7класс

Страница 74 номер 3.3.35, ГДЗ по алгебре за 7, 8 и 9 класс к задачнику Ткачевой

§3. Уравнения. 3.3. Квадратные уравнения. Страница 74. Номер 3.3.35
Задание / условие:

Решить уравнение (3.3.32—3.3.38): 1) \(2x^2 - 17x + 30 = 0\); 2) \(5x^2 - 48x + 91 = 0\); 3) \(6n^2 + 41n - 56 = 0\); 4) \(4m^2 + 29m - 63 = 0\); 5) \(3y^2 + 8y - 28 = 0\); 6) \(7y^2 - 12y - 27 = 0\); 7) \(12x^2 + 13x - 35 = 0\); 8) \(15x^2 + 22x - 48 = 0\).

Решение:

1) \(2x^2 - 17x + 30 = 0\); \[D = (-17)^2 - 4 \cdot 2 \cdot 30 = 289 - 240 = 49.\] \(D > 0\), \(x = \frac{17 \pm 7}{4}\), \(x_1 = \frac{10}{4} = 2{,}5\), \(x_2 = \frac{24}{4} = 6\);

2) \(5x^2 - 48x + 91 = 0\); \[D = (-48)^2 - 4 \cdot 5 \cdot 91 = 2304 - 1820 = 484.\] \(D > 0\), \(x = \frac{48 \pm 22}{10}\), \(x_1 = \frac{26}{10} = 2{,}6\), \(x_2 = \frac{70}{10} = 7\);

3) \(6n^2 + 41n - 56 = 0\); \[D = 41^2 - 4 \cdot 6 \cdot (-56) = 1681 + 1344 = 3025.\] \(D > 0\), \(n = \frac{-41 \pm 55}{12}\), \(n_1 = -\frac{96}{12} = -8\), \(n_2 = \frac{14}{12} = 1\frac{1}{6}\);

4) \(4m^2 + 29m - 63 = 0\); \[D = 29^2 - 4 \cdot 4 \cdot (-63) = 841 + 1008 = 1849.\] \(D > 0\), \(m = \frac{-29 \pm 43}{8}\), \(m_1 = -\frac{72}{8} = -9\), \(m_2 = \frac{14}{8} = 1{,}75\);

5) \(3y^2 + 8y - 28 = 0\); \[D = 8^2 - 4 \cdot 3 \cdot (-28) = 64 + 336 = 400.\] \(D > 0\), \(y = \frac{-8 \pm 20}{6}\), \(y_1 = -\frac{28}{6} = -4\frac{2}{3}\), \(y_2 = \frac{12}{6} = 2\);

6) \(7y^2 - 12y - 27 = 0\); \[D = (-12)^2 - 4 \cdot 7 \cdot (-27) = 144 + 756 = 900.\] \(D > 0\), \(y = \frac{12 \pm 30}{14}\), \(y_1 = -\frac{18}{14} = -1\frac{2}{7}\), \(y_2 = \frac{42}{14} = 3\);

7) \(12x^2 + 13x - 35 = 0\); \[D = 13^2 - 4 \cdot 12 \cdot (-35) = 169 + 1680 = 1849.\] \(D > 0\), \(x = \frac{-13 \pm 43}{24}\), \(x_1 = -\frac{56}{24} = -2\frac{1}{3}\), \(x_2 = \frac{30}{24} = 1{,}25\);

8) \(15x^2 + 22x - 48 = 0\); \[D = 22^2 - 4 \cdot 15 \cdot (-48) = 484 + 2880 = 3364.\] \(D > 0\), \(x = \frac{-22 \pm 58}{30}\), \(x_1 = -\frac{80}{30} = -2\frac{2}{3}\), \(x_2 = \frac{36}{30} = 1{,}2\).

Ответ: 1) \(2{,}5\); 6; 2) \(2{,}6\); 7; 3) \(-8\); \(1\frac{1}{6}\); 4) \(-9\); \(1{,}75\); 5) \(-4\frac{2}{3}\); 2; 6) \(-1\frac{2}{7}\); 3; 7) \(-2\frac{1}{3}\); \(1{,}25\); 8) \(-2\frac{2}{3}\); \(1{,}2\).

