Inequalities with radicals, absolute values and fractions

Text

Solve the following equations and inequalities:

• (a) $x-\sqrt{1-x}=-1$
• (b) $\sqrt{x+2}\le x$
• (c) $|x-2|\ge x$
• (d) $2+|x|=x+3$
• (e) $\dfrac{x (x^2+x+3)}{(x-2)^2}\le 0$
• (f) $\dfrac{1-x^2}{9-x^2}\le 0$

Solution

(a) $\sqrt{1-x}=x+1$ (with $x\le1$, $x\ge-1$). Squaring $x^2+3x=0\Rightarrow x=0$ or $x=-3$; only $x=0$ is acceptable.

(b) DC $x\ge-2$; need $x\ge0$. Squaring $(x-2)(x+1)\ge0$; with $x\ge0$, $x\ge2$.

(c) For $x\le2$: $2-x\ge x\Rightarrow x\le1$; for $x>2$ false. Solution $x\le1$.

(d) $|x|=x+1$: for $x\ge0$ impossible; for $x<0$, $x=-\tfrac12$.

(e) $x^2+x+3>0$ always and $(x-2)^2>0$ for $x\neq2$: the sign is that of $x$, so $\le0 \iff x\le0$.

(f) $\dfrac{(1-x)(1+x)}{(3-x)(3+x)}\le0$: sign study with breakpoints $\pm1,\pm3$ gives $-3<x\le-1 \ \vee\ 1\le x<3$.

$\boxed{\text{(a) } x{=}0;\ \text{(b) } x\ge2;\ \text{(c) } x\le1;\ \text{(d) } x{=}{-}\tfrac12;\ \text{(e) } x\le0;\ \text{(f) } -3<x\le-1\vee 1\le x<3}$