Страница 71 номер 3.3.10, ГДЗ по алгебре за 7, 8 и 9 класс к задачнику Ткачевой
Решить квадратное уравнение: 1) \(5x^2 + x - 4 = 0\); 2) \(4x^2 + x - 3 = 0\); 3) \(6y^2 - y - 22 = 0\); 4) \(3y^2 + y - 10 = 0\); 5) \(-4p^2 + 12p - 5 = 0\); 6) \(-6n^2 + n + 1 = 0\); 7) \(12x^2 - 3x - 15 = 0\); 8) \(10x^2 + 11x - 6 = 0\).
1) \(5x^2 + x - 4 = 0\): \(D = 1^2 - 4 \cdot 5 \cdot (-4) = 1 + 80 = 81\), \(D > 0\); \(x = \frac{-1 \pm \sqrt{81}}{10} = \frac{-1 \pm 9}{10}\), отсюда \(x_1 = -1\), \(x_2 = 0{,}8\);
2) \(4x^2 + x - 3 = 0\): \(D = 1^2 - 4 \cdot 4 \cdot (-3) = 1 + 48 = 49\), \(D > 0\); \(x = \frac{-1 \pm \sqrt{49}}{8} = \frac{-1 \pm 7}{8}\), отсюда \(x_1 = -1\), \(x_2 = \frac{3}{4}\);
3) \(6y^2 - y - 22 = 0\): \(D = (-1)^2 - 4 \cdot 6 \cdot (-22) = 1 + 528 = 529\), \(D > 0\); \(y = \frac{1 \pm \sqrt{529}}{12} = \frac{1 \pm 23}{12}\), отсюда \(y_1 = -1\frac{5}{6}\), \(y_2 = 2\);
4) \(3y^2 + y - 10 = 0\): \(D = 1^2 - 4 \cdot 3 \cdot (-10) = 1 + 120 = 121\), \(D > 0\); \(y = \frac{-1 \pm \sqrt{121}}{6} = \frac{-1 \pm 11}{6}\), отсюда \(y_1 = -2\), \(y_2 = 1\frac{2}{3}\);
5) \(-4p^2 + 12p - 5 = 0\); умножим обе части на \(-1\): \(4p^2 - 12p + 5 = 0\); \(D = (-12)^2 - 4 \cdot 4 \cdot 5 = 144 - 80 = 64\), \(D > 0\); \(p = \frac{12 \pm \sqrt{64}}{8} = \frac{12 \pm 8}{8}\), отсюда \(p_1 = 0{,}5\), \(p_2 = 2{,}5\);
6) \(-6n^2 + n + 1 = 0\); умножим обе части на \(-1\): \(6n^2 - n - 1 = 0\); \(D = (-1)^2 - 4 \cdot 6 \cdot (-1) = 1 + 24 = 25\), \(D > 0\); \(n = \frac{1 \pm \sqrt{25}}{12} = \frac{1 \pm 5}{12}\), отсюда \(n_1 = -\frac{1}{3}\), \(n_2 = \frac{1}{2}\);
7) \(12x^2 - 3x - 15 = 0\); разделим обе части на 3: \(4x^2 - x - 5 = 0\); \(D = (-1)^2 - 4 \cdot 4 \cdot (-5) = 1 + 80 = 81\), \(D > 0\); \(x = \frac{1 \pm \sqrt{81}}{8} = \frac{1 \pm 9}{8}\), отсюда \(x_1 = -1\), \(x_2 = 1{,}25\);
8) \(10x^2 + 11x - 6 = 0\): \(D = 11^2 - 4 \cdot 10 \cdot (-6) = 121 + 240 = 361\), \(D > 0\); \(x = \frac{-11 \pm \sqrt{361}}{20} = \frac{-11 \pm 19}{20}\), отсюда \(x_1 = -1{,}5\), \(x_2 = 0{,}4\).
Ответ: 1) \(-1\); \(0{,}8\); 2) \(-1\); \(\frac{3}{4}\); 3) \(-1\frac{5}{6}\); 2; 4) \(-2\); \(1\frac{2}{3}\); 5) \(0{,}5\); \(2{,}5\); 6) \(-\frac{1}{3}\); \(\frac{1}{2}\); 7) \(-1\); \(1{,}25\); 8) \(-1{,}5\); \(0{,}4\).
