Question

4. Решите систему неравенств: -x² + 4x + 12 < 0, (4 - 2(x + 1) >x-7 RauuuT epareCTRO​

Answer 1

$\displaystyle\bf\\\left \{ {{ - {x}^{2} + 4x + 12 \leqslant 0} \atop {4 - 2(x + 1) > x - 7 }} \right. \\ \displaystyle\bf\\\left \{ {{ {x}^{2} - 4x - 12 \geqslant 0 } \atop {4 - 2x - 2 - x > - 7 }} \right.$

Найдём корни квадратного трёхчлена по теореме Виета:

${x}^{2} + bx + c = 0 \\ x _{1} + x_{2} = - b \\ x _{1} x_{2} = c$

${x}^{2} - 4x - 12 = 0 \\ x_{1} + x_{2} = 4 \\ x_{1} x_{2} = - 12 \\ x_{1} = 6\\ x_{2} = - 2 \\ \\ {x}^{2} - 4x - 12 = (x - 6)(x + 2)$

$\displaystyle\bf\\\left \{ {{(x - 6)(x + 2) \geqslant 0} \atop { - 3x > - 9 }} \right. \\ \displaystyle\bf\\\left \{ {{(x - 6)(x + 2) \geqslant 0} \atop {x < 3 }} \right. \\ \displaystyle\bf\\\left \{ {{x\epsilon( -\propto; - 2]U[6; + \propto) } \atop {x \epsilon( - \propto;3)}} \right.$

Ответ: х ∈ ( - ∞ ; - 2 ]