Six irrational inequalities

Text

Solve the following inequalities:

• (a) $\sqrt{x+1}\ge x-1$
• (b) $\sqrt{3-2x}\le 9+2x$
• (c) $5\sqrt{5x-x^{2}}\ge 3+5x$
• (d) $\sqrt{x+3}\ge |x+3|$
• (e) $\dfrac{-1+x-\sqrt{x^{2}-9}}{16-x^{2}}\ge0$
• (f) $\sqrt{2+x}-\sqrt{4-x}\le 2$

Solution

(a) DC $x\ge-1$. If $x<1$ always true; if $x\ge1$, squaring $x(x-3)\le0\Rightarrow 1\le x\le3$. Union: $-1\le x\le 3$.

(b) DC $x\le\tfrac32$ and $x\ge-\tfrac92$. Squaring: $2x^2+19x+39\ge0$, zeros $-3,-\tfrac{13}{2}$; intersecting gives $-3\le x\le\tfrac32$.

(c) DC $0\le x\le5$; here $3+5x>0$. Squaring: $50x^2-95x+9\le0\Rightarrow \tfrac{1}{10}\le x\le\tfrac{9}{5}$.

(d) With $u=x+3\ge0$: $\sqrt u\ge u \iff 0\le u\le1 \iff -3\le x\le-2$.

(e) DC $x\le-3\vee x\ge3$, $x\neq\pm4$. The numerator is $\ge0$ only for $3\le x\le5$; the sign study with $16-x^2$ gives $x<-4 \ \vee\ 3\le x<4 \ \vee\ x\ge5$.

(f) DC $-2\le x\le4$. $\sqrt{2+x}\le2+\sqrt{4-x}$; squaring twice reduces to $x-3\le2\sqrt{4-x}$, true for $x<3$ and, for $x\ge3$, up to $x=1+2\sqrt2$. Solution $-2\le x\le 1+2\sqrt2$.

$\boxed{\text{(a) } [-1,3];\ \text{(b) } [-3,\tfrac32];\ \text{(c) } [\tfrac1{10},\tfrac95];\ \text{(d) } [-3,-2];\ \text{(e) } x<-4\vee 3\le x<4\vee x\ge5;\ \text{(f) } [-2,\,1+2\sqrt2]}$