Найдем корни квадратных уравнений, используя формулу \(x = \frac{-b \pm \sqrt{D}}{2a}\) где \(D = b^2 - 4ac\).
а) \(10x^2 - 3x - 0.4 = 0\) : a=10, b=-3, c=-0.4. \(D = (-3)^2 - 4*10*(-0.4) = 9 + 16 = 25\). \(x = \frac{3 \pm \sqrt{25}}{2*10}\). Корни: \(x_1 = \frac{3 - 5}{20} = -0.1\), \(x_2 = \frac{3 + 5}{20} = 0.4\)
б) \(7x^2 + 6x - 1 = 0\): a=7, b=6, c=-1. \(D = 6^2 - 4*7*(-1) = 36 + 28 = 64\). \(x = \frac{-6 \pm \sqrt{64}}{2*7}\). Корни: \(x_1 = \frac{-6 - 8}{14} = -1\), \(x_2 = \frac{-6 + 8}{14} = \frac{1}{7}\)
в) \(3x^2 - 4x + 2 = 0\): a=3, b=-4, c=2. \(D = (-4)^2 - 4*3*2 = 16 - 24 = -8\). Так как D < 0 то корней нет.
г) \(x^2 + 12 = 7x\) => \(x^2 - 7x + 12 = 0\) : a=1, b=-7, c=12. \(D = (-7)^2 - 4*1*12 = 49 - 48 = 1\). \(x = \frac{7 \pm \sqrt{1}}{2}\). Корни: \(x_1 = \frac{7 - 1}{2} = 3\), \(x_2 = \frac{7 + 1}{2} = 4\)
д) \(7y^2 + 5y = 2\) => \(7y^2 + 5y - 2 = 0\) : a=7, b=5, c=-2. \(D = 5^2 - 4*7*(-2) = 25 + 56 = 81\). \(y = \frac{-5 \pm \sqrt{81}}{2*7}\). Корни: \(y_1 = \frac{-5 - 9}{14} = -1\), \(y_2 = \frac{-5 + 9}{14} = \frac{2}{7}\)
е) \(1 + 8x = 9x^2\) => \(9x^2 - 8x - 1 = 0\) : a=9, b=-8, c=-1. \(D = (-8)^2 - 4*9*(-1) = 64 + 36 = 100\). \(x = \frac{8 \pm \sqrt{100}}{2*9}\). Корни: \(x_1 = \frac{8 - 10}{18} = -\frac{1}{9}\), \(x_2 = \frac{8 + 10}{18} = 1\)
Ответ:
a) x = -0.1, x = 0.4
б) x = -1, x = 1/7
в) нет корней
г) x = 3, x = 4
д) y = -1, y = 2/7
е) x = -1/9, x = 1