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Tugas mtk 2
1.
c. 𝑥2 − 5𝑥
− 24 = 0 ⟺ ( 𝑥 − 8) ( 𝑥 + 3) = 0 ⟺ 𝑥 − 8 = 0 V 𝑥 + 3 = 0 ⟺ 𝑥1 = 8 V 𝑥2 = −3 ⇔ HP = {8,−3} SOAL BAB II 1. Dengan menggunakan cara memfaktorkan , tentukanlah himpunan penyelesaian dari persamaan kuadrat berikut: a. 𝑥2 + 12𝑥 + 35 = 0 ⟺ ( 𝑥 + 7) ( 𝑥 + 5) = 0 ⟺ 𝑥 + 7 = 0 V 𝑥 + 5 = 0 ⟺ 𝑥1 = −7 V 𝑥2 = −5 ⇔ HP = {−7,−5} b. 𝑥2 − 13𝑥 + 42 = 0 ⟺ ( 𝑥 − 7) ( 𝑥 − 6) = 0 ⟺ 𝑥 − 7 = 0 V 𝑥 − 6 = 0 ⟺ 𝑥1 = 7 V 𝑥2 = 6 ⇔ HP = {7,6} 2. Dengan menggunakan cara rumus ABC,tentukanlah himpunan penyelesaian dari persamaan kuadrat berikut: a. 𝑥2 + 13𝑥 + 36 = 0 𝑥1,2 = −𝑏 ± √𝑏2 − 4𝑎𝑐 2𝑎 𝑥1,2 = −(13) ± √132 − 4 .1 .36 2 . 1 𝑥1,2 = −13 ± √169 − 144 2 𝑥1,2 = −13 ± √25 2 𝑥1,2 = −13 ± 5 2 𝑥1 = −13 + 5 2 = −8 2 = −4 V x2 = −13 − 5 2 = −18 2 = −9 b. 𝑥2 − 3𝑥 − 28 = 0 𝑥1,2 = −𝑏 ± √𝑏2 − 4𝑎𝑐 2𝑎 d. 𝑥2 − 3𝑥 − 54 = 0 ⟺ ( 𝑥 − 9) ( 𝑥 + 6) = 0 ⟺ 𝑥 − 9 = 0 V 𝑥 + 6 = 0 ⟺ 𝑥1 = 9 V 𝑥2 = −6 ⇔ HP = {9,−6}
2.
𝑥1,2 = −(−3) ±
√(−3)2 − 4.1. (−28) 2.1 𝑥1,2 = 3 ± √9 − (−112) 2 𝑥1,2 = 3 ± √121 2 𝑥1,2 = 3 ± 11 2 𝑥1 = 3 + 11 2 = 14 2 = 7 V x2 = 3 − 11 2 = −8 2 = −4 c. 𝑥2 + 2𝑥 + 10 = 0 𝑥1,2 = −𝑏 ± √𝑏2 − 4𝑎𝑐 2𝑎 𝑥1,2 = −(2) ± √(2)2 − 4.1.10 2.1 𝑥1,2 = −2 ± √4 − 40 2 𝑥1,2 = −2 ± √(−36) 2 𝑥1,2 = −2 ± 6i 2 𝑥1 = −2 + 6𝑖 2 = −1 + 3𝑖 V x2 = −2 − 6𝑖 2 = −1 − 3𝑖 d. 𝑥2 − 8𝑥 + 20 = 0 𝑥1,2 = −𝑏 ± √𝑏2 − 4𝑎𝑐 2𝑎 𝑥1,2 = −(−8) ± √(−8)2 − 4.1.20 2.1 𝑥1,2 = 8 ± √64 − 80 2 𝑥1,2 = 8 ± √−16 2 𝑥1,2 = 8 ± 4i 2 𝑥1 = 8 + 4𝑖 2 = −1 + 3𝑖 V x2 = −2 − 6𝑖 2 = −1 − 3𝑖
3.
