First- and second-degree inequalities

Text

Solve the following inequalities:

• (a) $-3x-2\le -2x+1$
• (b) $x^2+3x-4+2(x^2+6x+8)\ge 0$
• (c) $2x^2+1\ge 0$

Solution

(a) $-3x-2\le -2x+1 \implies -x\le 3 \implies x\ge -3$.

(b) Expanding: $x^2+3x-4+2x^2+12x+16\ge 0 \implies 3x^2+15x+12\ge 0 \implies x^2+5x+4\ge 0$. Factoring $(x+1)(x+4)\ge 0$ (upward parabola): the solution lies outside the roots $-4$ and $-1$, i.e. $x\le -4 \ \vee\ x\ge -1$.

(c) $2x^2+1\ge 0$: the sum of a square and $1$ is always positive, so the inequality holds for every $x$.

$\boxed{\text{(a) } x\ge -3;\quad \text{(b) } x\le -4 \ \vee\ x\ge -1;\quad \text{(c) } \forall x\in\mathbb{R}}$