Question

нужно срочно 20 баллов ​

Answer 1

$\displaystyle\bf\\\left \{ {{x - 4 > 0 } \atop { {x}^{2} - 2x - 15 < 0 }} \right. \\ \\ 1) \: x - 4 > 0 \\ x > 4 \\ \\ 2) \: {x}^{2} - 2x - 15 < 0 \\ {x}^{2} - 2x - 15 = 0 \\ po \: \: \: teoreme \: \: \: vieta \\ {x}^{2} + bx + c = 0\\ x_{1} + x_{2} = - b\\ x_{1} x_{2} = c \\ \\ x_{1} + x_{2} = 2\\ x_{1} x_{2} = - 15 \\ x_{1} = 5\\ x_{2} = - 3\\ \\ {ax}^{2} + bx + c = a(x - x_{1})(x - x_{2}) \\ {x}^{2} - 2x - 15 = (x + 3)(x -5 ) \\ \\ (x + 3)(x - 5) < 0 \\ + + + ( - 3) - - - (5) + + + \\ - 3 < x < 5 \\ \displaystyle\bf\\3) \: \left \{ {{x > 4} \atop { - 3 < x < 5 }} \right. \\ \\ otvet \: \: \: x \: \epsilon\: ( 4; \: 5)$