Question

СРОЧНО! решите неравенство

Answer 1

$- {x}^{2} + 3x + 4 \geqslant 0 \\ {x}^{2} - 3x - 4 \leqslant 0 \\ {x}^{2} - 3x - 4 = 0 \\ po \: \: \: teoreme \: \: \: vieta \\ {x}^{2} + bx + c = 0\\ x_{1} + x_{2} = - b\\ x_{1} x_{2} = c \\ x_{1}  +  x_{2} =  3\\ x_{1} x_{2} = - 4 \\ x_{1} =  - 1 \\ x_{2} =4 \\ {ax}^{2} + bx + c = a(x - x_{1})(x - x_{2}) \\ {x}^{2} - 3x - 4 = (x + 1)(x - 4) \\ \\( x + 1)(x - 4) \leqslant 0 \\ + + + + + [ - 1] - - - - - [4] + + + + + \\ x \: \epsilon \: [ - 1; \:4 ]$