Задание / условие:

Решить уравнение (3.3.32—3.3.38): 1) \(2x^2 - 17x + 30 = 0\); 2) \(5x^2 - 48x + 91 = 0\); 3) \(6n^2 + 41n - 56 = 0\); 4) \(4m^2 + 29m - 63 = 0\); 5) \(3y^2 + 8y - 28 = 0\); 6) \(7y^2 - 12y - 27 = 0\); 7) \(12x^2 + 13x - 35 = 0\); 8) \(15x^2 + 22x - 48 = 0\).

Решение:

1) \(2x^2 - 17x + 30 = 0\); \[D = (-17)^2 - 4 \cdot 2 \cdot 30 = 289 - 240 = 49.\] Дискриминант положителен, \(x = \frac{17 \pm 7}{4}\), \(x_1 = \frac{10}{4} = 2{,}5\), \(x_2 = \frac{24}{4} = 6\);

2) \(5x^2 - 48x + 91 = 0\); \[D = (-48)^2 - 4 \cdot 5 \cdot 91 = 2304 - 1820 = 484.\] Дискриминант положителен, \(x = \frac{48 \pm 22}{10}\), \(x_1 = \frac{26}{10} = 2{,}6\), \(x_2 = \frac{70}{10} = 7\);

3) \(6n^2 + 41n - 56 = 0\); \[D = 41^2 - 4 \cdot 6 \cdot (-56) = 1681 + 1344 = 3025.\] Дискриминант положителен, \(n = \frac{-41 \pm 55}{12}\), \(n_1 = -\frac{96}{12} = -8\), \(n_2 = \frac{14}{12} = 1\frac{1}{6}\);

4) \(4m^2 + 29m - 63 = 0\); \[D = 29^2 - 4 \cdot 4 \cdot (-63) = 841 + 1008 = 1849.\] Дискриминант положителен, \(m = \frac{-29 \pm 43}{8}\), \(m_1 = -\frac{72}{8} = -9\), \(m_2 = \frac{14}{8} = 1{,}75\);

5) \(3y^2 + 8y - 28 = 0\); \[D = 8^2 - 4 \cdot 3 \cdot (-28) = 64 + 336 = 400.\] Дискриминант положителен, \(y = \frac{-8 \pm 20}{6}\), \(y_1 = -\frac{28}{6} = -4\frac{2}{3}\), \(y_2 = \frac{12}{6} = 2\);

6) \(7y^2 - 12y - 27 = 0\); \[D = (-12)^2 - 4 \cdot 7 \cdot (-27) = 144 + 756 = 900.\] Дискриминант положителен, \(y = \frac{12 \pm 30}{14}\), \(y_1 = -\frac{18}{14} = -1\frac{2}{7}\), \(y_2 = \frac{42}{14} = 3\);

7) \(12x^2 + 13x - 35 = 0\); \[D = 13^2 - 4 \cdot 12 \cdot (-35) = 169 + 1680 = 1849.\] Дискриминант положителен, \(x = \frac{-13 \pm 43}{24}\), \(x_1 = -\frac{56}{24} = -2\frac{1}{3}\), \(x_2 = \frac{30}{24} = 1{,}25\);

8) \(15x^2 + 22x - 48 = 0\); \[D = 22^2 - 4 \cdot 15 \cdot (-48) = 484 + 2880 = 3364.\] Дискриминант положителен, \(x = \frac{-22 \pm 58}{30}\), \(x_1 = -\frac{80}{30} = -2\frac{2}{3}\), \(x_2 = \frac{36}{30} = 1{,}2\).

Ответ: 1) \(2{,}5\); 6; 2) \(2{,}6\); 7; 3) \(-8\); \(1\frac{1}{6}\); 4) \(-9\); \(1{,}75\); 5) \(-4\frac{2}{3}\); 2; 6) \(-1\frac{2}{7}\); 3; 7) \(-2\frac{1}{3}\); \(1{,}25\); 8) \(-2\frac{2}{3}\); \(1{,}2\).

Сообщить об ошибке

Не получилось открыть форму обратной связи.
Напишите нам: nqzva@cbzbtnyxn.zr