17А. Найдите все значения параметра a, при каждом из которых уравнение \(\left( {a + 6} \right)\,{x^2} + 2\left( {a-6} \right)\,x-2a + 6 = 0\) имеет два различных корня, модуль каждого из которых меньше 2.
Если \(a = -6\), то уравнение будет линейным, которое не может иметь два различных корня. Поэтому \(a = -6\) не подходит. Первый случай (см. рис. 1): \(\left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{a > -6,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{D > 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}}\\{-2 < {x_B} = -\frac{b}{{2a}} < 2,}\\{f\left( {-2} \right) > 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{f\left( 2 \right) > 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{a > -6,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{4{{\left( {a-6} \right)}^2}-4\left( {a + 6} \right)\left( {6-2a} \right) > 0,\,}\end{array}}\\{-2 < \frac{{6-a}}{{a + 6}} < 2,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{4a + 24-4a + 24-2a + 6 > 0,\,\,\,\,\,\,}\\{4a + 24 + 4a-24-2a + 6 > 0\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{a > -6,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{12{a^2}-24a > 0,\,\,\,}\end{array}}\\{\frac{{-3a-6}}{{a + 6}} < 0,\,\,\,\,\,\,\,\,\,\,}\\{\frac{{a + 18}}{{a + 6}} > 0,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\begin{array}{*{20}{c}}{54-2a > 0,}\\{6a + 6 > 0\,\,\,\,}\end{array}\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{a \in \left( {-6;\infty } \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{a\, \in \,\left( {-\infty ;0} \right) \cup \left( {2;\infty } \right),\,\,\,\,\,\,\,\,\,}\\{a\, \in \,\left( {-\infty ;-6} \right) \cup \left( {-2;\infty } \right),\,\,}\\{\begin{array}{*{20}{c}}{a\, \in \,\left( {-\infty ;-18} \right) \cup \left( {-6;\infty } \right),}\\{a \in \left( {-\infty ;\,27} \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{a \in \left( {-1;\,\infty } \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,a\, \in \,\left( {-1;0} \right) \cup \left( {2;27} \right).\) \(\left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{a < -6,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{D > 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}}\\{-2 < {x_B} = -\frac{b}{{2a}} < 2,}\\{f\left( {-2} \right) < 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{f\left( 2 \right) < 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{a < -6,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{4{{\left( {a-6} \right)}^2}-4\left( {a + 6} \right)\left( {6-2a} \right) > 0,\,}\end{array}}\\{-2 < \frac{{6-a}}{{a + 6}} < 2,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{4a + 24-4a + 24-2a + 6 < 0,\,\,\,\,\,\,}\\{4a + 24 + 4a-24-2a + 6 < 0\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{a < -6,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{12{a^2}-24a > 0,\,\,\,}\end{array}}\\{\frac{{-3a-6}}{{a + 6}} < 0,\,\,\,\,\,\,\,\,\,\,}\\{\frac{{a + 18}}{{a + 6}} > 0,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\begin{array}{*{20}{c}}{54-2a < 0,}\\{6a + 6 < 0\,\,\,\,}\end{array}\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{a \in \left( {-\infty ;-6} \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{a\, \in \,\left( {-\infty ;0} \right) \cup \left( {2;\infty } \right),\,\,\,\,\,\,\,\,\,}\\{a\, \in \,\left( {-\infty ;-6} \right) \cup \left( {-2;\infty } \right),\,\,}\\{\begin{array}{*{20}{c}}{a\, \in \,\left( {-\infty ;-18} \right) \cup \left( {-6;\infty } \right),}\\{a \in \left( {\,27;\infty } \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{a \in \left( {-\,\infty ;-1} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\emptyset .\) Таким образом, уравнение будет иметь два различных корня модуль каждого из которых меньше двух при \(a\, \in \,\left( {-1;0} \right) \cup \left( {2;27} \right).\) ОТВЕТ: \(\left( {-1;\,0} \right) \cup \left( {2;\,27} \right).\)
Если \(a \ne -6\), то графиком функции \(f\left( x \right) = \left( {a + 6} \right){x^2} + 2\left( {a-6} \right)x-2a + 6\) является парабола. Уравнение будет иметь два различных корня, модуль каждого из которых меньше 2, то есть принадлежат интервалу \(\,\left( {-2;2} \right)\), в следующих случаях:
Второй случай (см. рис. 2):