Question

Решить уравнения

x2+x-6=0
5x2+3x-8=0
5x2+14x-3=0

Answer 1

$1) {x}^{2} + x - 6 = 0 \\ x1x2 = - 6 \\ x1 + x2 = - 1 \\ x1 = - 3 \\ x2 = 2 \\ \\ 2)5 {x}^{2} + 3x - 8 = 0 \\ d = 9 + 160 = 169 \\ x1 = \frac{ - 3 + 13}{10} = 1 \\ x2 = - \frac{ - 3 - 13}{10} = - 1.6 \\ \\ 3)5 {x}^{2} + 14x - 3 = 0 \\ d = 196 + 60 = 256 \\ x1 = \frac{ - 14 + 16}{10} = 0.2 \\ x2 = \frac{ - 14 - 16}{10} = - 3$

Answer 2

#1

По Виету:

$x_1 + x_2 = -1, \, x_1 x_2 = -6 \Rightarrow x_1 = -3, x_2 = 2$

Через дискриминант:

$D = 1 - 4 \cdot (-6) = 25 \\ x_1 = \frac{-1 - 5}{2} = -3, \, x_2 = \frac{-1 + 5}{2} = 2$

#2

По Виету:

$5x^2 + 3x - 8 = 0 \Leftrightarrow x^2 + \frac{3}{5}x - \frac{8}{5}$

$x_1 + x_2 = -\frac{3}{5}, \, x_1 x_2 = -\frac{8}{5} \Rightarrow x_1 = 1, \, x_2 = -\frac{8}{5}$

Через дискриминант:

$D = 9 - 4 \cdot 5 \cdot (-8) = 169 \\ x_1 = \frac{-3 - 13}{10} = -\frac{8}{5}, \, x_2 = \frac{-3 + 13}{10} = 1$

#3

Через дискриминант:

$D = 14^2 - 4 \cdot 5 \cdot (-3) = 256 \\ x_1 = \frac{-14 - 16}{10} = -3, \, x_2 = \frac{-14 + 16}{10} = \frac{1}{5}$