Реши уравнения: 1) x²+x=0

Привет! Давай решим эти уравнения. 1) $x^2 + x = 0$ $x(x + 1) = 0$ $x_1 = 0$, $x_2 = -1$ 2) $x^2 - 2x - 2 = 0$ $D = (-2)^2 - 4 ">" 1 "*" (-2) = 4 + 8 = 12$ $x_{1,2} = \frac{-(-2) \pm \sqrt{12}}{2 "*" 1} = \frac{2 \pm 2\sqrt{3}}{2} = 1 \pm \sqrt{3}$ $x_1 = 1 + \sqrt{3} \approx 2.73$, $x_2 = 1 - \sqrt{3} \approx -0.73$ 3) $7x^2 - 4 = 0$ $7x^2 = 4$ $x^2 = \frac{4}{7}$ $x_{1,2} = \pm \sqrt{\frac{4}{7}} = \pm \frac{2}{\sqrt{7}} = \pm \frac{2\sqrt{7}}{7}$ $x_1 = \frac{2\sqrt{7}}{7} \approx 0.76$, $x_2 = -\frac{2\sqrt{7}}{7} \approx -0.76$ 4) $x^2 - 4x + 3 = 0$ $D = (-4)^2 - 4 "*" 1 "*" 3 = 16 - 12 = 4$ $x_{1,2} = \frac{-(-4) \pm \sqrt{4}}{2 "*" 1} = \frac{4 \pm 2}{2}$ $x_1 = \frac{4 + 2}{2} = 3$, $x_2 = \frac{4 - 2}{2} = 1$ 5) $5x = 3x^2$ $3x^2 - 5x = 0$ $x(3x - 5) = 0$ $x_1 = 0$, $x_2 = \frac{5}{3} \approx 1.67$ 6) $5x^2 + 14x - 3 = 0$ $D = 14^2 - 4 "*" 5 "*" (-3) = 196 + 60 = 256$ $x_{1,2} = \frac{-14 \pm \sqrt{256}}{2 "*" 5} = \frac{-14 \pm 16}{10}$ $x_1 = \frac{-14 + 16}{10} = \frac{2}{10} = 0.2$, $x_2 = \frac{-14 - 16}{10} = \frac{-30}{10} = -3$ 7) $3x^2 - x + 2 = 0$ $D = (-1)^2 - 4 "*" 3 "*" 2 = 1 - 24 = -23$ Так как дискриминант отрицательный, действительных корней нет. 8) $x^2 - 5x + 4 = 0$ $D = (-5)^2 - 4 "*" 1 "*" 4 = 25 - 16 = 9$ $x_{1,2} = \frac{-(-5) \pm \sqrt{9}}{2 "*" 1} = \frac{5 \pm 3}{2}$ $x_1 = \frac{5 + 3}{2} = 4$, $x_2 = \frac{5 - 3}{2} = 1$