решение
Конечно! Вот решения каждого из уравнений:
\(x^2 - 6x - 27 = 0\)
Используем формулу для квадратного уравнения \(ax^2 + bx + c = 0\): \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Здесь \(a = 1\), \(b = -6\), \(c = -27\).
\[ x = \frac{6 \pm \sqrt{(-6)^2 - 4 \times 1 \times (-27)}}{2 \times 1} \] \[ x = \frac{6 \pm \sqrt{36 + 108}}{2} \] \[ x = \frac{6 \pm \sqrt{144}}{2} \] \[ x = \frac{6 \pm 12}{2} \]
\[ x_1 = \frac{18}{2} = 9, \quad x_2 = \frac{-6}{2} = -3 \]
\(x^2 - x - 20 = 0\)
\(a = 1\), \(b = -1\), \(c = -20\).
\[ x = \frac{1 \pm \sqrt{1^2 - 4 \times 1 \times (-20)}}{2 \times 1} \] \[ x = \frac{1 \pm \sqrt{1 + 80}}{2} \] \[ x = \frac{1 \pm \sqrt{81}}{2} \] \[ x = \frac{1 \pm 9}{2} \]
\[ x_1 = \frac{10}{2} = 5, \quad x_2 = \frac{-8}{2} = -4 \]
\(3x^2 + 3x - 18 = 0\)
Сначала упростим: \(x^2 + x - 6 = 0\).
\(a = 1\), \(b = 1\), \(c = -6\).
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \times 1 \times (-6)}}{2 \times 1} \] \[ x = \frac{-1 \pm \sqrt{1 + 24}}{2} \] \[ x = \frac{-1 \pm \sqrt{25}}{2} \] \[ x = \frac{-1 \pm 5}{2} \]
\[ x_1 = \frac{4}{2} = 2, \quad x_2 = \frac{-6}{2} = -3 \]
\(3x^2 - 19x - 14 = 0\)
\(a = 3\), \(b = -19\), \(c = -14\).
\[ x = \frac{19 \pm \sqrt{(-19)^2 - 4 \times 3 \times (-14)}}{2 \times 3} \] \[ x = \frac{19 \pm \sqrt{361 + 168}}{6} \] \[ x = \frac{19 \pm \sqrt{529}}{6} \] \[ x = \frac{19 \pm 23}{6} \]
\[ x_1 = \frac{42}{6} = 7, \quad x_2 = \frac{-4}{6} = -\frac{2}{3} \]
\(2x^2 + 11x + 24 = 0\)
\(a = 2\), \(b = 11\), \(c = 24\).
\[ x = \frac{-11 \pm \sqrt{11^2 - 4 \times 2 \times 24}}{2 \times 2} \] \[ x = \frac{-11 \pm \sqrt{121 - 192}}{4} \] \[ x = \frac{-11 \pm \sqrt{-71}}{4} \]
Дискриминант отрицательный, действительных корней нет, но есть комплексные корни.
