Ответ:
Решение на фото.
Объяснение:
[tex]\displaystyle\bf\\x^{2} -2x-8 > 0\\\\x^{2} -2x-8=0\\\\D=(-2)^{2} -4\cdot(-8)=4+32=36=6^{2} \\\\\\x_{1} =\frac{2-6}{2} =-2\\\\\\x_{2} =\frac{2+6}{2} =4\\\\\\x^{2} -2x-8=(x+2)(x-4)\\\\\\x^{2} -2x-8 > 0\\\\(x+2)(x-4) > 0\\\\\\+ + + + + (-2) - - - - - (4) + + + + + \\\\\\Otvet \ : \ x\in(-\infty \ ; \ -2)\cup(4 \ ; \ +\infty)[/tex]
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Ответ:
Решение на фото.
Объяснение:
[tex]\displaystyle\bf\\x^{2} -2x-8 > 0\\\\x^{2} -2x-8=0\\\\D=(-2)^{2} -4\cdot(-8)=4+32=36=6^{2} \\\\\\x_{1} =\frac{2-6}{2} =-2\\\\\\x_{2} =\frac{2+6}{2} =4\\\\\\x^{2} -2x-8=(x+2)(x-4)\\\\\\x^{2} -2x-8 > 0\\\\(x+2)(x-4) > 0\\\\\\+ + + + + (-2) - - - - - (4) + + + + + \\\\\\Otvet \ : \ x\in(-\infty \ ; \ -2)\cup(4 \ ; \ +\infty)[/tex]