Задание 4, стр 60

Answers

a) $5 t^{2} - 11 t + 2 = 0$

D = b^{2} - 4 ac = 1 1^{2} - 4 \cdot 5 \cdot 2 = 121 - 40 = 81 .

t_{1} = \frac{- (- 11) + \sqrt{81}}{2 \cdot 5} = \frac{11 + 9}{10} = \frac{20}{10} = 2, t_{2} = \frac{- (- 11) - \sqrt{81}}{2 \cdot 5} = \frac{11 - 9}{10} = \frac{2}{10} = \frac{1}{5} .

b) $35 y^{2} + 2 y - 1 = 0$

\begin{aligned} D = 2^{2} - 4 \cdot 35 \cdot (- 1) = 4 + 140 = 144 \\ y_{1} = \frac{- 2 + \sqrt{144}}{35 \cdot 2} = \frac{- 2 + 12}{70} = \frac{10}{70} = \frac{1}{7}; y_{2} = \frac{- 2 - \sqrt{144}}{35 \cdot 2} = \frac{- 2 - 12}{70} = \frac{- 14}{70} = - \frac{1}{5} . \end{aligned}

c) $2 z^{2} + 7 z + 4 = 0$

\begin{aligned} D = 7^{2} - 2 \cdot 4 \cdot 4 = 49 - 32 = 17 \\ z_{1} = \frac{- 7 + \sqrt{17}}{4}; z_{2} = \frac{- 7 - \sqrt{17}}{4} . \end{aligned}

d) $x^{2} - x - 1 = 0$

\begin{aligned} D = 1^{2} + 1 \cdot 4 = 1 + 4 = 5 \\ x_{1} = \frac{1 + \sqrt{5}}{2}, x_{2} = \frac{1 - \sqrt{5}}{2} . \end{aligned}

gimli

2022-10-26 11:39