Предмет: Алгебра
Автор: sofipolishchuk2
Вопрос задан 7 лет назад

найти дискриминант и корни
119(3,4,5,6)
120(1,2,4)
125
срочно~

Приложения:

Ответы

Ответ дал: wsaspo

Ответ:

$119. \ 3)10 {x}^{2} - 9x + 2 = 0 \ d = {b}^{2} - 4ac = 81 - 80 = 1 \ x1 = frac{9 - 1}{2 times 10} = 0.4 \ x2 = frac{9 + 1}{2 times 10} = 0.5$

$4) : 21 {x}^{2} - 2x - 3 = 0 \ d = 4 + 252 = sqrt{256} = 16 \ x1 = frac{2 - 16}{2 times 21} = frac{ - 14}{42} = - frac{1}{3} \ x2 = frac{2 + 16}{42} = frac{3}{7}$

$5) : {x}^{2} + 8x - 13 = 0 \ d = 64 + 52 = sqrt{116} = 10.7 \ x1 = frac{ - 8 - 10.7}{2} = - 9.35 \ x2 = frac{ - 8 + 10.7}{2} = 1.35$

$6) : 2 {x}^{2} - 4x - 17 = 0 \ d = 16 + 136 = sqrt{152} = 12.3 \ x1 = frac{4 - 12.3}{4} = - 2.075 \ x2 = frac{4 + 12.3}{4} = 4.075$

$120. \ 1) : 3 {x}^{2} - 12x + 2x - 8 = 5 \ 3 {x}^{2} - 10x - 13 = 0 \ d = 100 + 156 = sqrt{256} = 16 \ x1 = frac{10 - 16}{6} = - 1 \ x2 = frac{10 + 16}{6} = 4.3= 4 times frac{1}{3}$

$2) : {x}^{2} - 2x + x - 2 - 4 {x}^{2} - 20x + 3x + 15 = {x}^{2} - 9x \ - 4 {x}^{2} - 9x + 13 = 0 : times ( - 1) \ 4 {x}^{2} + 9x - 13 = 0 \ d = 81 + 208 = sqrt{289} = 17 \ x1 = frac{ - 9 - 17}{8} = - 3.25 \ x2 = frac{ - 9 + 17}{8} = 1$

$4) : 27 {x}^{3 } + 1 - 27 {x}^{3} + 18 {x}^{2} - 6x + 4 = 16 {x}^{2} + 1 \ 2 {x}^{2} - 6x - 4 = 0 : : div 2 \ {x}^{2} - 3x - 2 = 0 \ d = 9 + 8 = sqrt{17} = 4.1 \ x1 = frac{3 + 4.1}{2} = 3.55 \ x2 = frac{3 - 4.1}{2} = - 0.55$