Question

ДАМ 100 БАЛЛОВ!!!
Помогите с ср пожалуйста срочноо

Answer 1

1)

1)

$5 {x}^{2} - 20 = 0 \\ {x}^{2} - 4 = 0 \\( x - 2)(x +2 ) = 0 \\ x_{1} =2 \\ x_{2} = - 2$

2)

${x}^{2} + 7x = 0 \\ x(x + 7) = 0 \\ x_{1} = 0\\ x_{2} = - 7$

3)

$3 {x}^{2} - 18 = 0 \\ {x}^{2} - 9 = 0 \\ (x - 3)(x + 3) = 0 \\ x_{1} =3 \\ x_{2} = - 3$

2)

1)

${x}^{2} + 5x - 14 = 0 \\ d = {5}^{2} - 4 \times 1 \times ( - 14) = \\ 25 + 56 = 81 \\ x_{1} = \frac{ - 5 + 9}{2} = \frac{4}{2} = 2 \\ x_{2} = \frac{ - 5 - 9}{2} = - \frac{14}{2} = - 7$

2)

${x}^{2} - 14x + 40 = \\ d = ( - 14) {}^{2} - 4 \times 1 \times 40 = \\ 196 - 160 = 36 \\ x_{1} = \frac{14 + 6}{2} = \frac{20}{2} = 10\\ x_{2} = \frac{14 - 6}{2} = \frac{8}{2} = 4$

3)

$3 {y}^{2} - 13y + 4 = 0 \\ d =( - 13) {}^{2} - 4 \times 3 \times 4 = \\ 169 - 48 = 121 \\ x_{1} = \frac{13 + 11}{3 \times 2} = \frac{24}{6} = 4 \\ x_{2} = \frac{13 - 11}{2 \times 3} = \frac{2}{6} = \frac{1}{3}$

4)

$12 {r}^{2} + r - 6 = 0 \\ d = {1}^{2} - 4 \times 12 \times ( - 6) = \\ 1+ 288 = 289 \\ r_{1} = \frac{ - 1 + 17}{2 \times 12} = \frac{16}{24} = \frac{2}{3} \\ r_{2} = \frac{ - 1 - 17}{2 \times 12} = - \frac{18}{24} = \\ - \frac{3}{4} = - 0.75$