Question

x в квадрате + 4х - 12 > 5x

Answer 1

$x^2 + 4x - 12 > 5x\x^2-x-12>0\x_1=4 ;x_2=-3$
///+///(-3)...-...(4)///+///=>x
$xin(-infty;-3)cup(4;+infty)$

Answer 2

x ² + 4х - 12 > 5x

x ² + 4х - 12 - 5x>,0

x ² -х - 12 > 0

x ² -х - 12 = 0

D = 1+48 =49

x₁= 4 ,  x₂ = -3

Рисунок во вложении

Ответ : х ∈ ( -∞ ; -3) υ (4; +∞)

 

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