Mixed equations and inequalities with radicals and absolute values

Text

Solve the following equations and inequalities:

• (a) $\sqrt{2x+5}=x-4$
• (b) $\sqrt{1+2x^2}\ge x-3$
• (c) $5\sqrt{1+x}\le 5+9x$
• (d) $\sqrt{2x-1}\ge \sqrt{12-x}$
• (e) $\dfrac{-1+2x-\sqrt{x^{2}-4}}{-3+x+\sqrt{x+5}}\le0$
• (f) $\dfrac{(-x^2-x-1)(x^2+2x+1)}{(x+2)^2(x-4)}\le 0$
• (g) $|x-3|-x+4=0$

Solution

(a) Need $x\ge4$; squaring $x^2-10x+11=0\Rightarrow x=5\pm\sqrt{14}$. Only $x=5+\sqrt{14}$ is acceptable.

(b) LHS defined and $\ge0$; if $x<3$ true; if $x\ge3$, $x^2+6x-8\ge0$ true. Solution $\forall x\in\mathbb{R}$.

(c) DC $x\ge-1$, $9x+5\ge0$. Squaring $x(81x+65)\ge0$; with the conditions, $x\ge0$.

(d) DC $x\ge\tfrac12$ and $x\le12$; squaring $x\ge\tfrac{13}{3}$. Solution $\tfrac{13}{3}\le x\le 12$.

(e) DC $x\le-2\vee x\ge2$ (and $x\ge-5$). Numerator $>0$ for $x\ge2$, $<0$ on $[-5,-2]$; denominator $>0$ for $x\ge2$, $<0$ on $[-5,-2]$. The quotient is always positive on the domain: no solution.

(f) $-x^2-x-1<0$ and $(x+1)^2\ge0$, $(x+2)^2>0$: the expression is $\le0 \iff \dfrac{(x+1)^2}{x-4}\ge0$, i.e. $x=-1$ or $x>4$.

(g) $|x-3|=x-4$ requires $x\ge4$, where $x-3=x-4$ is impossible: no solution.

$\boxed{\text{(a) } x{=}5{+}\sqrt{14};\ \text{(b) } \forall x;\ \text{(c) } x\ge0;\ \text{(d) } [\tfrac{13}{3},12];\ \text{(e) } \varnothing;\ \text{(f) } x{=}{-}1\vee x{>}4;\ \text{(g) } \varnothing}$