Задание 3, стр 60

Answers

a): $x^{2} - 8 x + 15 = 0$ $\begin{aligned} D = b^{2} - 4 ac = {(- 8)}^{2} - 4 \cdot 15 \cdot 1 = 64 - 60 = 4 = 2^{2} \\ x_{1} = \frac{- b + \sqrt{D}}{2 a} = \frac{8 + 2}{2} = 5 . x_{2} = \frac{- b - \sqrt{D}}{2 a} = \frac{8 - 2}{2} = 3 . \end{aligned}$
OTBET: $x \in {5; 3}$
b): $\begin{aligned} x^{2} + 14 x + 40 = 0 \\ D = 1 4^{2} - 40 \cdot 4 = 196 - 160 = 36 = 6^{2} \\ x_{1} = \frac{- 14 + 6}{2} = \frac{- 8}{2} = - 4; x_{2} = \frac{- 14 - 6}{2} = \frac{- 20}{2} = - 10 \end{aligned}$
ОТВЕТ: $x \in {- 4; - 10} .$
c): $x^{2} - 7 x + 6 = 0$
$D = {(- 7)}^{2} - 6 \cdot 4 = 49 - 24 = 25 = 5^{2}$
$x_{1} = \frac{7 - 5}{2} = \frac{2}{2} = 1; x_{2} = \frac{7 + 5}{2} = \frac{12}{2} = 6$ .
OTBET: $x \in {1; 6}$ .
d): $\begin{aligned} x^{2} - 11 x + 10 = 0 . \\ D = {(- 11)}^{2} - 10 \cdot 4 = 121 - 40 = 81 = 9^{2} \\ x_{1} = \frac{11 - 9}{2} = 1; x_{2} = \frac{- (- 11) + 9}{2} = \frac{11 + 9}{2} = 10 \\ OTBET: x \in {1; 10} . \end{aligned}$

gimli

2022-11-11 14:38