Question

$\frac{x^2-2x-8}{(x-1)^2} \leq 0$

Answer 1

$\frac{x^{2} - 2x - 8}{(x-1)^{2} } \leq 0 |*(x-1)^{2} \\ x^{2} -2x-8 \leq 0 \\ D=(-2)^2 - 4 * 1 * (-8) = 4 + 32 = 36, \sqrt{36} = 6 \\ x_{1} = \frac{-b+ \sqrt{D} }{2a} = \frac{2+6}{2} = 4 \\ x_2 = \frac{-b - \sqrt{D} }{2a} = \frac{2-6}{2} = -2$