Страница 120 номер 5.2.21, ГДЗ по алгебре за 7, 8 и 9 класс к задачнику Ткачевой
Нули функции \(y = x^2 + px + q\) — корни уравнения \(x^2 + px + q = 0\). По теореме Виета \(p = -(x_1 + x_2)\), \(q = x_1 x_2\).
1) \(x_1 + x_2 = 1 + (-3) = -2\), \(x_1 x_2 = 1 \cdot (-3) = -3\); значит, \(p = 2\), \(q = -3\).
2) \(x_1 + x_2 = -5 + 4 = -1\), \(x_1 x_2 = -5 \cdot 4 = -20\); значит, \(p = 1\), \(q = -20\).
3) \(x_1 + x_2 = -\sqrt{3} + 2\sqrt{3} = \sqrt{3}\), \(x_1 x_2 = -\sqrt{3} \cdot 2\sqrt{3} = -2 \cdot 3 = -6\); значит, \(p = -\sqrt{3}\), \(q = -6\).
4) \(x_1 + x_2 = -2\sqrt{2} + \sqrt{2} = -\sqrt{2}\), \(x_1 x_2 = -2\sqrt{2} \cdot \sqrt{2} = -2 \cdot 2 = -4\); значит, \(p = \sqrt{2}\), \(q = -4\).
5) \(x_1 + x_2 = 3\sqrt{5} + 2\sqrt{5} = 5\sqrt{5}\), \(x_1 x_2 = 3\sqrt{5} \cdot 2\sqrt{5} = 6 \cdot 5 = 30\); значит, \(p = -5\sqrt{5}\), \(q = 30\).
6) \(x_1 + x_2 = -\sqrt{6} + \left(-2\sqrt{6}\right) = -3\sqrt{6}\), \(x_1 x_2 = -\sqrt{6} \cdot \left(-2\sqrt{6}\right) = 2 \cdot 6 = 12\); значит, \(p = 3\sqrt{6}\), \(q = 12\).
7) \(x_1 + x_2 = \left(1 - \sqrt{2}\right) + \left(1 + \sqrt{2}\right) = 2\), \(x_1 x_2 = \left(1 - \sqrt{2}\right)\left(1 + \sqrt{2}\right) = 1 - 2 = -1\); значит, \(p = -2\), \(q = -1\).
8) \(x_1 + x_2 = \left(2 + \sqrt{3}\right) + \left(2 - \sqrt{3}\right) = 4\), \(x_1 x_2 = \left(2 + \sqrt{3}\right)\left(2 - \sqrt{3}\right) = 4 - 3 = 1\); значит, \(p = -4\), \(q = 1\).
Ответ: 1) \(p = 2\), \(q = -3\); 2) \(p = 1\), \(q = -20\); 3) \(p = -\sqrt{3}\), \(q = -6\); 4) \(p = \sqrt{2}\), \(q = -4\); 5) \(p = -5\sqrt{5}\), \(q = 30\); 6) \(p = 3\sqrt{6}\), \(q = 12\); 7) \(p = -2\), \(q = -1\); 8) \(p = -4\), \(q = 1\).
Нули функции \(y = x^2 + px + q\) — это корни уравнения \(x^2 + px + q = 0\). По теореме Виета сумма корней приведённого квадратного уравнения равна \(-p\), а произведение равно \(q\), поэтому \(p = -(x_1 + x_2)\), \(q = x_1 x_2\).
1) \(x_1 + x_2 = 1 + (-3) = -2\), \(x_1 x_2 = 1 \cdot (-3) = -3\); значит, \(p = 2\), \(q = -3\).
2) \(x_1 + x_2 = -5 + 4 = -1\), \(x_1 x_2 = -5 \cdot 4 = -20\); значит, \(p = 1\), \(q = -20\).
3) \(x_1 + x_2 = -\sqrt{3} + 2\sqrt{3} = \sqrt{3}\), \(x_1 x_2 = -\sqrt{3} \cdot 2\sqrt{3} = -2 \cdot 3 = -6\); значит, \(p = -\sqrt{3}\), \(q = -6\).
4) \(x_1 + x_2 = -2\sqrt{2} + \sqrt{2} = -\sqrt{2}\), \(x_1 x_2 = -2\sqrt{2} \cdot \sqrt{2} = -2 \cdot 2 = -4\); значит, \(p = \sqrt{2}\), \(q = -4\).
5) \(x_1 + x_2 = 3\sqrt{5} + 2\sqrt{5} = 5\sqrt{5}\), \(x_1 x_2 = 3\sqrt{5} \cdot 2\sqrt{5} = 6 \cdot 5 = 30\); значит, \(p = -5\sqrt{5}\), \(q = 30\).
6) \(x_1 + x_2 = -\sqrt{6} + \left(-2\sqrt{6}\right) = -3\sqrt{6}\), \(x_1 x_2 = -\sqrt{6} \cdot \left(-2\sqrt{6}\right) = 2 \cdot 6 = 12\); значит, \(p = 3\sqrt{6}\), \(q = 12\).
7) \(x_1 + x_2 = \left(1 - \sqrt{2}\right) + \left(1 + \sqrt{2}\right) = 2\), \(x_1 x_2 = \left(1 - \sqrt{2}\right)\left(1 + \sqrt{2}\right) = 1 - 2 = -1\); значит, \(p = -2\), \(q = -1\).
8) \(x_1 + x_2 = \left(2 + \sqrt{3}\right) + \left(2 - \sqrt{3}\right) = 4\), \(x_1 x_2 = \left(2 + \sqrt{3}\right)\left(2 - \sqrt{3}\right) = 4 - 3 = 1\); значит, \(p = -4\), \(q = 1\).
Ответ: 1) \(p = 2\), \(q = -3\); 2) \(p = 1\), \(q = -20\); 3) \(p = -\sqrt{3}\), \(q = -6\); 4) \(p = \sqrt{2}\), \(q = -4\); 5) \(p = -5\sqrt{5}\), \(q = 30\); 6) \(p = 3\sqrt{6}\), \(q = 12\); 7) \(p = -2\), \(q = -1\); 8) \(p = -4\), \(q = 1\).
