Question

{x²+4x+3<_0
{2x²+5x<0
СРОЧНО ДАЮ 100 БАЛОВ​

Answer 1

$\displaystyle\bf\\\left \{ {{ {x }^{2} + 4x + 3 \leqslant 0} \atop {2 {x}^{2} + 5x < 0}} \right. \\ \\ 1) \: {x}^{2} + 4x + 3 \leqslant 0 \\ {x}^{2} + 4x + 3 = 0 \\ \\ po \: \: \: teoreme \: \: \: vieta \\ {x}^{2} + bx + c = 0\\ x_{1} + x_{2} = - b\\ x_{1} x_{2} = c \\ \\ x_{1} + x_{2} = - 4\\ x_{1} x_{2} =3 \\ x_{1} = - 3 \\ x_{2} = - 1 \\ \\ {ax}^{2} + bx + c = a(x - x_{1})(x - x_{2}) \\ {x}^{2} + 4x + 3 = (x + 3)(x + 1) \\ \\ (x + 3)(x + 1) \leqslant 0 \\ + + + [ - 3] - - - [ - 1] + + + \\ - 3 \leqslant x \leqslant - 1 \\ \\ 2) \: 2 {x}^{2} + 5x < 0 \\ {x}^{2} + 2.5x < 0 \\ x(x + 2.5) < 0 \\ + + + ( - 2.5) - - - (0) + + + \\ - 2.5 < x < 0 \\ \displaystyle\bf\\3) \: \left \{ {{ - 3 \leqslant x \leqslant - 1} \atop { - 2.5 < x < 0 }} \right. \\ \\ x \: \epsilon \:( - 2.5 ; \: - 1]$