Absolute-value inequalities and sign study

Text

Solve the following inequalities:

• (a) $|x+1|+|x^2-1|\le 0$
• (b) $\dfrac{2x-1}{5x}-\dfrac{1-x}{2(x-1)}\le 4$
• (c) $\dfrac{(x+1)^2(x-4)^2}{2x^2+x+1}\le 0$
• (d) $\dfrac{2}{1-2x}-\dfrac{1}{2x}+4>0$

Solution

(a) Sum of two absolute values $\le 0$: possible only if both are zero. $|x+1|=0 \Rightarrow x=-1$ and $|x^2-1|=0 \Rightarrow x=\pm1$: the only common value is $x=-1$.

(b) Since $\dfrac{1-x}{2(x-1)}=-\dfrac12$ (for $x\neq 1$), the inequality becomes $\dfrac{2x-1}{5x}+\dfrac12\le 4 \implies \dfrac{2x-1}{5x}\le \dfrac72 \implies \dfrac{31x+2}{x}\ge 0$. Breakpoints $-\dfrac{2}{31}$ and $0$ (excluded), plus $x\neq 1$: solution $x\le -\dfrac{2}{31} \ \vee\ (x>0 \ \wedge\ x\neq 1)$.

(c) The denominator $2x^2+x+1$ has $\Delta<0$, always positive; the numerator $(x+1)^2(x-4)^2\ge 0$. The fraction is $\le 0$ only where the numerator vanishes: $x=-1 \ \vee\ x=4$.

(d) Reducing to the common denominator $2x(1-2x)$: $\dfrac{-16x^2+14x-1}{2x(1-2x)}>0$. Numerator zeros $x=\dfrac{7\pm\sqrt{33}}{16}$, denominator zero at $0$ and $\dfrac12$. Sign study: $x<0 \ \vee\ \dfrac{7-\sqrt{33}}{16}<x<\dfrac12 \ \vee\ x>\dfrac{7+\sqrt{33}}{16}$.

$\boxed{\begin{aligned} &\text{(a) } x=-1; \quad \text{(b) } x\le -\tfrac{2}{31} \vee (x>0 \wedge x\neq 1); \quad \text{(c) } x=-1 \vee x=4;\\ &\text{(d) } x<0 \vee \tfrac{7-\sqrt{33}}{16}<x<\tfrac12 \vee x>\tfrac{7+\sqrt{33}}{16}\end{aligned}}$