯2 = β6
BAB II PERSAMAAN DAN PERTIDAKSAMAAN KUADRAT
b. π₯2 β 3π₯ β 28 = 0
=> π₯1,2 =3 Β± 9 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± 4 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± 64 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
4. a. π₯2 + 14π₯ + 45 < 0
β π₯ + 9 π₯ + 5 < 0
β π₯ > β9 ππ‘ππ’ π₯ < β5
b. π₯2 β 15π₯ + 54 β€ 0
β π₯ β 9 π₯ + 6 β€ 0
β π₯ β₯ 6 ππ‘ππ’ π₯ β€ 9
c. π₯2 β 3π₯ β 10 > 0
β π₯ β 5 π₯ + 2 > 0
β π₯ < β2 ππ‘ππ’ π₯ > 5
π₯1 = 13 + 11
2= 7
π₯2 = 3 β 11
2= β4
π₯1 = β1 + 3π
π₯2 = β1 β 3π
π₯1 = 4 + 2π
π₯2 = 4 β 2π
π»π = π₯ β9 < π₯ < β5
π»π = π₯ 6 β€ π₯ β€ 9
π»π = π₯ π₯ < β2 ππ‘ππ’ π₯ > 5
BAB II PERSAMAAN DAN PERTIDAKSAMAAN KUADRAT
d. π₯2 + 5π₯ β 14 β₯ 0
β π₯ + 7 π₯ β 2 β₯ 0
β π₯ β€ β7 ππ‘ππ’ π₯ β₯ 2
5. a. π₯ + 3 = 5
π₯ + 3 2 = 52
π₯2 + 6π₯ + 9 = 25
π₯2 + 6π₯ β 16 = 0 π₯ + 8 π₯ β 2 = 0
π₯1 = β8 ππ‘ππ’ 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 ππ‘ππ’ π₯2 = β3
c. 2π₯ + 8 = 9
2π₯ + 8 2 = 92
4π₯2 + 32π₯ + 64 = 81
4π₯2 + 32π₯ β 17 = 0
=> π₯1,2 =β32 Β± β1024 + 272
4.2
=> π₯1,2 =β32 Β± 36
8
=> π₯1,2 =β8 Β± 9
2
π₯1 =β8 + 9
2=
1
2 ππ‘ππ’ π₯2 =
β8 + 9
2=
β17
2
d. 3π₯ β 4 = 5
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 = β1
3 ππ‘ππ’ π₯2 = 3
π»π = π₯ π₯ β€ β7 ππ‘ππ’ π₯ β₯ 2
BAB II PERSAMAAN DAN PERTIDAKSAMAAN KUADRAT
π»π = {π₯|β13
2< π₯ <
7
2, π₯ β π }
π»π = {π₯|β6
5< π₯ <
14
5, π₯ β π }
π»π = {π₯|π₯ β€ β7 ππ‘ππ’ π₯ β₯1
3}
π»π = {π₯|π₯ β€3
5 ππ‘ππ’ π₯ β₯ 1}
6. a. 2π₯ + 3 < 10
β10 < 2π₯ + 3 < 10 β10 β 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|
(2π₯ + 3)2 > (π₯ β 4)2
4π₯2 + 12π₯ + 9 > π₯2 β 8π₯ + 16
3π₯2 + 20π₯ β 7 > 0
=> π₯1,2 =β20Β± 400β(4.3.β7)
2.3
=> π₯1,2 =β20 Β± β400 + 84
6
=> π₯1,2 =β20 Β± β484
6
=> π₯1,2 =β10 Β± 11
3
π₯1 = β10 + 11
3 ππ‘ππ’ π₯2 =
β10 β 11
3= β7
d. 3π₯ β 2 β₯ |2π₯ β 1|
(3π₯ β 2)2 β₯ (2π₯ β 1)2
9π₯2 β 12π₯ + 4 β₯ 4π₯2 β 4π₯ + 1
5π₯2 β 8π₯ + 3 β₯ 0
β 5π₯ β 3 π₯ β 1 β₯ 0
π₯ β€3
5 ππ‘ππ’ π₯ β₯ 1