Решите методом интервала

x(x - 8) > (2x - 1)²

x² - 8x > 4x² - 4x + 1

x² - 8x - 4x² + 4x - 1 > 0

- 3x² - 4x - 1 > 0

3x² + 4x + 1 < 0

3x² + 4x + 1 = 0

D = 4² - 4 * 3 * 1 = 16 - 12 = 4 = 2²

$x_{1}=\frac{-4+2}{6}=-\frac{1}{3}\\\\x_{2} =\frac{-4-2}{6} =-1\\\\3x^{2}+4x+1=3(x+\frac{1}{3})(x+1)\\\\3(x+\frac{1}{3})(x+1)<0 \\\\(x+\frac{1}{3})(x+1)<0$

+ - +

___________₀____________₀____________

- 1 - 1/3

x ∈ (- 1 ; - 1/3)