Question

Решите уравнения: 1) х2 + 3x + 2 = 0
2) х2 - 2x – 3 = 0
3) – 2х2 - 10x – 8 = 0
4) x2 – 2х – 4 = 0
5) 3х2 - x+4 = 0
6) 9х2 + 12х + 4 = 0
7) 7х2 + 9x + 2 = 0​

Answer 1

1) (-2;-1)  2) (-1;3)  3) (11.5; 6.5) 4) ($-2\sqrt{5}$; $2\sqrt{5}$ )  5) (1; $\frac{4}{3}$)  6) ($-\frac{2}{3}$)  7) (-1 ; $-\frac{2}{7}$)

Пошаговое объяснение:

1) x² + 3x + 2= 0

D = 3² - 4 * 2 = 1

$x_{1,2} = \frac{- 3 +- \sqrt{1} }{2}$

$x_{1} = \frac{-3-1}{2}= -2\\$

$x_{2} =\frac{-3+1}{2}= -1$

2) x² - 2x - 3 =0

D = (-2)² - 4 * (-3) = 16

$x_{1,2} = \frac{2 +- \sqrt{16} }{2}$

$x_{1}= -1$

$x_{2} = 3$

3) -2x² - 10x - 8 = 0

D = (-10)² - 4 * (-2) * (-8) = 36

$x_{1,2} = \frac{10 +- \sqrt{36} }{2*(-2)}$

$x_{1}= 11.5$

$x_{2} = 6.5$

4) x² - 2x - 4 = 0

D = (-2)² - 4 * (-4) = 20

$x_{1,2} = \frac{2 +- \sqrt{4*5} }{2}$

$x_{1} = \frac{2 - 2\sqrt{5} }{2}= -2\sqrt{5}$

$x_{2} = \frac{2 + 2\sqrt{5} }{2}= 2\sqrt{5}$

5) 3x² - x + 4 = 0

D = 49

$x_{1} = \frac{6}{6} = 1$

$x_{2} = \frac{8}{6} = \frac{4}{3}$

6) 9x² + 12x + 4 = 0

D = 0

$x_{1} = \frac{-12}{18} = -\frac{2}{3}$

7) 7x² + 9x  + 2= 0

D = 25

$x_{1} = \frac{-14}{14} = -1$

$x_{2} = \frac{-4}{14} = -\frac{2}{7}$