Question

Решите квадратное неравенство х2 – 2x – 8> 0.
Ответ должен быть :6. Даю 30 баллов.!!!!!!!​

Answer 1

$\displaystyle \tt x^2-2x-8>0\\\displaystyle \tt x^2+2x-4x-8>0\\\displaystyle \tt x(x+2)-4(x+2)>0\\\displaystyle \tt (x+2)(x-4)>0\\\displaystyle \tt x+2=0\\\displaystyle \tt \underline{\bold{x_1=-2}}\\\displaystyle \tt x-4=0\\\displaystyle \tt \underline{\bold{x_2=4}}\\\\\displaystyle \tt \boxed{\bold{x\in (-\infty; \: -2)\cup (4; +\infty)}}$