Решить квадратное уравнение: 1) \(5x^2 + x - 4 = 0\); 2) \(4x^2 + x - 3 = 0\); 3) \(6y^2 - y - 22 = 0\); 4) \(3y^2 + y - 10 = 0\); 5) \(-4p^2 + 12p - 5 = 0\); 6) \(-6n^2 + n + 1 = 0\); 7) \(12x^2 - 3x - 15 = 0\); 8) \(10x^2 + 11x - 6 = 0\).
Вычислим дискриминант \(D = b^2 - 4ac\) и, если он неотрицателен, найдём корни по формуле \(x = \frac{-b \pm \sqrt{D}}{2a}\).
1) \(5x^2 + x - 4 = 0\): \(D = 1^2 - 4 \cdot 5 \cdot (-4) = 1 + 80 = 81\), \(D > 0\); \(x = \frac{-1 \pm \sqrt{81}}{10} = \frac{-1 \pm 9}{10}\), отсюда \(x_1 = -1\), \(x_2 = 0{,}8\);
2) \(4x^2 + x - 3 = 0\): \(D = 1^2 - 4 \cdot 4 \cdot (-3) = 1 + 48 = 49\), \(D > 0\); \(x = \frac{-1 \pm \sqrt{49}}{8} = \frac{-1 \pm 7}{8}\), отсюда \(x_1 = -1\), \(x_2 = \frac{3}{4}\);
3) \(6y^2 - y - 22 = 0\): \(D = (-1)^2 - 4 \cdot 6 \cdot (-22) = 1 + 528 = 529\), \(D > 0\); \(y = \frac{1 \pm \sqrt{529}}{12} = \frac{1 \pm 23}{12}\), отсюда \(y_1 = -1\frac{5}{6}\), \(y_2 = 2\);
4) \(3y^2 + y - 10 = 0\): \(D = 1^2 - 4 \cdot 3 \cdot (-10) = 1 + 120 = 121\), \(D > 0\); \(y = \frac{-1 \pm \sqrt{121}}{6} = \frac{-1 \pm 11}{6}\), отсюда \(y_1 = -2\), \(y_2 = 1\frac{2}{3}\);
5) \(-4p^2 + 12p - 5 = 0\); умножим обе части на \(-1\): \(4p^2 - 12p + 5 = 0\). Тогда \(D = (-12)^2 - 4 \cdot 4 \cdot 5 = 144 - 80 = 64\), \(D > 0\); \(p = \frac{12 \pm \sqrt{64}}{8} = \frac{12 \pm 8}{8}\), отсюда \(p_1 = 0{,}5\), \(p_2 = 2{,}5\);
6) \(-6n^2 + n + 1 = 0\); умножим обе части на \(-1\): \(6n^2 - n - 1 = 0\). Тогда \(D = (-1)^2 - 4 \cdot 6 \cdot (-1) = 1 + 24 = 25\), \(D > 0\); \(n = \frac{1 \pm \sqrt{25}}{12} = \frac{1 \pm 5}{12}\), отсюда \(n_1 = -\frac{1}{3}\), \(n_2 = \frac{1}{2}\);
7) \(12x^2 - 3x - 15 = 0\); разделим обе части на 3: \(4x^2 - x - 5 = 0\). Тогда \(D = (-1)^2 - 4 \cdot 4 \cdot (-5) = 1 + 80 = 81\), \(D > 0\); \(x = \frac{1 \pm \sqrt{81}}{8} = \frac{1 \pm 9}{8}\), отсюда \(x_1 = -1\), \(x_2 = 1{,}25\);
8) \(10x^2 + 11x - 6 = 0\): \(D = 11^2 - 4 \cdot 10 \cdot (-6) = 121 + 240 = 361\), \(D > 0\); \(x = \frac{-11 \pm \sqrt{361}}{20} = \frac{-11 \pm 19}{20}\), отсюда \(x_1 = -1{,}5\), \(x_2 = 0{,}4\).
Ответ: 1) \(-1\); \(0{,}8\); 2) \(-1\); \(\frac{3}{4}\); 3) \(-1\frac{5}{6}\); 2; 4) \(-2\); \(1\frac{2}{3}\); 5) \(0{,}5\); \(2{,}5\); 6) \(-\frac{1}{3}\); \(\frac{1}{2}\); 7) \(-1\); \(1{,}25\); 8) \(-1{,}5\); \(0{,}4\).
