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

Question 1

Question 2

Ответ:

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

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

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

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

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

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

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

Question 3

спасибо огромное!