3. Tentukanlah Himpunan
penyelesaian dari pertidaksamaan berikut: a. 𝑥2 + 4𝑥 + 45 < 0 ⇔ ( 𝑥 + 9) ( 𝑥 + 5) < 0 ⟺ 𝑥 + 9 < 0 V 𝑥 + 5 < 0 ⟺ 𝑥1 < −9 V 𝑥2 < −5 𝐻𝑃 = {𝑥 /−9 < 𝑥 < −5} b. 𝑥2 − 15𝑥 + 54 ≤ 0 ⇔ ( 𝑥 − 6) ( 𝑥 − 9) ≤ 0 ⟺ 𝑥 − 6 ≤ 0 V 𝑥 − 9 ≤ 0 ⟺ 𝑥1 ≤ 6 V 𝑥2 ≤ 9 𝐻𝑃 = {𝑥 / 6 ≤ 𝑥 ≤ 9} c. 𝑥2 − 3𝑥 − 10 > 0 ⇔ ( 𝑥 + 2) ( 𝑥 − 5) > 0 ⟺ 𝑥 + 2 > 0 V 𝑥 − 5 > 0 ⟺ 𝑥1 > −2 V 𝑥2 > 5 𝐻𝑃 = {𝑥 / x < −2 𝑎𝑡𝑎𝑢 𝑥 > 5} 4. Tentukanlah penyelesaian dari persamaan mutlak berikut: a. | 𝑥 + 3| = 5 √(𝑥 + 3)2 = 5 (𝑥 + 3)2 = 52 (𝑥 + 3)2 = 25 𝑥2 + 6𝑥 + 9 = 25 𝑥2 + 6𝑥 + 9 − 25 = 0 𝑥2 + 6𝑥 − 16 = 0 ( 𝑥 − 2) ( 𝑥 + 8) = 0 𝑥1 = 2 V x2 = −8 b. | 𝑥 − 4| = 7 √(𝑥 − 4)2 = 7 (𝑥 − 4)2 = 72 (𝑥 − 4)2 = 49 𝑥2 − 8𝑥 + 16 = 49 𝑥2 − 8𝑥 + 16 − 49 = 0 𝑥2 − 8𝑥 − 33 = 0 ( 𝑥 − 11) ( 𝑥 + 3) = 0 𝑥1 = 11 V x2 = −3 d. 𝑥2 + 5𝑥 − 14 ≥ 0 ⇔ ( 𝑥 + 7) ( 𝑥 − 2) ≥ 0 ⟺ 𝑥 + 7 ≥ 0 V 𝑥 − 2 ≥ 0 ⟺ 𝑥1 ≥ −7 V 𝑥2 ≥ 2 𝐻𝑃 = {𝑥 / x ≤ −7 𝑎𝑡𝑎𝑢 𝑥 ≥ 2}
4.
c. |2𝑥 +
8| = 9 √(2𝑥 + 8)2 = 9 (2𝑥 + 8)2 = 92 (2𝑥 + 8)2 = 81 4𝑥2 + 32𝑥 + 64 = 81 4𝑥2 + 32𝑥 + 64 − 81 = 0 4𝑥2 + 32𝑥 − 17 = 0 𝑥1,2 = −𝑏 ± √𝑏2 − 4𝑎𝑐 2𝑎 𝑥1,2 = −(32) ± √322 − 4 .4 .−17 2 .4 𝑥1,2 = −32 ± √1024 + 272 8 𝑥1,2 = −32 ± √1296 8 𝑥1,2 = −32 ± 36 8 𝑥1 = −8 + 9 2 = 1 2 V x2 = −8 − 9 2 = −17 2 d. |3𝑥 − 4| = 5 √(3𝑥 − 4)2 = 5 (3𝑥 − 4)2 = 52 (3𝑥 − 4)2 = 25 9𝑥2 − 24𝑥 + 16 = 25 9𝑥2 − 24𝑥 + 16 − 25 = 0 9𝑥2 − 24𝑥 − 9 = 0 : 3 3𝑥2 − 8𝑥 − 3 = 0 (3𝑥 + 1) ( 𝑥 − 3) = 0 3𝑥 + 1 = 0 V 𝑥 − 3 = 0 𝑥1 = − 1 3 V 𝑥2 = 3
5.
𝐻𝑝 = {
𝑥| −13 2 < 𝑥 < 7 2 , 𝑥 ∈ 𝑅 } 𝐻𝑝 = { 𝑥| −6 5 < 𝑥 < 14 5 , 𝑥 ∈ 𝑅 } 𝑥1,2 = −10 ± 11 3 𝑥1 = 1 3 𝑑𝑎𝑛 𝑥2 = −10 − 11 3 = −7 𝐻𝑝 = { 𝑥|𝑥 ≤ 3 5 𝑉 𝑥 ≥ 1} 𝐻𝑝 = { 𝑥|𝑥 ≤ 3 5 𝑉 𝑥 ≥ 1} 5. Tentukanlahhimpunanpenyelesaiandari pertidaksamaanmutlak berikut: a. |2𝑥 + 3| < 10 b. |5𝑥 − 4| ≤ 10 c. |2𝑥 + 3| > | 𝑥 − 4| d. |3𝑥 − 2| ≥ |2𝑥 − 1| a. |2𝑥 + 3| < 10 −10 < 2𝑥 + 3 < 10 −10 − 3 2 < 𝑥 < 10 − 3 2 −13 2 < 𝑥 < 7 2 b. |5𝑥 − 4| ≤ 10 −10 ≤ 5𝑥 − 4 ≤ 10 −10 + 4 5 < 𝑥 < 10 + 4 5 −6 5 < 𝑥 < 14 5 c. |2𝑥 + 3| > |𝑥 − 4| √(2𝑥 + 3)2 > √(𝑥 − 4)2 4𝑥2 + 12𝑥 + 9 > 𝑥2 − 8𝑥 + 16 3𝑥2 + 20𝑥 − 7 > 0 => 𝑥1,2 = −20± √400 −(4.3.−7) 2.3 => 𝑥1,2 = −20 ± √400+ 84 6 => 𝑥1,2 = −20 ± √484 6 c. |3𝑥 − 2| ≥ |2𝑥 − 1| √(3𝑥 − 2)2 ≥ √(2𝑥 − 1)2 9𝑥2 − 12𝑥 + 4 ≥ 4𝑥2 − 4𝑥 + 1 5𝑥2 − 8𝑥 + 3 ≥ 0 (5𝑥 − 3)( 𝑥 − 1) ≥ 0 𝑥 ≤ 3 5 𝑉 𝑥 ≥ 1
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