Question

Sa se rezolve inecuatiile:

a) x2-7x+12≥0;
b) –x2+4<0;
c) x2+x+1<0;
d)4x2-4x+1≥0.​

Answer 1

a)

$x^{2} -7x + 12 \geq 0\\(x-4)(x-3) \geq 0 \implies x \in (-\infty, 3]\cup[4, \infty)$

b)

$-x^{2} + 4 < 0 \iff x^{2} - 4 > 0 \iff (x-2)(x+2) > 0 \implies x \in (-\infty, -2)\cup(2, \infty)$

c)

$x^{2} + x + 1 < 0, \triangle = -1 -4(1)(1) = 1 -4 = -3 < 0 \implies x \in \O$

d)

$4x^{2}-4x +1 \geq 0 \iff (2x-1)^{2} \geq 0 \iff x \in \mathbb{R}$