Question

Знайдіть усі цілі розв'язки нерівності:
2x^2-5x+2<0 СРОЧНО! ПЖ

Answer 1

$2 {x}^{2} - 5x + 2 = 0 \\ a = 2 \\ b = - 5 \\ c = 2 \\ D = {b}^{2} - 4ac = ( - 5) {}^{2} - 4 \times 2 \times 2 = 25 - 16 = 9 \\ x_{1} = \frac{5 - 3}{2 \times 2} = \frac{2}{4} = 0.5 \\ x_{2} = \frac{5 + 3}{2 \times 2} = \frac{8}{4} = 2 \\ \\ {ax}^{2} + bx + c = a(x - x_{1})(x - x_{2}) \\ 2 {x}^{2} - 5x + 2 = 2(x - 0.5)(x - 2) \\ \\ 2 {x}^{2} - 5x + 2 > 0 \\ (x - 0.5)(x - 2) > 0 \\ + + + (0.5) - - - (2) + + + \\ x \: \epsilon\: (0.5; \: 2)$

Ответ: 1