Homepage › Solution manuals › Найма Гахраманова › Математика - 8 › Задание 6, стр 60
Решите уравнения различными способами:
1.x2 + 10x + 24 = 0 D = b2 − 4ac = 102 − 24 ⋅ 4 = 100 − 96 = 4 x1 = − 10 + 4 2 = − 10 + 2 2 = − 8 2 = −4 x2 = − 10 −4 2 = − 10 − 2 2 = − 12 2 = −6. 2.x2 + 6x + 8 = 0. D = 62 − 4 ⋅ 8 = 36 − 32 = 4 = 22 x1 = − 6 + 2 2 = − 4 2 = −2. x2 = − 6 − 2 2 = − 8 2 = −4. 3.x2 + 3x − 5 = 0 D = b2 − 4ac = 32 − 4 ⋅ (−5) ⋅ 2 = 9 + 40 = 49 = 72 x1 = − 3 + 7 2 ⋅ 2 = 4 4 = 1;x2 = −3 − 7 2 ⋅ 2 = − 10 4 = −2,5. 4.x2 − 4x = 3 x2 − 4x − 3 = 0. D = b2 − 4ac = (−4)2 − 4 ⋅ (−3) = 16 + 12 = 28. x1 = 4 + 28 2 = 2 + 7;x2 = 2 −7.
5. − 24 = −2y2 + 2y 0 = −2y2 + 2y + 24 − 2y2 + 2y + 24 = 0∣ : (−2) y2 − y − 12 = 0 D = b2 − 4ac = (−1)2 − 4 ⋅ 1 ⋅ (−12) = 1 + 48 = 49 x1 = 1 + 49 2 −1 + 7 2 = 8 2 = 4 x2 = 1 −49 2 = 1 − 7 2 = − 6 2 = −3 6. 9y2 − 12y − 5 = 0 D = (−12)2 − 4 ⋅ 9(−5) = 144 + 180 = 324 x1 = 12 + 374 92 = 12 + 18 18 = 30 18 = 112 18 = 12 3 x2 = 12 −324 92 = 12 − 18 18 = − 6 18 = −1 3 7. (3p − 5)(p + 1) = −4 3p2 − 5p + 3p − 5 = −4. 3p2 − 2p − 5 = −4. D = b2 − 4ac = (−2)2 − 4 ⋅ 3(−5) = 4 + 60 = 64. p1 = 2 + 64 32 = 2 + 8 6 = 10 6 ,p2 = 2 −64 3 = 2 − 8 32 = − 6 6 = −1 8. 3x(x − 2) = 24 3x2 − 6x = 24 3x2 − 6x − 24 = 0(: 3) x2 − 2x − 8 = 0 D = b2 − 4ac = (−2)2 − 4 ⋅ (−8) = 4 + 32 = 36. x1 = 2 + 6 2 = 8 2 = 4;x2 = 2 − 6 2 = − 4 2 = −2. 9. m(m + 10) = 2. m2 + 10m = 2 m2 + 10m − 2 = 0. D = 100 − 4 ⋅ (−2) = 100 + 8 = 108. m1 = − 10 + 108 2 = − 10 + 63 2 = −5 + 33. m2 = − 10 −108 2 = − 10 − 63 2 = −5 − 33.
