Second-degree and fractional inequalities

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Solve the following inequalities:

• (a) $x^2+x+1\ge 0$
• (b) $x^2-4x+4 \le 0$
• (c) $x^2+5x+6\le 0$
• (d) $\dfrac{-x^2+5x+6}{x^2+3x+2}>0$
• (e) $\dfrac{3}{2-z}+6-3z\le 10$

Solution

(a) $x^2+x+1$ has $\Delta=-3<0$ and upward concavity: always positive, so it holds for every $x$.

(b) $x^2-4x+4=(x-2)^2\le 0$: holds only where the square is zero, i.e. $x=2$.

(c) $x^2+5x+6=(x+2)(x+3)\le 0 \implies -3\le x\le -2$.

(d) $\dfrac{-x^2+5x+6}{x^2+3x+2}=\dfrac{-(x-6)(x+1)}{(x+1)(x+2)}=\dfrac{6-x}{x+2}$ (with $x\neq -1$). $\dfrac{6-x}{x+2}>0 \implies -2<x<6$, excluding $x=-1$: solution $-2<x<6 \ \wedge\ x\neq -1$.

(e) $\dfrac{3}{2-z}+6-3z-10\le 0 \implies \dfrac{3z^2-2z-5}{2-z}\le 0 \implies \dfrac{(3z-5)(z+1)}{2-z}\le 0$. Breakpoints $-1$, $\tfrac53$, $2$ (excluded). Sign study: $-1\le z\le \tfrac53 \ \vee\ z>2$.

$\boxed{\begin{aligned} &\text{(a) } \forall x\in\mathbb{R}; \quad \text{(b) } x=2; \quad \text{(c) } -3\le x\le -2;\\ &\text{(d) } -2<x<6 \wedge x\neq -1; \quad \text{(e) } -1\le z\le \tfrac53 \vee z>2\end{aligned}}$